ZC tSNK A,'aN l~t4& ELEMENTS OF GEOMETRY: CONTAINING THE FIRST SIX BOOKS OF E UCLID5 WITH A SUPPLEMENT ON TIlE QUADRATURE OF THE CIRCLE, AND THE GEOMETRY OF SOLIDS: TO WHICII AllS.ADDED ELEMENTS OF PLANE AND SPHERICAL TRIGONOMllETRY. BY JOHN PLAYFAIR, F.R.S. LOND. & EDIN. PROFESSOR OF NATURAL PHILOSOPHY, FORMERLY OF MATIIEMATICS, IN N'i'J.1; UNIVERSITY OF EDINBURGH. FROM THE LAST LONDON EDITION, ENLARGED. - NEW-YORK; VF 2SWI$ID BY,- -3i; H& IANNAY, COLLINS & CO., O. A. ROOB,.. t,' ER & W"HITE, AND G. & Ce ^' U: VILL., -It is a remarkable fact in the historyr of scienuce, ~hat the oldest book of Elementary Geometry is still -considered as the best, and that the writings of Ev-, -CLID, at the distance of two thousand years, continue-'to form the most approved introduction to the mathematicalsciences, This remarkable distinction the Greek Geometer owes not only to the elegance and correctness of his demonstrations, but to an arrangement most happily contrived for the purpose of ins struction,D —advantages which, when they reach a certain eminence, secure the works of an author.against the injuries of time more eflfectuaily tha even originality of invention. The elements of EuCLID, however, in passing through the hands of the ancient editors, during the decline of science, had suffered some diminution of their excellence, and much skill and learning have been employed by the modern mathematicians to deliver them from ble. mishes, which certainly did not enter into their origicnal composition. Of these mathelmaticians, Dr. Sa son, s- ihe may be accounted the last, has also been the most successful, and has left very little r'oow 16t the ingenuity of future editors to be exercised in, either by amending the text of EUCLID, or by improving the translations from it. Such being the merits of Dr. SiMsoN's edition, and the reception it has met with having been every way suitable, the work now offered to the public will perhaps appear unnecessary. And indeed, if the geometer just named had written with.a view of acfcommodating the Elements of EUCLID to the present state of the ma.thematical sciences, it is not likely that -n.y thing new in Elementary Geometry would have been soon attempted. But his design was different: it was his object to restore the writings of EUCLID t4 their original perfection, and to give them to Modern Europe as nearly as possible in the state wherein they mlade their first appearance in Ancient Greece. For this undertaking, nobody could be better qualified than Dr. SIMsoxN; who, to an accurate knowledge of the learned languages, and an indefatigable spirit of research, added a profound skill in the ancient Geometry, and an admiration. of it almost enthusiastic. Accordingly he not only restored the text of EuCLID wherever it had been corrupted, -but In.tsome cases removed imperfections that pro P R Fi A. C E;bFbiy beloingd tO tl e original worki, though his extreme partiality for his author never permIitted him. to suppose, that such honour could,all to the share either of himself; or of any other of the moderns. But, after al this was accoomplished, something still remained to be done, since., notwithstanding the acknowledged excellence of EUCLID's Elements, it could not be doubted, that some alterations might be made, that would acom-odate thenm better to a state of the mathem satical sciences. so much more improved and extended tlan. at the period when they were written. A.ccordingly the object of the edition now obrered to the public, is not so much to give to the writings of EUCLID the form which they originally had, as thLt wvicha m.ay at presen t render them most usefuii, One of the alterations made witi this view, respects the Doctrine of Propolrtioni, the method of treating which, as it is'lid down in- the fifth of E.v lJm, has great advantaC ges, accompanied with Con.siderable de'bets; of whichB, haowever, it must be ob served, that the advantages are essential, and the defects only accidental. To explain the nature of the former, requires a more minute examination hasn is ited to this plaee, and. must, thereforet be reserved for the Notes; but, in the mean time, it m-y be remarked, that no definition, except that of ERrCLmI has ever been- given, from which the properties of proportionals can be deduced by reasonings, which, at the same time that they are perfectly rigorous, are also simple and direct. As to the defects, the prolixness and obscurity, that have so often been complained of in the fifth BookS they seem to arise chiefly from the nature of the language employed, which being no other than that of ordinary discourse,.cannot express, without much tediousness and circumlocution, the relations of mathematical quantities, when taken in their utmost generality, and when no assistance can be received from diagrams. As it is plain, that the concise language of Algebra is directly calculated to remedy this inconvenience, I have endeavoured to introduce it here, in a very simple form however, and without changing the nature of the reasoning) or departing in any thing from the rigour of geometrical demonstration. By this means, the steps of the reasoning which were before far separated, are brought near to one another9 and the force of the whole is-so clearly and directly perceived, that I am persuaded no more difficultywill be found in understanding the propositions of the fifth Book, than those of any other of the Ele..ments. PIREI'AbCE Vii In the second BOok, also, some algebraic signs have been introduced, for the sake of representing more readily the addition and subtraction of the rece tangles on which the demonstrations depend. The use of such symbolical writing, in, translating from an original, where no symbols are used, cannot, I think. be regarded as an unwarrantable liberty for, if by that means the translation is not made into English, it is made into that universal language so much sought after in all the sciences, but destined, it would seem, to be enjoyed only by the mathematical. The alterations above mentioned are the nmost; material that have been attempted on the books of EvcLID. There are, however, a few others, which, though less considerable, it is hoped may in some; degree facilitate the study of the Elements. Such are those made on the definitions in the first Book, and particularly on that of a straight line. A new axiom is also introduced in the room of the 12th, for the purpose of demonstrating more easily some of the properties of parallel lines. In the third Bookl, the remarks concerning the angles made by a straight line, and'the circumference of a circle, are left out, as tending to perplex one who has advanced ao farther than the elemuents of the science, OTht, Vill PRtEi'ACE 27th, 28th, and 29th of the sixth are changed tbr easier and more simple propositions, which do not materially differ from them, and which answer exactlythe same purpose. Some propositions also have been added; but, for a fuller detail concerning these changes, I must refer to the Notes, in which several of the more difficult, or more interesting subjects of Elementary Geometry are treated at considerable length. 1The SUPPLEMENT no-w added to thle Six RBooks of EUCLID is arranged differently from what it was in the, first edition of these Elements, The First of the three Books, into which it is divivded, treats of the rectification and quadrature of" the circle, —subjects that are often omitted altogether in works of this kind. They are omitted, however, as I conceive, without any good reason, because, to measure the length of the simplest of all the curves -which Geonmetry treats of; and the space contained within it, are problems that certainly belong to the elements of the science, especially as they are vlot more difficult than other propositions which are usually admitted into them. When I speak ofthe rectifil catioii of the circle, or of measuring the length of the circumference, I must inot be supposed to mean, that a.St faight line is to be made eqn` t.fence exactiy —-a problem which, as;-: —,i ka'.own, Geometry has never been able to resolve. All that'is proposed is, to determine two straight lines that shall' differ very little from one another, not more, for instance,,than the four -hundred and ninety-seventh part of the diameter of the circle, and of which the one shall be greater than the circurmference of that circle, and the other less. In the same manner, the quadrature of the circle is performed only by approximation, or by finding two rectangles nearly equal to one another, one of them greater, and another'less9 than the space contained within the circlec In the second Book of the Supplement, which treats cof the intersection of Planes, I have departed as little as possible from EucLID'S method of considering the same subject in his eleventh Book. The demonstration of the fourth proposition is from LEGENDRE'S Elements of Geometry; that of the sixth is new1, asfar afar I'know; as is a-lso the solution ofthe problem in.the ninteenth proposition,-a probleml which, though -in itself extemely simpie, has been omitted by Euclid, and hardly ever treated of in an elementary form, by any geometer, -Wyith respect -to the Geometqry of Solids, in the I~/REFACJEtk third Book, 1 have departed fiom EucLu) altogether, -with a view of rendering it both shorter and more comprehensive. This, however, is not attempted by introducing a mnode of reasoning less rigorous than that of the Greek Geometer; for this would be to pay too dear even for the time that might there.. by be saved; but it is done chiefly by laying aside a certain rule, which, though it be not essential to the accuracy of deimonstration, EUCLID has thought it, proper, as much as possible, to observe. The rule referred to, is one which influences the arrangement of his propositions through the whole of the Elements, viz. That in the demonstration of a theorem, he never supposes any thing to be done, as any line to be drawn, or any figure to be constructed, the manner of doing which fhe has not previously explained. Now, the only use of this rule is to prevent the adltission of impossible or contradictory suppositions, which, no doubt, nmight lead into error: and it is a rule well calculated to answer that end, as it does not allow the existence of any thing to be supposed, unless the thing itself be actually exhibited~ But it is not always necessary to make use of this defence; for the existence of many things is obvious~ ly possible, anad very far fr~o implying a contradici o?,:ieitr~ -tA; t-.QC ts,at,2:ala}s exhibitina Q t. i tqr, may be altogether unklcnown. -Thus, it is plain, tliat on any given figure as a base, a solid may be colnstituted, or conceived to exist, equal in solid contentss to any given solid, (because a solid, whatever be its base, as its height may be indefinitely varied, is capable of all degrees of magnitude, fromn nothing upwards), and yet it may in many cases be a problem of extreme difficulty to assign the height of such a4 solid, and actually to exhibit it. Now, this very supposition, that on a given base a solid of a given magnitude may be constituted, is one of those, by the introduction of which, the Geometry of solids is much shortened, while all the real accuracy of the demonstrations is preserved; and therefore, to follow, as EUCLID has done, the rule that excludes this, and such like hypotheses, is to create artificial difficulties, and to embarrass geometrical investigation with more obstacles than the nature of things has thrown in its way. It is a rule, too, which cannot always be followed, and forom which even EucLID himself has been forced to depart, in more thlan. one instane.eO In-the Book, therefore, on the Properties -of Solids. which I now offer to the public, I have not sought to subject the demonstrations to the law just mentioned? and have never hesitated -to admlit the existence of' such solids, or suchl line as are cvidentrlv possible: though the manne' ofiactually describing them mInay not have been explained, In this way,' I have be&n enabled to offer that very refined artifice in geometrical' reasoning, to which we give the name of the Method of Exhaustions-s under a much simpler form than it appears in the 12th of EUCLID; and the spirit of the method may, I think, be best learned when it is thus disengaged from every thing not essential, That it mfay be the better understood, and because the demonstrations which require exhaustios:i ares:no doubt, the most difficult in the Elements, they are all conducted as nearly as possible in the same way, in the cases of the different solids, friom the pyramid to the sphere; The comparison of this last solid With the cylinder, concludes the last Book of the Supplement, and is a proposition that may not improperly be considered as terminating the elementary pa&t of Geometrvo The Book of the Data has been annexed to seven.tal editions of EUCLID's Elements, and particularly to Di. SIMSON'S, but in this it is omitted altogethe:. It is ornitted, however, not from any opinion of its being inl itselfuseless; but because it does not belong to this place, and is not often read by beginners. It confains the rudiments of what is properly called the O:mnetrical Analysis, and has itself an atnalyticW1 I ~E_'VA C,' Nit idrm; and fbr these reasons, I would willingly reserve its or rather a compend of it, for a separate work1 intended as an introduction to the study of that anal iysis; Ii explaining the elements of Plane and Spherical Trigonometry, there is not much new that can be attempted, or that will be expected by the intelligent reader. Except, perhaps, some new demonstratioyns: and some changes in the arrangement, these two treatises have, accordingly, no novelty to boast of The Plane Trigonometry is so divided, that the part of it that is barely sufficient for the resolution of Trio angles may be easily taught by itself: The method of constructing the Trigonometrical Tables is explain's ed, and a demonstration is added of those properties Of the sines and cosines of arches, which are the foundation of those applications of Trigonometry lately introduced, with so much, advantage, into the higher Geometry, In the Spherical Trigonometry, the rules for pre-. venting the ambiguity of the solutions, wherever it can be prevented, have been particularly attended to; and I have availed myself as mnucl as possible of that excellent abstract of the rules of this science which Dr. MASKELYNE has prefixed to TA,.YoR's T.-: bles Of Logarithms:. X~1V ERiFA(;E; An explanation of NAPIER'S very ingenious and useful rule of the- ('rcalattr Parts is haere added as an Appendixto Spberical Trigonometry. It has been objected to many of the writers on Elce mentary Geometry, and particularly to EuCLID, that they have been at great pains to prove the truth of many simple propositions, which every body is ready to admit, without any demonstration, and that thus they take up the time, and fatigue the attention of the student, to no purpose. To this objection, if there be any force in it, the present tretise: is certainly as much exposed as any other; for no attempt is here made to abridge the Elements, by considering as self-evident any thing that admits of being proved. Inldeed, those cwho make the objection just stated, do not seem to have reflected sufficiently on the end of Mathematical Demonstration, which is not: only to prove the truth of a certain proposition, but to shew its necessary connection with other propositions, and its dependance on them. The truths of Geometry are all necessarily connected with one another, and the system of such truths can never be rightly explained, unless that connection be accurately traced, wherever it exists. It is upon this that the beauty and peculiar excellence of the mathematical sciences depend. it is this. whicr, by preventing any one ~'B, EFCE, X V truth firomn being single and insulated, connects the different parts so firmly, that they must all stand, or all fall together. The demonstration, therefore, even of an obvious proposition, answers the purpose of connecting that proposition with others, and, ascertaining its place in the general system of mathematical truth. If, for example, it be alleged, that it is ieedless to demonstrate that any two sides of a tri~. angle are greater than the third; it may be replied, that this is lno doubt a truth which, w7ithout proof, most men will be inclined:to admit; but are we for that reason to account it of no consequence to know what the propositions are, which must cease to be true if this proposition were supposed to be false? Is it not useful to know, that unless it be true, that any two sides of a triangle are greater than the third, neither could it be true, that the greater side of every triangle is opposite to the greater angle, nor that the equal sides are opposite to equal angles, nor lastly, that things equal to the same thing are equal to one another? By a scientific mind this information will not be thought lightly of; and it is exactly that which wve receive fro m EUCLmD S demonstration. To all this it may be added, that the mind, especially when beginning to study the art of reasoning, xfjnn.t,b eaui\upovcd to re'ater erdv.rta.tge h'hanm inl XVI PREFACE, analysing those judgments,which, though they appear simple, are in reality complex, and capable of being distinguished into parts. No progress in ascending higher can be expected, till a regular habit of demonstration is thus acquired; it is much to be feared, that he who has leclined the trouble of tracing the connection between the proposition already quoted,, and those that are more simple, will not be very expert in tracing its connection with those that are more complex; and that, as he has not been carefiul in laying the foundation, he will never be successftal in raising the superstructure,:GOLLEGE OF EDINI3URGHI 'ELEM1ENTS OF Gw E 0 3V E T R Y6e BOOK I. )DEFINITIONS. 8' A POINT is that which has position, but: not magnitude'J, "8s Notes.) A line is length without breadth~' COROLLARtY. The extremities of a line are points. and the inter~ sections of one line with an, ther are also points s" I ff two lines are such that they cannot coincde irn any two points, "without coinciding altgether, each -of thein is called a straight "' line." " CoR. Hence two straight lines cannot inclose a space. Neither can,c two straight lines have a common segment; that is, they cannot " coincide in part, without colnciding altogethero" IV. A superficies is that which has only length and breadth. ~c Coa. The extremities of a superficies are lines; and the interseec' tions of one superficies with another are also lines.9 A plane superficies is that in which any two points being taken, the straight line between them lies wholly in that superficieso VI A plane rectilineal angle is the inclination of two straight lines to one another, which meet together, but are not in the same straight lineo T The defitlons mar ked with inverted comrnmas er diffire nt fro tlhne of' Eucidi 8 J/' __i_ Ni. B.' When several angles are at one point B, any one of themt is expressed by three letters, of which the letter that is at the vertex of the angle, that is, at the point in which the straight lines that - contain the angle meet one another, is put between the other two. letters, and one of these two is somewhere upon one of those straight, lines, and the other upon the other line: Thus the angle which'is contained by the straight lines, AB, CB, is named, the angle' ABC, or CBA; that-which is contained by AB, BD, is named the' angle ABD, or DBA; and that which is contained by BD, CB, is' called the angle DBC, or CBD; but, if there be only one angle at 6 a point, it may be expressed by a letter placed at that point, as the angle at E.' VII. When" a straight line standing on another straight line makes the adjacent angles equal to one another, each of the angles is called a right angle; and the straight line which stands on the other is called a perpendicular to it. VIIl. An obtus.e angle is that which is greater than a right angle, IX. As acute angle is that which is less than a right angle. X. A figure is that which is inclosed by one or more boundaries.-Ths vord-area denotes the quantity of space contained in a figure, w~ithou:.ny refere.nce to the n ra ofe oq h i?, or!l,?,s bnnih A 4.P-r, OF GWOMETRY. B11100K XL A circle is a plane figure contained by one line, which is called t h circumference, and is such that all straight lines drawn from a certain point within the figure to the circumference, are equal to one another. And this point is called the centre of the circlee XIIL A diameter of a circle is a straight line drawn through the centre, and terminated both ways by the circumference. A semicircle is the figure contained by a diameter and the part of.the circumference cut off by the diameter. XV. Rectilineal figures are those which are contained by straight lino, XVIL Trilateral figures, or triangles, by three straight lines XVII. Quadrilateral, by four straight lines. XVIII. M3!ultilateral figures, or polygons, -by more than four straight lineso Of three sided figures, an' equilateral triangle is that which has thre. equal sides. XX. An isosceles triangle is that which has only two sides equals XXI, A scalene triangle, is that which has three unequal sideSo XXII. rA right angled riangle is that which has a right angle~ XXIM an obtuse angied triange, i' that whiclh hs ait obtuse angi'~ XXIVo An acute angled triangle, is that which has three acute angles$ XXV. Of four sided figures, a square is that which has all its sides equa] and all its angles right angles. XXVI. An oblong, is that which has all its angles right angles, but has not all its sides equal. A rhombus, is that which has all its sides equal, but its angles are not xight angles. /,,' / XV III A rhomboidt, a tat vhich has -ts opposite sides equal to one snot'hey' but all its sides are not equal, nor its angles right angles. All other four sided figures besides these, are called Trapeziumso Parallel straight lines, are such as are in the same plane, and whicb, being produced ever so far both ways, do not meet. O:$ GEOM~ETR~Y. BOOK 1.' POSTULATEo LaT' it be granted that a straight line may be drawn from any one point to any other point. II. That a terminated straight line may be produced to any length in a straight line. II., And that a circle may be described fromn any centre, at any distance from that centre. AXIOMS. THINGS which are equal to the same thing are equal to one another. IL. If equals be added to equals, the wholes are equal. II!. If equals be taken from equals, the remainders are equals. IV. If equals be added to unequals, the wholes are unequal. V. Jf equals be taken from unequals, the remainders are unequal. VI. Things which are doubles of the same thing, are equal to one another. VII, TIhiihgs which are halves of the same thing, are equal to one another, VIII. Magnitudes which coincide with one another, that is, which exactly fill the same space, are equal to one another. Ix. The whole is greater than its part. All right angles are equal to one another. XI. "2 Two straight lines which intersect one another, cannot be both pa"rallel to the same straight Uine.'" LtE1ENTS PROPOSITION I. PROBLEMo To describe an equilateral triangle upon a givenfinite straight line. Let AB be the given straight line; it is required to describe amn equilateral triangle upon it. From the centre A, at the distance AB, describe (3. Postulate) the circle BCD, and from the centre B, at the distance BA, describe the circle ACE; and from the point C, in iD B which the circles cut one -another, draw the straight lines / (1. Post.) CA, CB to the points A, B; ABC is an equilateral triangle. Because the point A is the centre of the circle BCD, AC is equal (1 1. Definition) to AB; and because the point B is the centre of the circle ACE, BC is equal to AB: But it has been proved that CA is equal to AB; therefore CA, CB are each of them equal to AB; now things which are equal to the same are equal to one another, (1. Axiom); therefore CA is equal to CB; wherefore CA, AB, CB are equal to one another; and-the triangle ABC is therefore equilateral, and it is described upon the given straight line AB. Which was required to be done. PROP. II. PROB. from a given point to draw a straight line equal to a given straight line. Let A be the given point, and BC the given straight line; it is required to draw, from the point A, a straight line equal to BC. From the point A to B draw (1. Post.) the straight line AB; and upon it describe (1. 1.) the equilateral triangle DAB, and produce (2. Post.) thie straightlines DA, BD, to E and F; from the centre B, at the distance BC, describe (t:. Post.) the D circle CGH, and from the centre D, at the distance DG, describe the circle GKL. C AL is equal to BC. Because the point B is the. centre of the circle CGH, BC is equal (11. Def.) to BG; and because D is the centre of the circle GKL, DL is equal to DG, and DA DD, parts of them, are equal; therefore OF GEOMETRY. BOOBK 1'; the remainder AL is equal to the remainder (3. Ax.) BG: But it has been shewn that BC is equal to BG; wherefore AL and BC are each of them equal to BG; and things that are equal to the same are equal to one another; therefore the straight line AL is equal to BCo Wherefore, from the given point A, a straight line AL has been drawn equal to the given straight line BC. Which was to be done. PROP. IIl. PROB. From the greater of two given straight lines to cut oF a part equal to the less. Let AB and C be the two given straight lines, whereofAB is'the great er. It is required to cut off from AB, I) the greater, a part equal to C, the less. From the point A draw ('2. 1.) the straight line AD equal to C; and friom C the centre A, and at the distance AD, describe (3. Post.) the circle DEF; and because A is the centre of the circle DEF, AE is equal to AD; but the straight line C is likewise equal to AD; whence AE and C are each of them equal to AD; wherefore the straight line AE is equal to (I. Ax.) C, and from AB the greater of two straight lines, a part AE has been cut off equal to C the less. Which was to be done. PROP. IV. THEOREMo.Vf two triangles have two sides of the one equal to two st.tes of the oth2cr% each to each; and have likew-ise the angles contai-zed by those sides equal to one another, their bases, or third sides, Jhall be equal; anfd the areas of the triangles shall be equal; and Kleir other angles shocl be equal, each to each, vizo those to whizch the equal sides are opposite.* Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two sides DE, DVF each to each, viz. AB to DE, and AC to DF; and let the angle BAC be also A ) equal to the angle i EDF: then shall the, base BC be equal to " / the base EF; and the triangle ABC to the tri- / angle DEF; and the i i other angles, to which,the equal sides are op- B C B F * The three conclusions in this eniancwaton are.ae briefly exrressed by sayin%:.:.T~f ~j''~ YZ54 46.~ 87_lt~dy P/,c eWi -.. 24+ S~LEMENTb posite, shall be equal, each-to each, viz, the angle ABC to the angll DEF, and the angle AC.B to DFE. For, if the triangle ABC be applied to the triangle DEF, so that the point A may be on D, and the straight line AB upon DE 4 the point B shall coincide with the point E, because AB is equal to DE and AB coinciding with DE, AC shall coincide with DF, because the angle BAC is equal to the angle EDF; wherefore also the point C shall coincide with the point?F, because NC is equal to DF ~ But the point B coincides with the point E; wherefore the base BC shall coincide with the base EF (cor def. 3.), and shall be equal to it. Therefore also the whole triangle ABC shall coincide with the whole triangle DEF, so that the spaces which they contain or their areas are equal; and the remaining angles of the one shall coincide with the remaining angles of theil other, and be equal to them, viz. the angle ABC to the angle DEF. and the angle ACB to the angle DFE. Therefore, if two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise the angles contains ed by those sides equal to one another; their bases shall be equal, and their areas shall be equal, and their other angles, to which the equal sides are opposite, shall be equal, each to each. Which was to be demonstrated. PROP. V. TIlEOR. TiPhe angles at the base of an Isosceles hiang(le are equal to one another' and if the equal sides be produced, the angles eupon the other side of the base shall also be eqznal. Let ABC be an isosceles triangle, of which the side AB is equal to AC, and let the straight lines AB, AC be produced to D and E, the angle ABC shall be equal to the angle ACB, and the angle CBD to the angle BCE. In BD take any point F, and from AE the greater cut off AG equal (3. 1.) to AF, the less, and join FC. GB. Because AF is equal to AG, and to AC, the two sides PA, AC are equal to the two GA, AB, each to each; and they contain the ~angle FAG common to the two trian- gles, AFC, AGB; therefore the base FPC is equal (4. 1.) to the base GB and the triangle 1FC to the triangle / \ kGB; and the remaining angles of \ the one are equal (4. 1.) to the re/ maining angles of the other, each to cach, to which the equal sides are opposite, viz. the angle ACF to the angle ABG, and the angle AFC to the / ( angle AGB: And because the whole, AF is equal to the whole AG, and the part AB to the part AC; the remain- j/. der BF shall be equal (3, Ax.) to the OF GEOMETRY. BOOK 00 5'etnainder CG; and FC was prove(d to be, equall to GB, d.fierfre the two sides BF, FC are equal to the two -G, GB, each to each;l u but the angle BFC is equal to the angle CGB; whleefr ore the t:iarg-les BFC, CGGB are equal (3. ), and their remaining anogles are eqial, to which the equal sides are opposite; thererore the angle F1L3 is equal to the angle GUB, and the angle BCl( to the angle t-v,.,, S Now, since it has been demonstrated, that the whole angle A B(G is equal tfo the whole ACF, and th.e part CBG to the part BCF, the remaining angle ABC is therefore equal to the rernaining angle AUB, whwi-!h are tihe angles at the base of the trianale A.BC A.nd it has also beern prvod that the angle FBC is equal to the anile -GC.B, which ale the angles upon the other side of the base. Tlmerefore, the angles at the base, &c. Q. E. P. COROI.LAnRY Hence every equilateral triangle is aso ecquiangularo PROP. VI. THEOR. two angles of a triantgle be equal to orte anotnh.er, the srde.S wzhichI Sub.tend, or are opposite to the m, are also ejual to one,neolther. Let ABC be a triangle having the anpgle ABC equlal to the angle ACB; the side AB is also equal to time side AC. For, if AB be not equal to AC, one of them is greater than thze other::Let AB be the greater, and frorn i4t cut (3. I,) off 3DB equal to AC the less, and join DC; therefore, be- cause in the tri.angles DBC, ACB, DB is equal to AC, and BC common to both, the /.}wo sides DB, BC are equal to the two AC, CB, each to each; but the angle DBC is also equal to the angle ACB' therefore the baseo DC is equal to the base AB, and the area / of the triangle DBC is equal to that of the triangle (4. 1.) ACB, the less to the greater; / w-hich is absurd. Therefore, A13 is rlot rn- / e-qual to AC, that is, it is equal- to it. VW7.here- -.-___ fore, if two angles, zc. Q. E. D. CoQP. I-ence every equtiangular triangale is also e';iilatea-'2, 7POP. V1 T! 0? Upo2 tile sace base, a,.d on the saine side of it, there cannot be two!r i? angles, that hare their sides whiich are terminated in one cixtrenlity of the base equal 1o one aiother, o.nd likewise those wehich arwe tCerminated?i the other extremity, eqget!lZ to ao.e anotllere Let there be two triangles ARC, ADB, upon the same base A.3 and upon the same side of it, which hiave their sides CA, DA, terminated In A equal to one another th en their side,, CIB, DIB, telrminaitel int B, cannot be equal to one another. Join CD, and if possible let CB /. be equal to DB; then, in the case: in which the vertex of each of the triangles is without the other triangle, because AC is equal to AD, thle angle AC D is equal (5. 1.) to the / angle A.DC: But theangle ACD is / / greater than the angle BCD; therce fore the angle A DC is greater also / than BCD; much more then is the // angle BDC greater than the angle' - -- BCD. Again, because CB is equal to DB, the angle BDC is equat (5. 1.) to the angle BCD; but it has been demonstrated to be greater than it; which is impossible. But if one of the vertices, as _' ),- be within the other triangle i ACB; produce AC, AD to E, F; therefore, because AC is /x'./ -equal to A D in the triangle ACD, the angles ECD, FDC upon the other side of the base CD are equal (5. 1.) to one another, but the angle ECD is greater.... thabn the angle BCD; wherefore 3 the angle FDC is likewise greater than BCD; much more then is the angle BDC greater than the angle BCD. Again, because CB is equal to DB, the angle BDC is equal (5. 1.)' to the angle BCD; but BDC has been proved to be greater than the same BCD; which is impossible. The case in which the vertex of one triangle is upon a side of the other, needs no demonstration, Theretore, upon the same base, and on the same side of it, theri cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise those which are terminated in the other extremity equal to one another..E. D. PROP. VII. THEOt.o If two triangles have two sides of the one equal to t, o sides of the othe r, each to each, and have likewise their bases equal; the angle which is contained by the two sides of the one shall be equal to the angle contained by the two sides of the other. Let ABC, DEF be two triangles having the two sides AB, AC?4 equal to the two sides DE, DF, each to eaeh, viz. AB to DE, and A (C to DF; and also the base BC equal to the base EFo The angolf BA3 is equal to the angle EDF. For, if the triangle ABC be applied to the triangle DEF, so that lthe point B be on l, and the straight line BC upon CF; thie point C shall also coincide with the point F. )becauese B3C is eIqual to E' therefWore BC coiinx5cidhu vwth Ea 3 A Wa-d. AC shal' coincide with ED) and DF; for, if BA, and C do not coinrede with.ED andl FD, but have a different situation as EG an]d FG; then, upon thlle sale ase EiF, and upon the same qde of it, there can be two tri'anies E DF,:EGF, that have their sides whiclh are terminate d in one e xtremity of the base equal to one another, and likewise their sides terminated in the otherextremity; but this is impossible (70 1.) therefore, if the base BC coincides with the base EF, the sides BA, AC cannot but coincide with the sides ED, DlF; whereffore likewise the angle BAC coincides with the angle EDF, and is equal (8. Ax.) to ito Therefore if two trtiargles, &c. Q. E. D PROP. IX. PROB. 7b bisect a give~n rectilineal anmgle, that is, to divide it into two equal angles. Let BAC be the given rectilineal angle, it is reaquired to bisect ite Take any point D in AB, and firom AC cut (8. i.) off AE equal to AD; join DE, and upon it describe (1. 1.) an equilateral trianglle Dl;'t l/ T then join AF; tile straighnt line AF bisects the angle BAC. Because AD is equal to AE, and AF /,AV'; the two sides DA, AF, are equal to tle two sides EA AF, each to each;. but the base DF is also equal to the / base EF; therefore th angle DAl-i is.equal (8. 1.) to the angle EAF?: wherefore the given rectilineal angle BAC is bisected by the straight line A83 F'Which was to be doheo -e IR eC '-ROP. X. PROIB Po bisect a gaienfstile stra ijht line, that zs, to divide it ilnb itwo equal parts. Let AB be, the given straig.ht line; it is required to divide it into two eqiual parts. Describe (1. 1.) upon it an equilateral triangle ABC, and bisect (9. 1. i the anlgle, A<(.B by the straight line CD. AB is cut into two bqual parts in the point D. Because AC is equal to CB, and CD common to the two triangles ACD, BCD: the two sides AC, CD, are C equal to the two BC, CD, each to each; but the angle ACD is also equal to the angle BCD1; therefore the base AD is equal {o the base (4. 1.) DB, and the straioht linb AB is divided into two equal parts in the point DS Which was to be done, A D B PtROP. XI. PPROB. To draw a st'cilmth line at right angles to a given straight lineSJfro,; a given point.iaL that line. Let AB eo a given straight line, and C a point given in it; it is required to draw a straight line from the point C at right angles to AB. Take any point D in AC, and (,3. 1.) make CE equal to CDQ arid upon DE describe (1. 1.) the bquilateral triangle DFE, and join FC; the straight line FC iaawvn frry;i the giVen p:int C, is at right angles to th e giveli Straight line A B. BecalJse DC is equal to CE \:nd FC cormmn to the two tri angles DCF,.:EC F, the two sides DC, CGF are equal to the two 0C, CpF, each to each; but the base DF is also.qilal to the base EF- therefore the angle DCF is equal (8. 1.) to the angle ECF; and they are adjacent angles. But, when the adjacent angles which one straight line makes with another straight line are equal teo one another, cach of them is called a right (7. def.) angle; therefore each of the angles DCF, ECF, is a right angleo WhIekefbre, from the given point C, in the given straight line AB, FC has been o drawn at right angles to AWB3 Which was to be done. 0 V G E 0`1ti' tR. BOOG 0 0:PROP. X.Io PROB. To dv'ara a str'i lZigt ine perpendicularci- to a given straight line, of an unlim-nited length,fromn a given point awithout it. Let AB be a given straight line,:which may be produced to any l-angth both ways, and let C be a point without it. It is required to draw a straight line perpendi- C cular to AB frum the point C. Take any point D upon the sther side of AB, and fronm the centre C, at the distance CD] ) describe (3. Post.) the circle EGF meeting AB inF, G: and X-, -_ " " bisect (10. 1.) FG in H. and join CF, CH, CG; thestraight line CLI, drawn from the given point C, is perpendicular to the giveo straight line AB. Because FH is equal to HG, and HC common to the two triangles FHC, GHC, the two sides FI-I, -IC are equal to the two G-H, HC, each to each; but the base CF is also equal (I 1. Def. 1.) to the base CG; therefore the angle CHF is equal (8. 1.) to the angle CHG; and they are adjacent angles; now when a straight line standing on a straight line makes the adjacent angles equal to one another, each of them is a right angle, and the straight line which stands upon the other is called a perpendicular to it; therefore fiorm the given point C a perpendicular CLH has been drawn to the given straight line ABWhich was to be done. PROP. Xti. THEOG, The angles which one straight line makes w:ith anotlher upon one side ojit, are eilher two right angles, or are together equal to two right an.. gles. Let the straight line AB make with CD., upon one side of it the angles CBA, ABD; these are either two right angles, or are together equal to two right angles. For, if the angle CBA be equal to ABD, each of them is a right angle (Def. 7.); but, if not, from the point B draw BE at right an. - A, I/ X 1,' :ELEMENTS gles (11. 1.) to C'D; therefore the angles CBE, EBD are two right: angles. Now, the angle CBE is equal to the two angles CBA, ABE together; add tihe angle EBD to each of these equals, and the two angles CBE, EBD will be equal (2. Ax. ) to the three CBA,.&BE EBD. Again, the angle DBA is equal to the two angles DBE, EBA; add to each of these equals the angle ABC then will the two angles DBA, XBC be equal to the three angles DBE, EBA, ABC; but the angles CBE, EBD have been demonstrated to be equal to the same three angles; and things that are equal to the same are equal (1. Ax.) to one another; therefire the angles CBE, EBD are equal to the angles DB:A, A:7C; but CBEi, EBD, are two right angles; therefore DBA, ABC are together equal to two right angles. Wherefore whert a straight line, &c. Q, E. D. PROP. XIV. THEORo af, at a point in a straight line, tewo other straight lines, upon the opposite sides of it, make the adjacent angles together equal to two right angles, these zwo straight lines are in one and the same straight lineo At the point B in the straight line AB, let the twoV straight lines / BC, BD upon the opposite sides of AB, make the adjacent angles ABC, ABD equal together to two right angles. BD is in the same straight line with CB. / For if BD be not in the same / straight line with CB, let BE be -_, — in the same straight line with it; C -, therefore, because the straight line AB makes angles with the straight line CBE, upon one side of it, the angles ABC, ABE are together equal (13. 1.) to two right angles; but the angles ABC, ABD are likewise together equal to two righit angles: thereibre the angles CBA, ABE are equal to the angies CBA, ABD-; Take avay the common angle ABC, and the remaining angle ABE is eqalol (3. Ax.) to the remaining angle ABD, the less to the greater, which is impossible; therefore BE is not in the same straight line with PC. And in like manner, it may be demonstrated, that no other can be in the same straight line with it but BD, which therefore is in the same straight line with CB. VWherefore, ifat a point, &c. Q. ED. PROP. XV. THEORo If two straight lines cut one another, the vertical, or opposite angles are eqzualt Let the two straight lines AB, CDI cut one another iri the point E the angle AEC' shall be equal to the angle DEB, and CEB to AED, OF (:EOiVETRY. 3BOOK l. 8 For the angles COEA, AED, which the straight line AE makes with the straight line CD, are together equal (13. 1.) to two right angles; and the angles AET), DEB, C which the straight line DE makes with the straight line AB, are also together equal (13. 1.) to two right angles; A therefore the two angles CEA, AED are equal to the two AED, DEB; Take away the common angle AED, D and the remainingangle CEA is equal (3. Ax.) to the remaining angle DEB. In the same manner it may be demonstrated that the angles CEB, AED) are equal. Therey fore, if two straight lines, &c. Q. E. D. CoR. 1. From this it is manifest, that if two straight lines cut one another, the angles which they make at the point of their intersection, are together equal to four right angles. CoR. 2. And hence, all the angles made by any number of straigh. lines meeting in one point, are together equal to four right angles. PROP. XVL THEORT Yj one side of a triangle be p2roduced, the exterior angle is greater thaws either of the interior, and opposite angles. Let ABC be a triangle, and let its side BC be produced to D, the exterior angle ACD is greater than either of the interior opposite ano gles CBA, BAC. Bisect (10. 1.) AC in E, join A F BE and produce it to F, and / make EF equal to BE; join also FC, and produce AC to G. Because AE is equal to EC, / and BE to EF; AE, EB are equal to CE, EF, each to each; and the angle AEB is equal:(15. 1.) to the angle CEF, because / they are vertical angles; there- "ore the base AB is equal (4. 1.) to the base CF, and the triangle AEB to the triangle CPEF, and i the remaining angles to the remaininrg angles each to each, to which the equal sides are opposite; wherefore the angle BAE is equal to the angle ECF; but the angle ECD is greater than the angle ECF; therefore the angle ECD, that is ACD, is greater than BAE o In the same manner. if the side iLEMENTS BC be bisected, it may be demonstrated that the angle BCG, that is (15. 1.), the angle AC1D, is greater than the angle ABC. Therefore f one side, &c. Q. E. D. PROP. XVII. THEOR. jny two anCgles of a triangle are to)gether less thqua twzo right angllces Let ABC be any triangle; any two of its angles together are less than two right anoles. Produce BC to ID; and because ACD is -the exterior angle of the triangle ABC, ACD is greater (16. 1.) than the interior and op~ / posite angle AB'; to each of these add the angle ACB; therefore the angles ACID, ACB are greater than the angles ABC, -,- ACB; but ACD, ACB are to- ~:;: gether equal (13. 1.) to two right angies: therefore the angles ABCU BCA are less than two right angles. In. like manner, it may be dqnonstrated, that BAC, ACB, as also, CAB, ABC, are less than two ~'ight angles. Therefore, any two angles, &c. Q E. D. 7PROP" XVTi. TtiEOR, The gr'eater7 side of every tr'iazngle fihs the greater angle opposite to il Let ABC be a triangle of which the side AC is greater than the side AB; the angle ABC is also greate-r dhan the angle BCA. N From AC, which is greater that f AB, cut off (3. l ) AD equal to AB,,ad join BD nd join BD: and because ADB is the exterior angle of tho triangle p C BDC, it is greater (16. 1.) than the interior and opposite angle DCB; but ADB is equal (5, 1.) to ABD, because the side AB is equal to the side AD; therefore the angle ABD is likewise greater than the angle ACB; wherefore much more is the angle ABC greater than ~CBP Therefore the greater side &9e. Q. El2 [,}.T .OF GEOiMETR'. BOOKlo L PLQOP. XIX. THEOR, ~iThe gre*ater angle of every trianlgle is subtended by the g.'reate, s de, o0? has the greater side o)posite to it. Let ABC be a triangle, of which the angle ABC is greater than the angle BCA; the side AC is likewise greater than the side AB., For, if it be not greater, A.C must eitherbe equal to AB, or less than it; it is not equal, because then the angle.BC would be equal (5.. I.) to the angle ACB; but it is not; therefore AC is not equal to AB; neither is it less; because then the angle ABC,would:be less (1$8. 1.) than the angle - ACB; but it is not; therefore the side AC is not less than AB; -and it has been shewn that it is not equaL,to AB; therefore AC is greater -than AB. WItherefore th.,e greatst,angle, &c Q E. D. PROP. XX. THEOLL Any twosides ef a triangle are together greater than the third suidc Let ABC be a triangle; any two sides of it togethecr are greater -than the third side, viz. the sides BA, AC greater than the side BC; and AB, BC greater than AC; and B.C, CA greater than ABE Produce BA to the point D,.and make (3. I.) AD equal to AC; and join DU. A / Because DA is equal to AC, — he angle ADC is likewise equal / \ (5. 1.) to ACD;:but the angle _jJ~ BUD is greater than the angle _ ACD; therefore the angle BC- D is greater than the angle AD:C; and because the angle BUCD of- the triangle DCB -is greater than its angle BDC, and that the greater (19. 1.) side is opposite to the greater' angle: therefore the side DB is,reater than the sidie BC; -but.I)I is equal to BA and AC.togeth.1er; therefore BA and AC togetther.are greater than BC. In the same m uannerit may be demtlonstrated, tha the sides AB, BC are greater than CAX, and BC, CA greater than AyB Therefore any two sides, &c. Q. E, I) PROP. XXI. THEOWR Ifffromz the ends of one side of a triangle, there be dra:wn tIwo straigali lines to a point within the triangle, these two lines shall be less than, te other two sides of the triangle, but shall contain agreater angles Let the two straight lines BD, CD be drawn from B, C, the ends of the side BC of the triangle ABC, to the point D within it; BD and DC are less than the other two sides BA, AC of the triangle, but contain an angle BDC greater than the angle BAC. Produce BD to E; and because two sides of a triangle (20. 1.) are greater than the third side, the two sides BA, AE of the triangle ABE are greater than BE. To each of these add EC; therefore the sides BA, AC are greater than BE, ECo Again, because the two sides CE, ED, of the triangle CED are greater than CD, if DB be added to each, the sides CE, EB, will be greater than CD, DB; but it has been shewn that BA, AC are greater that BE, EC; much more then are BA, AC greater than B D, DC. Again, because the exterior angle of a triangle (16. 1.) is greate' than the interior and opposite angle, the exterior angle BDC of the triangle CDE is greater than CED; for the same reason, the exterior angle CEB of the triangle ABE is greater than BAC; and it has been demonstrated that the angle BDDC is greater than the angle CAEB much more then is the angle BDC greater than the angle BACo Therefore, if friom the ends of, &c Q. E. D. PROP. XXIIo PROB. To const nuct a trianzte of ewhich the sides shall be equal to three give;, straight lines; but any two whatever of these lines must be greaten than the third (20. 1.) Let A, B, C be the three given straight lines, of which any two whatever are greater than the thild, viz. A and B greater than C v A and C greater than B; and B and C than A. It is required to make a triangle of which the sides shall be equal to A, B, C, each tor eacht Take a straight line DE, terminated at the point 1D ),., but unlimited towards E, and make (3. 1.) DF equal to A, FG to B, and GHl centre F, at the distance FD, describe (3. Post.) the circle DKL; and from the centre G, at the distance GH, describe (3. A Post. ) another circle HI, K; and join KE, KG; the tri- B - angle IKFG has its sides equal to the three straight C lines, A, B, C. Because the point F is the centre of the circle DKL, FD is equal (11. Def.) to FK; but FD is equal to the straight line A; therefore FK is equal to A: Again, because G is the centre of the circle LKH, GH is equal (1. Def.) to GK; but GH is equal to C; there. fore, also GK is equal to C; and FG is equal to B; therefore the three straight lines KF, FG, GK, are equal to the three A, B, C: And therefore the triangle KFG has its three sides KF, pG, GK equal to the three given straight lines, A, B, C. Which was to be done. PROP. XXIII. PROD. Zit a given point in a given straight line, to make a rectilineal angle equal to a given rectiliaeal angle. Let AB be the given straight line, and A the given point in it, and PDCE the given rectilineal angle; it is required to make an angle at the given point A in the given straight line AB, that shall be C( A equal to the given rectilineal angle DCE. Take in CD, CE any points D, E, and join DE; and make (22. 1.) the triangle AFG, the sides of which shall be I AE equal to the three straight / B lines, CD, DE, CE, so that / CD be elual to AF, CE fo AG, and DE to FG; and be.. /:ause DC, CE are equal to B FA, AG, each to each, and the base )DE to the base FG: the angle DI(E is equal (8. 1.) to the angle ES:LEMENrS'FAG. Trlibre, at the given point A il the given straight line At, the angle FAG is made equal to the given rectilineal angle DCE, Which was to be done. PROP. XXIV. THEOR4 Jf tuwo triangles haive two sides of the one equal to two sides of the other; each to each, beut the canrle contained by the tzwo sides of the one great* er than the angle contained by the two sides of the other; the base of that zzhih. has the greater angle shall be greater than the base of the other. Let ABC, DEF be two triangles which have the two sides AB, AC 6qual to the two DE, DF each to each, viz. AB equal to DE, and AC to DF; but the angle BAC greater than the angle EDF; the base BC is also greater than the base EF. Of the two sides DE, DF, let DE be the side which is not greater zhan the other, and at the point D, in the straight line DE make (23. 1.) the angle IEDG equal to the angle BAC: and make DG equal (3. 1.) to AC, or DF, and join EG, GF. Because AB is equal to DE, and AC to DG, the two sides BA, AC iie equal to the two ED, DG, each to eacht and the angle BAC is equal to the angle EDiGD, therefore the 2 D base BC is equal (4. io) to the base EG; and because:D:G is equal to iDF, the angle DFG is equal (5, 1.) to the angle DGF; but the angle DGF is greater than the anmle _!.o EGF; therefore theC angle DFG is great F er than EGF; and much more is the angle EFG greater than the angle EGF; and because the angle EFG of the triangle EFG is greater than its angle EGF, and because the greater (9. 1.) side is opposite to the greater angle, the side EGG is greater than the side EF; but EG is equal to BC; and therefore also Bt is greater than EF. Therefore, if two triangles, &c. Q. E. ED. PROP. XXV. THEOR wf itwo triangles have two sides of the one equal to two sides of the other& each. to eachl, but, the base of the one greater than the base of the other; the angle contained by the sides of that thich has the greater base, shall be greater than the angle contained by the sides of the other. Let A1JC, DEF be two triangles which have the two sides AB. AC: ~equai to the tiwo sides DE, DF, each to each, viz. AB equal to DE, and AC to DF: but let the base CB be greater than the base EF, the angle BACi is likewise greater than the angle EDF. For, if it be not greater, it must either be equal to it, or less; but the angle BAC is not equal to the angle EDF, because then the base BCG would be equal,4. 1.) A ) to EF; but it is not; there fore the angle BAC is n't equal to the angle EDF; neither is it less; because then the base BC would be less (24. 1.) than the base \ EF; but it is not; there fore the angle BAC is not i "\ less than the angle EDF' and it was shewn that it is C not equal to it: therefore ] [C F the angle BAC is greater than the angle EDF. Wherefore, if two triangles,&c. Q.E. D. PROP. XXVI. THEORG f two triangles hadve two acngles of the one equal to two angles of the other, each to each; and one side equal to one side, vi2. either the side adjacent to the equal angles, or the sides opposite to the equal angles in each; then sholl the other sides be equal, each to each; and also the third angle of the one to the'third angle of the other. Let ABC, DEF be two A triangles which have the angles ABC, BCA equal G to the angles DEF, EAED, viz. ABC to DEF, and BCA to EFD'; also oneside equal to one side; and first, let those sides be equal which are adja- cent to the angles that \ _ are equal in the two tri- C EgF angles, viz. BC to EF; the other sides shall be equal, each to each, viz. AB to DE, and AC to DF; and the third angle BAC to the third angle EDF. For, if AB be not equal to DE, one of them must be the greater. Let AB be the greater of the two, and make BG equal to DE, and join GC; therefore, because BG is equal to DE, and BC to EF, the two sides GB, BC are equal to the two DE, EF, each to each; and the angle GBC is equal to the angle DEF; therefore the base GC is equal (4. 1.) to the base DF, and the triangle GBC to the triangle E LI ijl l EN It 6 DEF, and the otther angles to the other angles, each to each, to whict'i the equal sides are opposite; therefore the angle GCB is equal to the angle DFE, but 1DFIE is, by the hypothesis, equal to the angle BCA; wherefore also the angle BCG is equal to the antgle BCA, the.ess to the greater, which is in;possible; therefore AB is not unequal to DE, that is, it is equal to it; and BC is equal to EF; therefore the two AB, BC are equal to the two DE, EF, each to each; and the angle ABC is equal to the angle DEF; therefore the base AC is equal ( i. 1.) to the base DF, and the angle BAC to the angle QIDY Next, let the sides which are opposite to equal angles in each ri angle be equal to one an- IX other, Viz. AB to DE; likewise in this case, the other sides shall be equal, 1, AC to DF, and UC to EF; and also the third angle BAC to the third EDF. DI_ For, if BC be not equal to EFi', le BO be the greater of athem, and make BH equal to EF, and join AH; and because BH is equal to IEF, and AB to: E; the two 3AB, BH are equal to the two DE, EF. each to each; and they contain equal angles; therefore (4 1.) the base AH is equal to the base DF, and the triangle ABH to the trian. gle DEF, and the other angles are equal, each to each, to which the equal sides are opposite; therefore the angle BHA is equal to the angle EFD; but EFD is equal to the angle BCA; therefore also the angle BHA is equal to the angle B. A, that s, the exterior angle BHA of the triangle AHC is equal to its interior aud opposite angle BCA,'which is impossible ( 16. to); wherefore BC is not unequal to EF, that is, it is equal to it; and AB is equal to DE therefore the two AB, BC are equal to the two DE, EF, each to each; and they contain -equal angles; w'here fore batse AC is eqlal to tthe base FD, and t.he third angle BA C to the third angle EDF. Therefo:re. if tvo tori a gles, &:c. EQ.' D PP.0P XXVITL THE0X a. st'asIraiht iinejT flitg uIpoon gwo other str aight 1ines -akes the alte't.na e angles equal to one alother, these two straight lines are parallel~ Let the straight line EF, which falls upon the two straight lines AB, CD make the alternate angles AEF, EFD equal to one another; AB is parallel to CD.; VFor, if it be not parallel, AB and CD being produced shall imeet either towards B, I), or towards A, C; let them be produced and meet towards B. -) in the rp)oint G4 therefore GEF is ai triangle, and OF GEiOOMI ETELY. BOOK i( its exterior angle AEF is greater (16. 1.) than the islteri6r and oppop site angle. EFG; but it is also equal to it, which is impossible: therefore, AB A and CD being produced, do not meet towards B, D. In like manner it may be demtonstrated that they do not meet towards A, C; but those straight lines uwhich meet neither way, though produced ever so far, are paralleli (30. Def.) to one another. AB therefore is parallel to CD. Where~ fore, if a straight line, &c. Q.E. D. PROP. XXVl1I. THEOR. if a straight line falling upon two other straight'lines makes the cxterio:? angle equal to the interior and opposite upon the samne side of the line; or makes the interior angles upon the same side together equal to two right angles; the two straight lines are p,-Trallel to one another. Let the straight line EF, which falls upon the two straight lines AB, CD, make the exterior angle EGB equal to GHD, the interior and opposite angle upon the, same side; or let it make the interior angles on the same A side BGH, GHD together _ - equal to two right angles; AB is parallel to CD. Because the angle EGB'is equal to:the angle GHID, and C = - also (15. 1. ) to the angle AGH, the angle AGH is equal to the angle GHD; and they are the alternate angles; therefore AB is parallel (27. 1.) to CD. Again, because the angles BGH, GHD are eqlal (By Hyp ) to right angles, and AG H, BGH, are also equal (13. 1.) to two right angles, the -angles AGH, BGHI are equal to the angles BGH, GHD: Take away the common angle BGH; therefore the remaining angle AGH is equal to the remaining angle GHDi; and they are alternate angles; therefore AB is parallel to CD. Wherefore, if a straight line, &c. Q. E. Do PROP. XXIX. TIIEOR. ~' a straight linefall upon two parallel straight lines, it makes the alte2;" nate angles equal to one another. an3d the exterior angle equal to the interior and opposite upon the same side; and likewise the t'o interior angles upon the same side together equal to two right angles. Let the straight line EF fall upon the parallel straight lines AB, CD; the alternate angles AGHI GHlD are equal to one another;: and i4s LLELEMENTS the exterior angle EGB is equal to the interior and opposite, upwo the same side, GHD; and the two interior angles BGH, GHD upon the same side are together equal to two right angles. For if AGHI be not equal to GI-ID, let KG be drawn making the angle KGH equal to GI-ID, and produce KG to L; then KL will be parallel to CD (27. 1.); but E AB is also parallel to CD; therefore two straight lines are drawn through the same._ _ _ _ point G, parallel to CD, and \ yet not coinciding with one another, which is impossible C (11. Ax.) The angles AGH, - - GHD therefore are not unequal, that is, they are equal to one another. Now, the angle EGB is equal to AGH (15. 1.); and AGH is proved to be equal to GHD; therefore EGB is likewise equal to GHID; add to each of these the angle BGH; therefore the angles EGB, BG-H are equal to the angles BGH, GHD; but EGB, BGH are equal (13. 1.) to two right angles; therefore also BGH, GHID are equal to two light angles. Wherefore, if a straight line, &c. Q. E. D. COR.. If two lines KL and CD make, with EF, the two angles KGH, GHC together less than two right angles, KG and CH will meet on the side of EF on which the two angles are that are less than two right angles. For, if not, KL and CD are either parallel, or they meet on the;other side of EF; but they are not parallel; for the angles KGH, -GHC would then be equal to two right angles. Neither do they meet on the other side of EF; for the angles LGH, GHD would then be two angles of a triangle, and less than two right angles; but this is impossible; forthe four angles KGH, HGL, CHG, GHD are together equal to four right angles (13.o 1.) of which the two, KGIH, Cl1G are by.supposition less than two right angles; therefore the other two, HGL, GIlD are greater than two right angles. Therefore since iL and CD are not parallel, and since they do not tmeet towards L and Dt, -hey must meet if produced towards K and C, PROP. X XX THLIEORo Strai,ght lines uahich are parasllel to the same straight line are paralie:, to onle anc ther. Let AB, —CD, be achah of them parallel to EFR; AB is also parallel:to CD, Letthe straight line G HK cut AB, EF, CD; and because GHK cnts the parallel straight lines AB, E~., the angle:AGH is equa-.. ~ ua -~ m, ~-.~. OF GEOMETRY. BOOK 1. 4A?29.'.) to the angle GHF. A. gain, because- the straight line GK cuts the parallel straight A __i ___ lines EF, CD, the angle GHEF is equal (29. 1.) to the angle GKD: and it was shewn that thp angle AGK is equal to the angle GHF.; therefore also A GK is equal to GKD; and they are -. alternate angles; therefore AB is parallel (27. 1.) to CD. Wherefore straight lines, &c. Q. E. D. PROP. XXXJI PROB. -To draw a straight line through a given poin:t parallel to a given straight line. Let A be the given point, and BC the given straight line, it is rDI quired to draw a straight line through the point A, parallel to A the straight line BC. In BC take any point D, and join AD; and at the point A,. in the straight line AD, make: I C (23. 1.) the angle DAE equal to the angle ADC; and produc~e the straight line EA to F. Because the straight line AD, which meets the two straight lines BC, EF, makes the alternate angles EA AD, ADC equal to one another, EF is parallel (27. 1.) to BC. Therefore the straight line PEAF is.drawn through the given point A parallel to the given straight lnja BC. Which was to be done. PROP. XXXIL. THEORo i a side of any triangle be produced, the exterior angle is equal to the two itnterior and opposite angles; and the three interior angles of every triangle are equal to two right angles. Let ABC be a triangle, and let one of its sides BC be produced to ]D; the exterior angle ACD is equal to the two interior and opposite angles CAB, ABC; and the three interior angles of the triangle, viz. AI$C, BCA, CAB, are together equal to two right angles. Through the point C draw CE parallel (31. 1.) to the straight line AB; and because AB is parallel to CE, and AC meets them, the alternate angles BAC, ACE are equal (29. L.) Again, because AB is B parallel to CLf, n ld fl flls upoa them, thde exterior angie ECD,E equal tot the irl1i a:nd opposite angle ABC, but the angle ACL ywas shevni to hb eqotial i' the ancgle BAC; fheref'ore the whole exteri-. or tItngle ACD is equal to tlhe tx:o interlio' and opposite angles CAB, ABC; to the.age aongl es add tthe anlfle AC B, and the;angles ACD, ACB arei equarl tio tnll e e anrl, C4BA, BAC, CB; but the angles ACD AC B are, eqooidl (;to two vc rigit angles; therefore also) the angles CBA, BAR,;aCB.re equal to tiwo right angles, %Wherefore, if side of a triai, &co Q c. E. Do Cooo o. All the interior angles of any reetilineal figure are equal to twice as n any right angles as the figure has sides, wanting four right anglesi,,For any rectilineal figure ABCDE can be divided in-to as many trni angles as the figure has sides, by drawing straight lines from a point,; F within the fi,-ure to each\ of its angles. And, ty hle preceding proposition, all the angles of these tri angles are equal to twice as -many right anuies as there are triangles, that is, as." there are sides of the figur; anld the same angles are equal to the angles of the figure, together witl the angles at the point F, which is the commoin ier- tex of the triangles; that is, (2 Coor, l5..) together with f,nur right angleso Therefore, twice as many right angles as the figure has sides, are equal to all - the angles of the figure, together with four eaght angles, that is, the angles of the figure ae equal to twice as many right angles is the figure has sides, wanting foiur Cor. 2. All the exterior angles of a ectielin fignre are, iP ther equal to finr righlit atges, Becaultse every interiri angle ABC, with its adja-./' cennt exterior A BI) is e f tqua (13. 1. ) to two right / angles; ther efre all tihe interior, together with all l — the exteario< an; es of the II fiure, are equal to twice as many rigtt analees as there are sides of the figure; that is, by the fores.... going corollary, they are equal to all the interior angles of the figure, to. / gether with four tight angles hlereforeall the exterior aingles ai tqu1 te B.r fright anglea, PROP. XXXII. THEOR. Vie straight lines which jinm the extr. entities jf wo evq'cal and pa'alleid strazght lines, towarts the same parts, are also themselves eqeutl and parallel. Iect AB, CD9 be equal and parallel straight lines, and ioined towards the same parts by the straight lines AC, BD1; AC, B U are also equal and parallel. Join BC; and because AB is pa-. iallel to CD, and BC imets thmn, -:he alternate angles ABC, BCD are equal (29. 1.); and because AB is equal to CT), and 1GC co n; on to tflhe / two triangles ABC, t)CB, the two <'sides AB,-BC are equal to the two. 1. DC, CB; and -the angle ABC is cequal to thet>e afn{h'..1oBD; therefore the base AC is' equal (4. 1.) to the ba!se; BD, and thet tlrianle A BC to the triangle BCD), and the other arnniest t(, the otter angles (4 1.) each to each, to which the equal sides are o))ooqi l; tilelefore the angle ACB is equal to the angle CBI); and 11cause the straight line.BC meets the two straight lines AC, BD, and mnkuss the:!iternate angles, ACB, CBD equal to one another, AC is parallel (27. 1.e to BD; and it was shewn to be equal to it. Therefore straight lines, &c, Q.l o D. PROP. XXXIV. THEOR. hlie opposite sides and angles of a paorallelogra:,l arte equall to one ano.. ther, and the diamneter bisects it that is, t-ia'es'it into.two eqaln parts. "N. B. A Parallelra is foursider figure, of which the opposite sides are parallel; and the diameter is the straighlt line joinitg twvo of its opposite ana'les. Let AC )B be a paraliellogram, of whjich C is a diame(ter; tleb opposite sides and angles of the figure re equal to one tlanotbher andi the diamneter BC bisects it. Because AB is parall to CD, and A' i3C meets them, the alternate angles ABC, BCD are equal i'29. 1) to one ancother; and because AC is psrallsl to BID, and BC meets theve, the alternate arngles aCB, CBD ace eual (29. 1.) to one another; wherefore thte two triangles AiSC, CBD have two angles ABC,, BCA in one, equal to two angles BCD, CBD in tlhe other, each to each, and the side BC, which is adjacent to these equal angles, common to the two triangles.; therefore their other sides air, equal, each to eacll and the third atigle of the one to the third angle of the other (2k. i.) viz. the side AB to the side CD, and AC to BD, and the angle BAC bqual to the angle BDC. And because the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB, the whole angle ABD is equal to the whole angle ACD: And the angle BAC has been shewn to be equal to the angle BDC; therefore the opposite sides and angles of a parallelogram are equal to one another; also, its diameter bisects it: for AB being equal to CD, and BC common, the two AB, ]BG air equal to the two DC, CB, each to each; now the angle ABC is equal to the angle BCD; therefore the triangle ABC is equal (4. 1.) to the triangle BCD, and the diameter BC'diVides the parallelogramn ACDB into two equal parts. Theiefore, &c. Q. E. D, PROP. XXXV. THIEOR. Paralleloggramins upon the same base anad between the same parallels, anr?equal to one another. (SEE TaE 2d AND 3d FIGURES.) Let the parallelograms ABCD, EBCF be upon the same base BC, and between the same parallels AF, BC; the patallelogram ABCD iz bqual to the parallelogramn EBCF. If' the sides AD, DF of the paral A. D ieiograms ABCD, DBCF opposite to the base BC be terminated in the same point D; it is plain that each of the parallelograms is double (34. 1.) of the triangle BDC; and they are therefore equal to one another. B C But, if the sides AD, EF, opposite to the base BC of the parallelo: jrams ABCD, EBCF, be not terminated in the same point; then, because ABCD is a parallelogram, AD is equal (34. 1.) to BC; for the same reason EF is equal to BC; wherefore AD is equal (1. Ax.) to EF; and DE is common; therefore the whole, or the remainder, AE is equal (2. or 3. Ax.) to the whole, or the remainder DF; now AB is also equal to DC; therefore the two EA, AB are equal to the two:A D.A ED DF B C b3 C t3') DC, reach to each; but the exterior angle FDC is equal (29. 1.) t* the interior EAB, wherefore the base EB is equal to the base FC, OF tGE61ETRYE iBOOK.0 4L arid the triangle EAB (4. 1.) to the triangle FDC. Take tle triahgle FDC from the trapezium ABCF, and from the same trapezium take the triangle EAB; the remainders will then be equal (3. Ax.), that is, thile parallelogram ABCD is equal to the parallelogram EBCF. Therefore, parallelograms upon the same base, &c.. E. D. PROP. XXXVI. T-lEOR. ra'rallelogram-s p2on equal bases, and between the same parallels. are equal to one another~ Let ABCD, EFGH be parallelograms upon equal bases BC, FGA and between the same parallels AH, BG; the parallelo-D E il gram ABCD is equal to Join BE, C-I; and because BC is equal to PG, and FG to (34. 1.) EL-, BC / is equal to E H; and they B are parallels, and joined too wards thie same parts by the straight lines BE, CH: But straight lines which join equal and parallel straight lines towards the same parts, are themselves equal and parallel (33. 1.); therefore EB, CI are both equal and parallel, and EBCH is a parallelogram; and it is equal (35. 1.) to ABCD, because it is upon the same base BC, and between the same parallels BC, AHl; For the like reason, the parallelogram EFGH is equal to the same EBCH: Therefore also the parallelogram ABCD is equal to EFGHo. Wherefore, parallelograms, &c; Q. E. D. PROP. XXXVIL THEOR. triangles upon the same base, and between the same parallels, are equal to one aisotterO Let the triangles ABC, DBC be upon the same base BC, and between the same parallels, AD E - BC: The triangle ABC, is / equal to the triangle DBC. Produce AD both ways to the / points E, F, and through B draw / (31. 1.) BE parallel to CA; and / through C draw CF parallel to BD: Therefore, each of the 3 C figures EBCA,DBCF isa parallelogram-; and DBCA is equal (35. 1.) to DBCF, because they are gplon the same base BC, and between the same parallels BC. EF Mj 0 s EN TiS but the triangle -BC is the. half of' h parallelogra t' 1BE A, becatiuse the diameter AB bisects (31. 1.) it; and the t'iangtle DBC is the half -of tihe parallelogram 1)Bi )', because the diameter D)C bisects it; and the halves of equal things are equal. (7; Ax.}; the efitoe tile triangle ABC is equal to tile triangle D1B. TVherefbre triangles, &c. Q.. E.0 D PROP. XXXVILo THEOE. Triangles,utpo2 equal bases, rnd beween the same parallels, are eqlai to one another. Let the triangles ABYC, DItF he upon equal bases BC, EF, and between the same. parlallels BF, AD; The t riangle ABC is equal to the triangle ) E'. Produce AD both ways to the points G, H, and thro'ghc B draw Bv parallel (31. L) to CA, and through F draw-FH1 parallel to ED Then each of the figaUres. GIBCA,EFFM is a paadl _ _ angle ABC is equal to the triangle DE F; therefore also the triangle DEF is equal to the triangle GEF, the greater to the less, which is impossible; Therefore AG is not parallel to BF; and in the same manner it may be demo nstrated that tLhele is nt other parallel to it bUt AD; AD is therefore parallel t o BF her e equ- al tr iangle &c. Q. E. D.8 Pi ROP. XL T HEOR ~f a paarallelogram and a triangle be spon the samne Lnse, anid. betwcee?, The same parallel; the parallelogram is doublt of the triagtleo Let the parallelogram.ABCUD and the triangle EBC be upon the sarne base BC and between the same pa-,allels BC, AE; the parallelogram ABCD a. is double of the triLngle EBCe Join AC; then the tianae ABC is equal (37. 1.) to the triangle EBC, because they \ are upon the saime base BC, and between K / the same parallels BC, A Ee But tle partal \ - ///~ lelogram ABCD is double (34. t.) of the/ triangle'AC, because s the diameter AC divides it into two equal parts;s where fore ABCD is also double,fthe rtiangie P' YIBUC. Therefore, if a l)realle)ogtas &c. Q.. Px1:'C. X. PRXOB7 2'o' describe a pa'railelogrcsra that shall cc cO'Ma tl' a'veil trictngfie ala, have one of its angles eqcal to a givet,,t rcctilincel angle. Let ABC be te -given triangle and thD e 1 iveu:;e! tiin.nial l agle-. It is required to describe a parallelogram that shall be equal to the given triangle ABC, and have one of its angles equal to D. Bisect (10. 1.) BC in E, join AE, and at the point E in the straight line EC make(23. 1.) the angle CEF equal to D; and through A draw (31. I.) AG parallel to BC, and through C draw CG (31. 1.) parallel to EF: Therefore F CG F: G is a parallelogram: And because - BE is equal to EC, the triangle ABE is likewise equal (38. 1.) to the triangle AEC, since they are upon equal bases BE, EC, and between the same parallels BC AG; therefore the triangle ABC \ is double of the triangle AEC.o D And the parallelogram FECG is B'E C likewise double (41. 1.) of the triangle AEC, because it is upon the same base, and between the same parallels: Therefore the parallelogram FECG is equal to the triangle ABC, and it has one of its angles -CEF equal to the given angle D; Wherefore there has been described a parallelogram FEGG equal to a given triangle ABC, having one of its angles CEF equal to the given angle D. Which was to be done. PROP. XLIII. THIEOR. ithe complements of the parallelograms which are about the diameter ofany parallelogram, are equal to one another. Let ABCD be a parallelogram of which the diameter is AC; let EH, FG be the parallelograms about AC, that is, through which AC passes, and let B K, KD be the other parallelograms, which make up the whole figure ABCD, and are there- fore called the complements: The L _ _ complement BK is equal to the com / plement KD. I / Because ABCD is a parallelogram i and AC its diameter, the triangle ABC is equal (34. 1.) to the triangle ADC: And because EKHA is a pa- B Crallelogram, and-AK its diameter, the triangle AEK is equal to the triangle AHK: For the same reason, the, triangle KGC is equal to the triangle KFC. Then, because the triangle AEK is equal to the triangle AHKE, and the triangle KGC to the triangle KFC; the triangle AEI, together with the triangle KGCR is equal to the triangle AHK, together with the triangle KFC: Brut the whole triangle ABC is equal to the whole A DC; therefore the re.rmaining complement BK is equal to the remaining complement KIM,:hereforeF, the comnlemenats, &x9;. e_ T'r 01F GEOMETRY. BOOK i. 4 PROP. XLLV. PROB.:'To a given straight line to apply a parallelogram, Which shall be equat to a given triangle, and have one of its angles equal to a given reccti lineal angle. Let AB be the given straight line, and C the given triangle, and D the given rectilineal angle. It is required to apply to the straight line AB a parallelogram equal to the triangle C, and having an angle equal o De. Make (42. 1o) the parallelogram BEFG equal to the triangle Ive \j/ i j U, having the angle EBG equal to the angle D, and the side BL in the same straight line with AB: produce FG to H, and through A di':y,(31. 1.) All parallel to BG or EF, and join HB. Then because the straight line HF falls upon the parallels AHl, E.T, the angles AHIF, HFE, are together equal (29. 1.) to two right angles; waherefore the angles BHF, HFE are less than two right angles: But straight lines which with another straight line make the interior angles, upon tho same side less than two right angles, do meet if produced (Cor. 29. 1.): Therefore HB, FE will meet, if produced;- let them meet in., and through K draw KL parallel to EA or FH, and produce 1-A, GB to the points L, M: Then HLKF is a parallelogram, of which the diamee ter is HK, and AG, ME are the parallelograms about HEK; and LB, BF are the complements; therefore LB is equal (43. 1.) to BF: but BF is equal to the triangle C; wherefore LB is equal to the triangle C; and because the angle GBE is equal (1 5. 1.) to the angle ABIA and likewvise to the angle D; the angle AliM is equal to the angle D - Therefore the parallelogram LB, which is applied to the straight line AB, is equal to the triangle C, and has the angle ABM equal to tlhe angle D W: hich was to be done. PROP. XLV. PROB. To describe a parallelogram e qual to a given rectilineacf igur c' atd hav.. ing an angle eqtal to a given rectilineal angle. Let ABCD be the given rectilineal figure, and E the given rectili, neal angle. It is required to describe a parallelogrlam eoual to A-B. CL an.d having an angle equal to Ee g~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ e,f Join DB, and describe (42. 1.) the parallelogram FI equai tho thl triangle ADB, and having the angle HKF equal to the angle E; and to the straight line GIH (44. 1,) apply the parallelogram GM equal to the triangle DBC, having the angle GIHM equal to the angle E. And because the angle E is equal to each ofthe angles FKH, GHM, the angle FKH is equal to GHM; add to each of these the angle KHG; theres fore the angles FKH, KHG are equal to the angles KHG, GHM; but FKH, KHG are equal (29. 1.) to two right angles; therefore also kHG, GHM are equal to two right angles: and because at the point 3D.F II L // / / / / \ ~ / / H in the straight line Gi, the tHwo staight lines RH, -iM, upon tile opposite sides of GH, malke the adjacent angles equal to two right an gles, K(H, is in the same straight line (14. 1.) withriT. And be. cause the straight line HG meets the parallels KM, FG, the alternate angles MHG, HGF are equal (29. 1.); add to each ofthese the anm gle HGL; therefore the angles MHG, HGL, are equal to the angles HGF, 1HGI: But the angles 1MUG, HGL, are equal (29. 1.) to t.wo aight angles; wherefore also the angles HGF, HiGL, are equal to two right angles, and FG is therefore in the same straight line with GLo And because KF is parallel to HIG, and HG to ML, KF is parallel (30, 1.) to ML; but KM, FL are parallels: wherefore KFLM is a parallelogram. And because the triangle ABD is equal to the parallelogram H F, and the triangle DBC to the parallelogram GM, the whole rectilineal figure ABC D is equal to the whole parallelogram KFLM; therefore the parallelogram KFLM has been described equal to the given recti. lineal figure ABCD, having the angle FKM equal to tie given angle El. Which was to be done. CoR. From this it is manifest how- to a given straight line to apply a parallelogramn, %which shall have an angle equal to a given rectilia neal angle, and shall be equal to a given rectilineal figure, viz. by applying (44. 1.) to the given straight line a parallelogram equal to the first triangle ABD, and having arn angle equal to the given angle. PROP. XLVI. PROB. To describe a sqvare upoL n a given stra'ig Uine Let AB be the given straight line: t i s required to describe a square upon AD. From the point A draw ( 1. 1.) AC at right angles to AB; and make (3. 1.) AD equal to AB, and tihrugh the oisnt D draw DE: P-, OF (GEO01.iTRYo BOO8K I fsalie <3 1. 1.) to AB, and through B draw BE parallel to AD; therefore ADEB is a parallelogram; whence AB is equal (34. 1.) to DE, and AD to BE; but BA is equal to AD C therefore the four straight lines BA, AD, DE, EB are equal to one another, and the parallelogram ADEB is equilateral; it is likewise -rectangular; for the straight line AD meeting the parallels, AB, DE, makes the angles BAD, ADE equal (29. i.) to two right angles; but BAD is a right angle; therefore also ADE is a right angle now the opposite angles of parallelograms are equal (34. 1.); therefore each of the opposite angles ABE, BED is a right anr AB gle; wherefore the figure ADEB is rectangular, and it has been demonstrated that it is equilateral; it is therefore a square, and it is described upon the given straight line AB.; Which was to be done. CoR. Hlence every parallelogram that has one right angle has alW its angles right angles. PROP. XLVII. THEOR. in any right angled triangle, the square which is described upon the side subtending the right angle, is equal to the squares described upon the sides which contain the right angle. Let ABC be a right angled triangle having the right angle BAC; ile square described upon the side BC is equal to the squares deserib; ed upon BA, AC. On BC describe (46. 1L) the square BDEC, and on BA, AC the squares GB, HG; and through A draw (31. 1.) AL parallel to BD or CE, and join AD, FC; then, because eachof the angles BACG BAG is a right angle (25. def.), the two straight lines AC, AG upon the opposite sides of AB, make with it at the point A the adja cent angles equal to two right an-e gles; therefore CA isin the same straight line (14., 1 ) with AG; for the same reason, AB and All - are in the same straight line. Now because the angle DBC is equal to the angle FBA, each of B C them being a right angle, adding to each the angle ABC, the whole angle DBA will be equal (2. Ax.) to the whole FBC; and because the two sides AB, BD, are equal to the two FB, BC each to eaceh, and theangle 1DBA equal the jLE',I ENi C angle FBC, therefore the base AD is equal (4. 1.) to the base FCi: and the triangle ABD to the triangle FBC. But the parallelogram BL is double (41 1.) of the triangle ABID, because they are upon the same base, BD, and between the same parallels, BD, AL; and the square GB is double ofthe triangle BFC, because these also are upon the same base FB, and between the same parallels FB, GC. Now the doubles of -equals are equal (6. Ax.) to one another; therefore the parallelogram BL is equal to the square GB: And in the sarne manner, by joining AE, BK, it is demonstrated that the parallelogram CL is equal to the square I-IC. Therefore, the whole square BDECI is equal to the two squares GB, HC; and the square BDEC is described upon the straight line BC, and the squares GB, HIC upon BA, AC: wherefore the square upon the side BC is equal to the squares upon the sides BIA, AC. Therefore, in any right angled triangle, &e. qE. D PROP. XLVIII. TIlEOR. If the square described upon one of the sides of a triangle, be equal to the squares described upon the other two sides of it; the angle con.l tained by these tizo sides is a right angle. If the square described upon BC, one of the sides of the triangle ABC, be equal to the squares upon the other sides BA, AC, the angle 3BAC is a right angle. From the point A draw (11..1.) AD at right angles to AC, and mSake AD equal to BA, and join DC. Then because DA is equal to AB, the square of DA is equal to the square of AB; To each of these add the square of / i AC; therefore the squares of DA, AC are /. equal to the squares of BA, AC. But the / square of DC is equal (47. 1.) to the squares A. of DA, AC, because DAC is a right angle / and the square of Bi, by hypothesis, is equal / to the. squares of BA, AC; therefore, the square of DC is equal to the square of BC; B C and therefore also the side DC is equal to the side BC. And because the side DA is equal to AB, and AC toinmmon to the two triangles DAC. BAC, and the base DC likewise equal tothle base BC, the'angle DAC is equal (8. 1.) to the angle BAC,; But DAC is a right angle; therefore also BAC is a right angle. Therefore if the square, &c. Q. ED ELEMENTS OF G EO M ET R BOOK I1. DEFINITIONS. I. _EVERY right angled parallelogram, or rectangle, is said to be con~ tained by any two of the straight lines which are about one of the right angles. " Thus the right angled parallelogram AC is called the rectangle " contained by AD and DC, or by AD and AB, &c. For the sake of " brevity, instead of the rectangle contained by AD and DC, we shall " simply say the rectangle AD.DC, placing a point between the two "sides of the rectangle. Also, instead of the square of a line, for " instance, of AD, we shall frequently in what follows write AD2." " The sign + placed between the names of two magnitudes, signi-' ies that those magnitudes are to be added together, and the sign -' placed between them, signifies that the latter is to be taken away "from the former." " The sign = signifies, that the things between which it is placed s are equal to one another." II. In every parallelogram, any of the parallelograms about a din- A i ameter, together with the twol complements, is called agne - Go mon. "Thus the parallelo- "gram HG, together wither with the i "'complements AF, F C, is the. "gnomon of the parallelogram: "AC. Thisgnomon may also, 6 for the sake of brevity, be.3 O "called the gnomon AGTK or ( .3'} _:6 the straight line AC must be within (?2 3.) the same circle, which is \ absurd: Therefore a circle cannot touch another on the outside in more than one'B noint: and it has been shewn, that a cirle cannot touch another on the inside in more than one point. Therefore, one circle, &c. Q. E, P. PROP. XIV. TWEOR.o 2'qua! strtighlt lines in a circle are equally distant frorn the centre; Ca:d; thiose whrich are eqyzully disttantJrom the centre, are equal to onPe another, Let the straight lines AB, CD, in the circle ABDC, be equal to one'another; they are equally distait fr'om tlhe centre. T'ake E the centre of the circle ABDC, and fr om it draw EF, Ew - perpendiculars to AB, CD; join AE and EC. Then, bocause the straight line ElB passing through thie ce ntre, uis tile straight line AB, which does hnot pass through the cenflre at i'ght an 7. gles, it also bisects (3. 3.) it: Where- At fore AF is equal to FB, and AB double F AF,. For the same reason, C D is double f CG: But AB is equal to CD therefore A'_'is equal to CG: And be\- ~ ause AE is equal to EC, the square do AE~ is equal to the square of EC: Now B D Jqa squares of AF, FE are equal (47. 1.) tb thoe square of At[B, because thie angle AFE is a fright anle; andfor the Eke reas'fi the squares of F4IG, GC are equal to the square of EC: Therefo>re tl.e squares of AF, FE are equal to the squares of CG, GE, of which the square of AF is equal to the square of CG, bc, cause ~AF;is equal to (G; therefire the remaining square of FE is equal t th e remaiing square of EG, and the strailght line EF is there, fore equal teo!:G:; But qtraight lines in a circle are said to be equally distant from the centre6 -w J.,en the perpendiculars drawn to them frone hp 9centrte re equal (3, DeI 3.) T herefoe AB, GC are, cqu!lHY dis tant tfrorn fh'e,entre. fNext, if the sttaight lines Ak3, t0D be equally dittant firom the e-itt:s, that is, if FE be equal to EG, -AB is equal to CD, For, the samdiie a-enstructioa being made, it may, a-s before, be dernoistrated, that AB is double clf AF, and CD double of' C, and that the sqtiares -of EF, FA are equal to the squares of EG, GC; of which the square -of FE is,equal to the square of EG. because FE is equal to EG; Ihlherefore the remaining square of A F is equal to the remaining square -of:.CG; and thestraight line AF is theTetbre equal to CUG: But AB,s double of AF, and CD double of CG wherefore AB is equal to CI)Do Therefore equal straighit fines-, &c. Q.E. D PROP XV. THEOR. iThe dia'meter,`zs,:'s.e greatest strai~gz lixne in a circle; aind of all others.Chat which is n ie-a'r to the centre'is always greater than on9e,or-e re-;rote; and: the greater. nearer tb the centre thaTl the less. Let ABCD be a circle,:of which the di A ameter is AD, and the centre.-&; and )e:,.- - BC be nearer to the centre than -.6s; AD is.greater ihan any straight line BC which E/ is not a diameter, and BC greater than!:G. Fromr the centre draw El-, EK perpen diculars to BC, FG, and join EB. EC, EF \:,and because AE is equal to EB, and ED to iEC, AD is equal to'EB, EC: But E'B, EC ia're greater (20. 1.) than BC; wherefore, -1 lblAD is greater thatn BC. .rid, because BC is nearer to the, centre than iF, EH is less 4. Det: 3.) than I'i.2; But, as was demonstrated in the preceding, BC is double of 13A, aoJl F double of IK, and the squares of EiH, HB are equal to t.ho. squares of EK, IK, of' whci the square of EHl is leos thani the squtare of 2EK, hecatse LE is less than EK; thereI'ore the squaire of f~Il is greater thlan th;, sqluare ol'FK, and the straigh t hline BH Greeater tharl Fr a;mc thecefforo BC is -greater than FG.' Next, let BC be greater thara.; BC is nearer to the centre tha.l FiG: that is, the sam.e co n: tItraction being made, EHl is less than EKr Because BC is greater than FG, BLt likewise is greater than KF: but the squares of B'l, IEI are equal to the squares of FK KiF, of whicl the square of BHl is greater than the square of 2IK, because BH is greater than t1; therefore the square of ELH is less than the sqttuai' of e, and thle straight lia E loss tn E than 1; Wherefrl 8r, T;he diamceter, &c. Q. 5. D PROP. XVI. THEO 0 2'4c straight line d;Z;awn at right angles to the diarme'ler e'a circle,jfrro,...he extremituy of it, falls witvhout lhe circle; a(.zd no straghi't line caI be drawn between that straight line and the circurtference, frono the extremity of the diani;ter, so as not to cut the. circle. Let ABC be a circle, the centre of which is D, and thle diameter AB: and let AE be drawn frorm A perpendicular to AB, A E shall fill without the circle. In AL take any point F, join DF and.et DF meet the circle inl C. Because D AF is a right angle, it is greater than the angle AFD (32. 1); but the greater angle of any triangle is subtended by the greater side (19. 1.), therefore D3F is greater than DBA now DA is equal to DC,- therefore DF is greater than DC, and the point Fi is therefore without the circle. And F is any point whatever in the 3 >.1 line AE, therefore AE falls without tihe circle. Again. between the sttraight line -AE and the circumference, no straight line can be drawn from the point A, -which does not cut thle circle, L-et AG be dra;wn in the angle T)AE'i ,76 ELJM~1i3EhAl from D draw DH at right angles to AG; and because the angle DIHA is a right angle, and the anfgle I)AH less than a C right angle, the side DII of the triangle DAH is less than the side'DA (19. i.). The point H, therefore, is within the circle. and therefore the straight line AG cuts the circle. B Con. From this it is manifest, that the straight linte which is drawn at right an- F gles to the diameter of a circle from the extremity of it, touches the circle; and that it touches it only in one point; because, if it did meet the circle in two, it would tall within it (2. 3.). Also it is evident that there can be but one straight line which touches the circle in the same point. PROPo XVIlo PROB.'To draw a straight line from a given point either without or iZ the ctrh cuiryfereace, which shall touch a given circle. First, Let A be a given point witbout the given circle BCD; it is required to draw a straight line front which shall touch the circle. Find 1. 3. ) the centre F, of the circle, and join AE; and from the -entre E, at the distance EA, describe the circle AFG; from the point D draw ( 11 1.) DF at right angles to EA, join EBF, and draw AB. AB touches the circle BCDo Because E is the centre of the cir. cles BCD, AFG, ~EA is equal to EF1U, and E to - B; lheretore the two / sides AE. EB areequal tc the etwo FE, El), and they coltain the angle at E common to the two tri,inales AEB, IFED; t COererei tihe base DF is eqtual / to the base NB,,and the triangle FED to the triangle A E B, and the other angles to the other angles (4. 1.); There\ bore theo angle IEBA is equal to the cangle EDF; but E DF is a right angle, wherefore EBA is a right angle; and EB is a line drawn friom the centre: but a straight line drawn from tlhe extremity of a diameter, at right angles to it, touches the circle (Cor. 16t. 3.) ~ Therefore AB touches the circle; and is drawn from the oiven point A. ~hich was to be done. But if the given point be in the circumference of the circle, as the point D, draw DE to the centre E, and DF. at right angles to DE; D.F touches the circle (Cor. 16. 3) OF GEOIMETR' Y. BOOK).'I. PROP. XVIII. TIIEOR1.i a straight line touch a circle, the strtigiht line draw7 from the centre to the point of contact, is perpendicular to the lile touchirg the circle. Let the straight line DE touch the circle ABC in the point C; take the centre F, and draw the straight line FC: FC is perpendicular to DE. For, if it be not, from the point F draw FBG perpendicular to DE; and because FGC is a right angle, GCF must be ( 17. 1.) an acute a13 gle; and to the greater angle the great- er (19. 1.) side is opposite: Therefore FC is greater than FG; but FC is equal to FB; therefore FB is greater than FG, the less than the greater, which is inr- H possible; wherefore FG is not perpendicular to DE: In the same manner it may be shewn, that no other line but, FC can be perpendicular to DE; FC is therefore perpendicular to DE. There\ fore, if a straight line, &c. Q. E.. PROP. XIX. THEOR. If a straight line touch a circle, andfrom the point of contact a straight line be drawn at right angles to the touching line, the centre of the circle is in that lineo Let the straight line DE touch the circle ABC, in C, and from eC let CA be drawn at right angles to DE; the centre of the circle is in CA. For, if not, let F be the centre, if possible, and join CF. Becauss DE touches the circle ABC, and FC is drawn from the centie to the point of A contact, FC is perpendicular (13. 3.) to DE; therefore FCE is a right angle; But ACE is also a right angle; therefore the angle FCE is equal to the angle ACE, the less to the greater, which is im- B possible; Wherefore F islnot the centre of the circle ABC: In the same manner it may be shewn, that no other point which is not in CA, is the centre; that is, D ~ the centre is in CA. Therefore, if a straight line, &c, Q, E. E? 1E E,'N TS FOP. XXI. THEOR. hXe a.gLle at the centre of a circlb is' dom. bte of the antgle at a/i circmtjr. ence, upon the same base, that is, u pon the sam.e part ofFthe circuljfesr eilce. Let ABC be a circle, and BDC an angie at the centre, and BA6C atn angle at the circumference which have the same circumference BC ior the base; the angle BDC is double of the angle BAC. First, let D, the centre or the circle, be within the angle BAC, and join AD, and produce it to E: Because DA is cqall to DB, the angle DAB is equal (5. 1.) to the angle DBA: therefore the angles DAB, DBA together s..i are double of the angle DAB; but the an- gle BDE isequal(3.l.)to the angle-s DA B, \ DOBA; therefore also'the angle BDFE is double of the angle DAM1; For the same reason, the angle EDC is double of the angle DAC: Therefore the whole angle/ B3DC is double of the -whole angle BAC. Again, let D, the centre of the circle., be- without the angle BAC; and join AD. " and produce it to E. It may be demonstrated, as in the first case, that the an- \ gle EDC is double of the anivle DAC5, and that EDB a part of the first, is double of DAB, a part of the other; there- / fore the remaining angle BDC is double \ of the remaining angle BAC. rhere- ore the angle at the centre, &c. Q.L.D. PROP. XXI. THEOi-E. T.lae angles in the salme segment (J a circle are equal to one anotherl.. Let ABCOD be a circle, and B D, BEED angles in t'he same segment BAED: The angles BAD, BED, are equal to one another..e - Take F the centre of the circle AB~CD: And, first, let tre sengment BA e, t /k |r ea er than a semricircle, and join BF, FD: And because the ar.le BFD is at the cei- / tre, and the angle BAD at the circumference, both having the sanme part of the,: /.. circumference, viz. BC1), for their base; therefore the angle BFl) is double (20. 3.) of theo emg1e BADP { *,r tht, same rea - (3 uO GEVOMETYIt BOOK Il 0 itO -on, the angie BFD is double of the angle BED; Tlhaeforoe the a}ni gle BAD is equal to the angle BED. But, if the segment BAED be not greater than a se.icircle, let BAD, B-ED he an- - gles in it; these also are equal to one another. Draw AF to the centre, and produce /. - to C, and join CE: Therefore the seg- \ i ment BADC is greater than a semicircle; a-nd.the angles in it BAC, BEC are equal, by the first case: For the same reason, because CBED is greater than a selmicir- cle, the angles CAD, CED are equal; Therefore the whole angle BAD is equal to the whole angle BED. Wherefore the angles in the samne segt~ inent, &c. Q. E. D. PROP,. XXII. THIEOR. hfie opposite angties of any quadrilateral figure described irz a circle, are together equal to two right angles. LektABCD be a quadrilateral figure in the circle ABCD; any two of its opposite angles are together equal to two right angles. Join AC, BD. rThe angle CAB is equal (21. 3.) to the angie CI)B, because they are in the same se,, ment BADC, and the angle ACB is equal to the angle ADB3, because they are in C the same segment ADCB; therefore the whole angle AD.C is equal to the angles:CAB, ACB: To each of these equals add the angle ABC; and the an(les ABC, ADC, are equal to the angles ABC, CAB,' 3 BCA. But ABC, CAB, BCA are equal to two right angles (32. 1.).; therefore also the angles ABC, ADC are equal to two right angles: In the sanme manner, the angles BAD, DCB may be shewn to be equal to two right angles. Therefore the opposite,a rp gles, &;c. Q. E. D. PROP. XXIIL THI EOR. tUjon the same sti aight iine, and uponl the sae side of it, there cannot be twosimilar segments of circles, not coinciding zwith one another. If it be possible, let the two similar segments of circles, viz. ACB, AkDB, be upon the same side of the same straight line AB, not.coin ciding with one another; then, because the circles ACB, ADB, cut ~one another in the two points A, B, they cannot e.ut one another il r iA. $367 ~ELEMENTiS any other point (10. 3.): one of the sega 1)j ments mnilst therefra'e fall within the other /7. let ACB fall within A DB, draw the straight line BCD, and join CA, DA: and because the segment ACBR is similar to the segment ADB, and similar segments of circles contain (9. deC. 3.) equal angles, the angle ~- ACB is equal to the angle ADB, the extenor to the interior. which is impossible (16. l.). Therefore, there cannot be two sirnilar soegment"s of circles upion the same side of the same line:, which do not coincDide,. E. D. PROP. XXIV. THEOR. Similar segmients of circles upon equal straight lines are equal to one another. Let AEB, CF'D be similar segments of circles upon the equal straight lines AB, CD,; the segment AEB is equal to the segment C;FDo For, if' the segment AEB be applied to the segment CFD, so as the point A be on C, and E -the straight line AB upon CD, the point B shall coincide with the point D, because - AB is equal to CD: B C D'herefore the straight line AB coinciding with CD, the segment AEIB must (03 3.) coincide with the segment CFD, and therefore is equa! to it. Wherefore, similar segments, &c. Q. E. D. PROP. XXV. PROB. J segment of a circle being given, to describe the circle oJfwhich it is the segment. Let IBC be t!,he given segmoent of a circle; it is required to de, scribe the circle of whicht it is the segment. Bisect (10. 1.) AC in D, and from the point D draw (1l. 1.) DB at right angles to AC, and joiin AB: First, let the angle:s ABD BAD be equal to one another; then th.e straight line BD is equal (6. i.) to DA,~ and therefore to DC; and because the three straight lines IBA, DB, DC, are all equal; D is the centre of the cikclo ('2. 3.); fron, the centre D, at the distance of anlly,f e three DA, DB, DC, describe a circle; this shall pass through the other points; and the circle of which ABC is a segment is described; and because the centre D is in AC, the segment ABC is a semicircle. Next let the angles ABD, BAD be unequal; at the point A, in the straight!ine AB make (23. I.) the angle BAE equal to the angle ABD, andl produce BD if neces OF GEOiETitRY. BOOK iI. a A D C A; ) Sary, to E, and join EC: and because the angle ABE is equal to theangle BAE, the straight line BE is equal (6. 1.) to EA: and because AD is equal to DC, and Dn V. common to the triangles ADE, (CDE, the two sides AD, DE are equaltto the two CD, DK, each to each; and th.e angle ADE is equal to the angle CDE, for each of them is a right anm gie; therefore the base A.E is equal (4. I.) to tht base EC': b AIt IA vras shewn to be equal to EBI, mwherefore also BE is equal to EOC: and the three straight lines AE, EB, EC are therefore eantl'to ono anotheor; wherefore (o. 3. E is the centre o~ the ci -' rm the ceontre E', at the distance of any of the tlhree AE, E4,f'}', dcaribo' a circle, this shall pass through the other points: and the circle oCf swhich ABgi is a segment is described: also, it is evident, that if the angjle A)bD be greater than the angle BAD, the cetane E falls Without the segiment ABC, which thetreforIe is less than a semicircle; but if the angle ABD be less than BAO, the centre E ftills within the segment ABC, which is therefore greater than a semicircle Wherefore, a segment of a circle being given, the circle is described of which it is a segment. Which was to be'done. PROP. XXVI. THiEOR. Ln equal circles, equal angles.stand u.Wuon equal arches,'hether they be Ut the centres or circu oferences.Let A:~C, DEF be equal circles, and the equal angles BGC, EHF, at their centres, and BAC, EDF at their cireumferences: the arch I3KC is equal to the arch EiLLF. Join BC, EF; and because the circles ABC, DEF are equal, the straight lines dawn fi-om their centres are eoual: therefore the two &A /3 /: / .ides BG, GIC, are equal to tile two EH, HF; and tile angie ai G is equal to the angle at H4; thereftre the base BC is equal (4. 1.) to thle base EF: and because the angle at A is equal to the angle at ID, tha segment BAC is similar (9. def. 3.) to the segment EDF; and they are upon equal sti'aight lines BC,'EF but sinmilar segrnients of circles upon equal straight lines are equal (24;;3.) to one anlother, therefore, the segment BA is equal to the segmnent DF: but the whole circle ABC is equal to the whole DEF; therefore the remaining segment BKC is equal to the remaining segment ELF, and the arch BKC to the arch ELF. Wherefore in equal circles, &ce Q. E. D. PROP. XXVII. THEOR0.I equal circles, the angles hldch stand upon equal arches are equal to one another, cwhether they be at the centres or circuvnferenceso Let the angles BGC, EHIF at the centres, and BAC, EDF at the ciri tumferences of the equal circles ABC, DE.F stand upon the equal arches BC, EF; the angle BGC is equal to the angle EHF, and the angle BAC to the angle EDF. If the angle BGC be equal to the angle EHF, it is manifest(20. 3.) that the angle BAC is also equal to EDF. But, if not, one of them is the greater: let BGC be the greater, and at the point G, in the straight line BG, make the angle (23. 1.) BGK equal to the angle EHF. And because equal angles stand upon equal arches (26. 3.), when they are at the centre, the arch o:K is equal to the arch F: but EF is equal to BC; therefore also BK is equal to BC, the less to the greater, which is impossible. Therefore the angle BGC is not unequal to the;angle EHFf; that is, it is equal to it: and the angle, at A is half the angle BGC, and the angle at D half of tie angle EHF; therefore tho iangle at A is equal to the angle at D. Wherefore, in equal circles, &e. Q. E. D. PROP. XXVIII. THIEOR. In equial circles, equal straight lines cut off equal arches, the greater equaE to the greater, and, the less to the less. Let ABC, DEF be equal circles, and BC, EF equal straight lines in. OF GE 0AiE TY* 13 BOOK Mj. them, which cut off. the two greater arches BAC, El)F, and ti e two less BGC, EHIF: the greater BAC is equal to the greater EDF, and the less BGC to the less EHF. Take (. 3.) K, L, the centres of the circles, and join BK, KC, EL, IJF; and.because the circles are equal, the straight lines from their centres are equal; therefore BK, RC are equal to EL, LF; but the base BC is also equal to the base EF; therefore the angle BKC is equal (8. 1.) to the angle ELF: and equal angles stand upon equal (26. 3.) arches, when they are at the centres; therefore the arch BGC is equal to the arch iEEF.; But the whole circle ABC is equal to the whole EDF; the remaining part, therefore, of the circumference, viz. BACG, is equal to the remaining part EDF. Therefore, in equal circles, &CQ E.:D. PROP. XXIX. THEOR. in equal circles equal arches are subtended by equal straitgft lines~ Let ABC, DIEF be equal circles, and let the arches BGC, EHtF also be equal; find join BC, EF: the straight line BC is equal to the straight line EF. Take (1. 3.) K, L, the centres of the circles, and join BK, KC, EL, LF: and because the arch BGC is equal to the arch EHF, the angle BKC - is equal (27. 3.) to the angle ELF: also because the circles ABC, DEF are equal, their radii are equal: therefore BK, KC are K E L: LEMENTS equal to EL, LF: and they contain equal angles; therefore the base BC is equal (4. 1.) to the base EF. Therefore, in equal circles, &co Q. E. D. PROP XXX. PROB-. To bisect a given arch, that is, to divide it into two equal partst, Let ADB be the given arch; it is required to bisect it. Join AB, and bisect (10. 1.) it in C; from the point C draw CD at right angles to AB, and join AD, DB: the arch A iB is bisected in the point D. Because AC is equal to CB, and CD common to the triangle ACD, BCD, the two sides AC, CD are equal to the two BC, CD; and the angle ACD is equal to the, angle BCD, because each of themn is a right angle; therefore the base ADl is equal (4 1.) to the base BD. But equal straight'ines cut off equal (28. 3~) A C arches, the greater equal to the greater, and the less to the less; and AD, DB are each of them less than a semicircle, because DC passes thr6ugh the centre (Cor. 1. 3.) 3 wherefore the arch AD is equal to the arch D)B; and therefore the given arch ADB is bisected in D. Which was to be done, PROP. XXXI. THEOR. In a circle, the angle in a semicircle is a right angle; but the angle in a' segment greater than at semicircle is less than a right angle; and the angie in, a segment less than a semicircle is greater than ti right angle; Let ABCD'be a circle, of which the diameter is BC, and centre 7; draw CA dividing the circle into segments ABC, ADC, and join BA, AD DC; the angle in the semicircle BAC is a right angle; -and the angle-in the segment. ABC, which is greater than a semicirele, is less than a right angle; and the angle in the segment ADC, which is less -than a semicircle, is greater than a right angle. Join AE, and produce BA to F; and because BE is equal to EA, the angle E AB is equal (6. 1.) to,BA: also, because AE is equal to EC, the angle EAC is equal to ECA; where- A fore the whole angle BAC is equal to the two angles ABC, ACB. But FAC, the exterior angle of the triangle ABC, is also equal (32. 1.) to the two angles ABC, ACB; therefore the angle BAC BE C is equal to the angle FAC, and each of them is therefore a right (7. def. 1.) angle; wherefore the angle BAC in a semicirele is a right anlglg OF GEOMETRY. BOOK 1l.'td' And because the two angles ABC, BAC of the triangle ABC are together less (17. 1.) than two right angles, and BAC is a right angle. ABC must be less than a right angle; and therefore the angle in a segment ABC, greater than a semicircle, is less than a right angle. Also because ABCD is a quadrilateral figure in a circle, any tNvo of its opposite angles are equal ('2. 3.) to two right angles; therefore the angles ABC, ADC are equal to two rigtlt angles; and ABC is less than a right angle: wherefoire the other ADC is greater than a right angle. Therefore, in a circle, &c. Q. E D. Cor, From this it is manifest, that if one angle orfa triangle be equal to the other two, it is a right angle, because the angle adjacent,to it is equal to the same two; and when the adjacent angles are equal,:they are right angles. PROP. XXXII. THEOR. a straight line touch a circle, and fron the point of contact a straighlE line be drawn cutting, the circle, the angles made by this line with the line which touches thetcircle, shall be equal to the angles in the alternlate segmlents of the circle. Let the straight line EFl touch the circle ABCD in B, and from the point B let the straight line BD be drawn cutting the circle: The angles which BD makes with the touching line EF shall be equal to the angles in the alternate segments of the:circle; that is, the angle FBD is equal to the angle which is in the segment DAB, and the an: gle DBE to the angle in the segment BCD. From the point B draw (11. 1.) BA at right angles to EF, and take any point C in the arch BD, and join AD, DC, CB; and because the straight line EF touches the circle ABCD in the point B, and BA is drawn at right angles to the touching linle, from the point of contact B, the centre of the circle is ( I 9. 3.) in BA; therefore the angle ADB, in A a semicircle, is a right (3 1. 3.) angle, and consequently the other two angles,.BAD, ABD are equal (32. I.) to a right anfle: but ABF is likewisea right angle; therefore the angle A BF is equal to the angles BAD, ABDl:.take from these equals the common angle ABD; and there will remain the angle DBF equal to the angle BAD, which is in the alternate seg- - ment of the circle. And because ABCD is a quadrilateral figure in,a circle, the opposite angles BAD, BCD are equal (22. 3.) to two right angles; therefore the angles DBF, DBE, being likewise equal (13. 1.) to two right angles, are equal to the angles RBAD, BCD and DBFI, h3s 9~~~~~'"""''"". ELEIEi.NTS been proved equal to BAD: therefore the remaining angle DDBE Ai. equal to the angle BCD in the alternate segmnent of the circle. Wherefore, if a straight line, &c. Q. ED. PROP. XXXIII. PROB. Upos.a giveen straight linle to describe a segmrent of a circle, containnilng an angle eqztsl to a given rectilineal angle. Let AB be the given straight line, and the angle at C the given reck tilineal angle; it is required to describe upon the given straight line AB a segment of a circle, containing an angle equal to the angle C. First, let the angle at C be a right angle; bisect (10. 1.) AB in F4. and fiom the centre F, at the distance FB, describe the semnicircle AHB; the angle ALIB being in a se- / micircle is (31. 3.) equal to the / right angle at C. But if the angle C be not.a aright A B angle at the point A, in tbhe straight line AB. make (23. 1.) the angle BAD equal to the angle C, and ft.om the point A draw (l..) AE at right angles to AD; bisect (10. 1.) AB in F, and / from F draw (11. 1.) FG at / right angles to AB, and joia GB:E; Then because AF is equal to FB, and FG common to the triangles AFG, C BFG, the two sides AF, FG are equal to the two BF, FG; but the angle AFG is also equal to the angle BFG; therefore the babe AG is.equal (4. 1.) to the base GB; and the circle described from the centre G, at the distance GA, shall pass through the point B; let this be the circle; AHB: And because from the point A the extremity of the diameter AE, AD is drawn at right angles to AE, therefore AI) (Cor..,16. 3.) touches the circle; and C because AB, drawn fiom the point / 3 of contact A, cuts the circle, the angle DAB is equal to the allgle in the alternate segment Al-B (32. 3.); but the angle DAB is equal to the angle C, therefhre also the angle C is equal to the - angle in the segment. AI.!SB u - -I.'J.t.:i o t.-t[.lo Bk% fi. &1. ~.'Aherefbre, upon the givcn straight'no ABi the segment AHB of g circle is described whichll contains an angie eqiual to the given nDgl at C. Which was to be doneo PROP. XXXIV-o gROB. LrCob 1 O'a segmeltnt f om a given circle which shall cntauin at 4alng& equtal to a given re tilineal angile Let ABC be the given circle, and D the given rectilineal angle; "s required to cut off a segment from the circle ABC that shall con" tain an angle equal to to tl e aige D.0 Draw (17. 3. ) the straight line IF touching the circle ABC in the point B, and at the point B, in the. straight line BF, make (;'i3.i) I, the anrle FBC equal to the an'le, h; th ere fore, because the stra - Iht line EF touches the circle A BC and B;C is drawn from the poin f cont-act B, the angle FB s / equal (32 3.) to the a1n rg1 in 1e / alternate segrmeit BAC; but t he 2 angle FBC is equal to the angle - D' thereftore the angle in the - saement BAC is equalto tneO 5& inga D wherebore the segment B3A is cut off fiom tie given circle AB g containing an angle equa.l to t!h! given angle.Do Whichl vas to be done.'1HgRO!. ZXXwo THE)R.~ ti two stcra'il.t lines 7aitt i a circle cut one anoroher, the'ecatmrie con., taiotetdby the se-rtts C ofnc oaf teins is eqpual to the'rectangle con~ ttined by the l scrten!s oi.t.e othero Let the two straight lines AC, B D, within the, circle ABCD, cut one another in the point E: the rectangle contained by AE, EC is equal to the rectang!e contained by BE, ED. If AC, BD pass each of them through the & centre, so that E ius the centre, it is evident /,.'hat AE, EC, B, ED, being i oequal, the rectangle AE.EC is lEkewkse equal to the..... rectanale BEIOE"D. B But let one of thern, -D, pass througl-h the \ centre, and cut the other AC, which does notpass through the centre, at rirht angles in t-e,?it, F; then. if Bll) he bis.ected in F. F is the. ntrem r -o, th a ci OaELEMENTS ABCD; join AF: and because BD, which passesthrough tile cent%,& cuts the straight line AC, which does not pass through the centre at right angles, in E, AE, EC are equal (3 3.) to one another; and because the straivht line BD is cut into two equal parts in the point;F, and into two unequal, in the point E, BE.ED (5. 2.)+ EF =2 FB = AF2. But AF2 = AE2 -+ AE~+EF2, and taking EF2 from each, BE ED AE2.= AE EC.o Next, Let BD, which passes through the centre, cut the other AC, which does not pass through the centre, in E, but not at right angles: then, as before. if BD be bisected in F, F is the centre of the circle. Join Al, and from F draw (12. I.) FG perpendicular to AC; there. -' forte AG is equal (3. 3.) to GC; wherefore.j AE.EC -- (5. 2.) EG2-=AG, and adding. GFa2 to both, AE.EC4- EG+G3- F2=AG2 -- GF. Now EGd -4- GF2 = EF2, and AG2+GF2= AF2; thereibre, AE.EC+EF=2AF2=FB2. But FPB -B E.E D- (5. 2.) E F2,therefore AE. EC+EF2-BE.ED+EF2,and taking EF2 from both, AE.EC-SEo.ED. Lastly, Let neither of the straight lines AC, BD. pass through the centre: take the centre F, arnd through E, the(_ Intersection of the straigh-t lines AC, DB, draw the diameter GEFH: and because, as has been shown, AE.EC=GE.EH, and BE ED =GE.EH; therefore AE.EC = BE.EDo. Wherefore, if two straight lines, &c. Q. E. DJ P OP. X;XXV$ I THEOR, if from any point,'without a circle two sirnight lines be drawl:, one i: rhich cuts the circle, and the,,ther touches it; the rectangle contain;, ed.by the. whole line which, cuts the circlf, and the part oJ it withou&t' the circle, is equal to the squlre of the line',hich touches it. Let RDbe any point without the circle ABC, and DCA, DB two, straight lines drawn fiom it, of which DCA cuts the circle, and DP. touches it jeA-:sR'ctanglc.ADJC. is equal to tlhe, square of PB. 0 F G L 0, 3i!. S ( ( i K5' l;ither DCA passes through the ce ntre, or,t does not; first, Let it pass through the cen. tre E, and join EB; therefore the angle..; j EBD is a right (18. 3.) angle: and beC a1/ the straight line AC is bisected in E, and pre- duced to the point 09, AI. DC - FC2... -- ED (6: 2.), But EC 1-B.i tlheefote AD' DC -- EBs ED2. Now EDW- ( 7. 1.) EB2+BD2, because E:-;D is a right angle; therefbore AD DC -, and C+:aE taking EB2 from each, AD. 2D. BD_. I 1 But, if DCA does not pass through the, cerent4e of the circe ABU, take, (i. 3.' t'he centre E, and draw EF perpendicular 1 72.'.) to AC, and join EB, EC, ED: and be. cause the straight line EF, vwhich passes.; th-rough the centre, cuts the straight line d: AC, Which does not pass through the cen- /1 tre, at right angles, it likewise bisects (3..) -t; therefobre AF is equal to FC; and be- f..... cause the straight line AC is bisected ina F \ and produced to D (6. 9.), AD.DCA-FC I- F D; add FE 2 to both, then AD.DC+-FU 1C2: // +FE-=FD" +~FE2. But (47. 1) )EfC2=FC" F: FE2, and ESD2-= FD 2 + FE! because DFE is a right angle; therefore AD.'DC-~ iEC'2-ED2. Now, because EB ) is a right angle, ED'-=EB2+BD':EC'+BD', andi therefore, AD.DC+EC2=EC2+BD, and.AD. D-C=BD. Wherefore,ifffi om any poaint, &c. Q. E. -D. Con. If fiom any point without a c ircle there be drawn two straight lines cutting it, as AB, AC, the rectangles contained by tie whole lines and the parts of them wvt}tout the circle, are equal to one another, vi.z, BA. AE=CA.AF; for each of these rect- angles is equal to the square of the straight 7::ne AD, which- touches the c.rci-ml PROP. JXXXVi. TIEORt fjfromr a poinit ilthout a c;rcle thIcIe be drsazn: two straight lines, one ozahic1h cuts the circle, and the other meets it; if the rectangle.containd ed by the whole line, which cuts the circle,,atnd the part of it withou' the ci.?cle, be equal to th e squtare of Mh "lne w ehich meets it, the lin'e.:hici mneets shftll looch othe circle. Let any point D be taken without the cikele ABC, and fronm it le& two straight lines DCA and DB be drawn, of which DCA cuts the circle, and DB meets it; if the rectanEgle AD.DC be equal to the square of DB, DB touches the circle. Draw (17. 3.) the straight line DE touching the circle ABC; find the cientre F, and join FE, FB, FD; then FED is a right (18. 3.) an: gle: and because DE touches the circle ABC, and DCA cuts it, the rectangle AD.DC is equal (36. 3.) to the.square of DE; but the recto angrle AD.DC is, by hypothesis, equal to the square of DB: therefore the square of DE is equal to the square of DB and the straight line DE equal to the straight line DEB: but FE is equal to FB, wherefore DE, EF are equal to DB, BF; and the bfase FD is commnin to the two triangles DEF, DBF; the'efore the angle DEF is equal (S'. 1. to theangle DBF; and DEF is a right angle, therefore also DBF is a right angle: but FB, if produced, is a 9 diamieter, and the straight line Wthich is drawn at right angles to a diameter, from the extremnity of it, touches (16. 3.) the circle: there tfore DB touches the circle ABC. Where-'ore, if from a point; &'c Q. o \ / (N. DEFIN-ITON T A REC'rILI,7,AL figure is said to be inscribed in another rectilinea~ figure, when all the angles of the inscribed figure are upon the sides of the figure itl which it -.....is inscribed, each upon each. 1.B~~~~~~~~~o~~"- ji in like mannerl, a figure is said to be described about another figure, when all the sides of the -... circumscribed figure pass throu.gh- the angular points of the figure about wvhich it is described, each through'achD A rectilineal figure is said to be inscribed in a circle, when all the angles of the inscribed figure are upon the circumfrcren cc of th-e, circle. / A rectilineal figure is said to be described,.. about a circl(e, when each side of the cir cumrscribed figure touches the cireumfcr r " ence of the circle.! n like ma. a In like manner, ia circle is said to be inscrixbc ed in a rectiliaeal figure, whern the circum,- i ference of the circle touches each side of,he figure, A circle is said to be described about a recti lineal figure, when the circumference of. the circle passes through all the angular points of the figure about which it is described. A straight line is said to be placed in a circle, when the extremities of` it are in the circumnference of the circle. PROP.. PROB. In a given circle to place a straight line equal to a given straight linec: not greater than the diamreter of the circle. TLet ABC be the given circle, and D the given straight line, not greater than the diameter of the circle. Draw BC the diameter of the cir A ale ABC; then, if BC is equal to D, the thing required is done; for in the circle ABC a straight line BC is phlced equal to D: But, if it is C not, BC is greater that D; make / CE equal (3. 1.) to D, and from the centre CG at the distance CE__ describe the circle AEF, and join CA: Therefore, because C is the T-. —. centre of the circle ALF, CA is equal to CF; but D is equal to CE; therefore D is equal to CA Wherefore, in the circle ABC, a straight line is placed, equal to the given straight line D, which is not greater than the diameter of the circle. Which wasto be done. PROP. Ii. PROB. in a given circle to inscribe a triangle equiangular to a given trianglZo Let ABC be the given circle, and DEF the given triangle; it is required to inscribe in the circle ABC a triangle equiangular to the triangle DEF. Draw (17. 3.) the straight line GAH touching the circle in the point A, and at the point A, in the straight line AHll, make (-23. 1.) the angle HAC equal to the angle DEF-: and at the point A, in the straight line }Fl G-E OiM'ETYt. B3-OOICK IV L 4G, make the angles' GAB eo qual to the angle DFE,andjoin G BC. Thereforo, because > HAG touches the circle ABC 7 and AC is drawn fiom the/ point of contact, the angle ID HiC-I. is-equal (32. 3.) to the angle _ABC in the alternate segment of the circle: But HAC is equal to the angle,, DEF; therefore also the an- 4 gle ABC is equal to DEF; ibr the same reason, the angle ACB is equal to the angle DFE; theres fore the remaining angle BAC is equal (32. 1.) to the remaining angle EDY- Wherefore the triangle ABC is equiangularto the triangle DEF, and it is inscribed in the circle ABC. Which was to be doneO PROP. Ill. PROB.::bout a given circle to describe a triagle equiangular to a given trU' angle~ Let ABC be the given circle, and DEF the given triangle it is re' quired to describe a triangle about the circle ABC equiangular to the Irianale DEF. Produce EF both ways to the points G, H, and find the centre K of the circle ABC, and from it draw any straight line EKB; at the poQ.nt K in the straight line KB, make (23. I.) the angle BKA equal to the angle [D )'G, and the alngle BK C, equal to the angle D FH; and through the points A, B, C, draw the straight lines LA M, MBN,.NCLC touching ( 7. 3.) the circle ABC: Therefore, because LM, MN. NE touch the circle ABC in the points A, B, C, to vwhich fronm the centre are drawn KA., KB, KC, the angles at the points A, B, C, are right (18. 3.) angles. And because the four angles of the quadrilateral figure AMBiK are equal to four right angles, for it can be divided into Jl-wo triangles; and because two of them, KkAM, KBJi are right angle.? ID /?\,7 \!t\/,-:, i — - -jCT/f i~ LELMi ENTS ihe other two AKB, AMB are equal to two right angies: But the a, — gles DEG, 1)EF are likewise equal (13. 1.) totwo right angles; therefore theangles AKB, AIMB are equal to the angles DEG, DEF, of ~which AKB is equal to DEG; wherefore the remnaining angle AMB is equal to the rermaining angle DEF. In like manner, the angle LMNmay be demon-trated to be equal to DFE; and therefore the remaiilm ing angle MLN i; equal (32. 1.) to the remaining angle EDF: Wherefore the triangle LMN is equiangular to the triangle D EF And it i>;. described about the circle ABC. Which was to be done. PROP. IV. PROB. To inscribe a circle in, a given triacngle, Let the given triangle he ABC; it is required to inscribe a circle in ABC. Bisect (9. 1.) the ngles ABC, BCA by the straight lines BD, CD meeting one another in the point D, fiom which draw (12. 1.) DE, DF? )DG perpendiculars to AB, -C7, CA. J Tihen because the angle E B D is equal to the angle FBD, the angle ABC be-. ing bisected by BD; and because the right angle BED, is equal to the right angle BFD, the two triangles EBD, // FBD have two angles of the one equal to two angles of the other; and the side PD, which is opposite to one of the equal angles in feach,; is common to L both; iherefore their other sides are B - equal (26. 1.); wherefore DE is equal to DF. For the samne reason, DG is equal DF; therefore the three straight lines DE, DF, DG, are equal to one another, and the circle described firom the centre D, at the distance of any of them, wilI pass through the extriemit:ies of the other two, and will touch the 0traight lines AB, BC, CA, because the angles at the points E, F, G, are right angles, and the straight line which is drawn fromr the extremitry of a diameter at right angles to it, toulhes (Cor. 16. 3 ) the circle. Thercfore the straight lines AB, B3C, CA, do each of them touch the circle, and the circle E G is inscribed in the triangle ABeC.Which was to be done. PROP. V. PROB. To desc-ribe a circle about a g'iven; t7riantglc Let the given triangle be ABC; it is required to describe a c;rele about ABC. Bisect (LO. 1.) A1B, AC iqn the poits D, DE, and from these points draw iv F4 EF'at right auglcT (61 1,, to AB, AC,; BF, EF produced1 '\ t 5t' 8;' \ J?A.~,. / -,,',v. \/.w;'ii meet one ancther; fa r, t dey do net ~aet, dtey are, para sl e vyheretore, AB, A, which are at right angles to them, are paralle, wlhich is absurd; Let them melet in F, and join FA; also, if'the point: be not in BC, join BF, GF. then, because AD is eq?, a. to B, and'DF common, and at itgnt anaoles to AB, the base AF is equal (4. L) io the base FB. In like manerIr it mayv be shewn that UE is equal to A A; andltherefore BF is eq uill to FC ad f; FA B, FC are equal tq one another; wherefore the circle described firorm the centre, a$ the distance of one of thesm, ill pass through the ext'eremilies of the otlher two, and be described about'te thrian;'le ABC, which was to be doneCoo. When the centre o(f th. e ile falls within the triange, each of its angles is less thran a rigtt angle, each of them being in a seag mgent Treater than a selicircle; but when the centre is in one of?.he sides of the triancgie the anule {.pposite to this Sida, being in a sell mnicircle, is a rihtI angle - and if the centre falls withoul the triangle, the angle opposite to the siide heyond ihichl it is, being in a segment; less than a sermicircle, is greater thnll a righ3t angle. Wheirefore, if fth given treingle be acute angled, the cant'e of tlh cirle alis withit, At: if it be a right anigled triangle, the cetPe, is an i the2 sde Oeprosite to the right angle; anl ifit be an obtuse angled b'ian(le, the centre alls without 4heh triangle, beyond the sade pposit/e Ato the obtuse angle, PR OPe PROBo'.o inscribe a s.quare it a given circgc. Let ABCD be the givan cirCre; t; is required to inscribe a squav. i. n ABC D. Draw the di-ameters AC, BD at right angies to on.e another, and joil AB, BC, C), D A; bccausea BE' is equal to ED, E beinog the cetre, and because EA is at right angles to BD and common to the triangles ABE, ADE, E- -Y t.e base BA is equal (4.')o 4 lhe b) aas' AD; and, for the same reason, BC, CD /' P.1c each of them, equal tco BP or AD / therefore the quadrilaterlal figure ATBD A. -.s equilateral. It i is als-o rctangulal - for the straight iiao 3 D being a daia.'- 7 /, of the,. circle AB, i,',) i ai s('.ic, /.-c; vlhrefere the angle BAD i s a rigl h,.I ( 1; F;, angle; rh sa4e moraso n ht E ia *-,t.... _. of the angles ABC, BCD, CDA is a rigtht angie; therefore the qua>d.. rilateral figure ABCD is rectangular, and it has been shewn to be equio lateral; therefore it is a square; and it is inscribed in the cir'le ABCD. Which was to be done. PROPR V~Ii Pi{OBG 2O describe a square aboPt a given cicl'e. Let ABCD be the given circle; it is required to describe a squaat about it. Draw two diameters AC, BD of the circle ABCD, at right angles to one another, and through the points A, B, C, D draw (17. 3.) FG, GH. HiK, KF touching the circle; and because FG touches the circle ABCD, and EA is drawn firom the centre E to the point of contact A, the angles at A are right ( 18. 3.) angles; for the same reason, the angles at the points B, C, D are right angles; and because the angle AEB is a right angle, as likewise is EBG, GiH is parallel (28. I.) to AC; for the same reason, G AC is parallel to FK, and in like mnan- G A F ner, GF, HIK may each of them be de monstrated to be parallel to BED; therefore the figures GK, GC, AK, FB, BK are parallelograms; and GF is therefore equal (34. l.) to HK, and G' — to FK; -and because A'_ is equal to BD, and also to each of the two GH, FK; and BD to each of the two GF, I __. HK: GH, FK are each of them equal to GF or'HK; therefore the quadrilateral figure FGIK is equilateral. It is also rectangular; for, GBEA being a parallelogram, and AEB a right angle, AGB (34. 1) is likewise a right angle In the same manner, it may be shewn that the angles at H, K, F are right anales; therefore the quadrilfteral figure FGHK is rectangular; and wt was demonstrated to be equilateral; therefore it is a square; and it is described aboat the circle ABCD. Which was to be done. PROP.o LfIo PROB. To inscribe a circle i.' a given square. Let ABC D be the given square it is required to inscibe a, cir~e iA ABCD. Bisect (10. 1.) each of the sides AB, AD, in the points F, E,,and through E draw (31. 1.) EH parallel to AB or DC, and through.li draw FK parallel to AD or BC - therefore each of the figures, AL:, KB, AH, LID, AG, GC, BG, GD is a parallelogram, and their oppo — site sides:e equal (34, 1o) anl beas8e. hat A1 se er~, to B: and that Aig is the alf of AD, and AF the half of AB, Al-i is equal to "!AF; wherefbore the sides opposite to these are etqual, viz. FG to GE; in the sacae manners, it may he dernonstrated, that G- GIK ae' each of them equal to FG. or (E C tete r fore the four straight lines, iC, C-F, GU, Z GK, are equal to one aRother; and thce circle described from the centro G.C, at the distance of one of them, will pass through the extremities of the other three; and wvill also touch the str aight jines A) U PC CD, D_, because the angles at the points E, F, 11, K, are rifht (29. 1.) angles, and because the straight line which is drawn. from the extremity of a diameter at righit I - ('~ angles to it, touches the circle (16. a.) therefore each of the straight. lines AB, BC, CD, DA touches the circle, whiclh s tiher ece nstforib ed in thle sqall:a ABCD, Whlllh was to be done., PROPo TX. PROBe lSb describe a cir2cle nbout a gc ien sqli sceo Let ABCD be thle given square; it is required to describe a circie'Zbout it. Join AC, B D, cutting one another in E; and because DA is equal to AB, and AC common to the triangles DAC, B AC, the two sides DA, AC are equal to the two B A, AC, and the base DC is equal to the base BC; wherefore the angle DAC is equal (8. 1.) to the angle B.C, and the an- A gle DAB is bisected by the straight line AC In the same manner, it may be dem,nstra, ed, that the angles ABC, BC D, CDA are | severally bisected by the straight lines BI,, _.AC; thereflore, because the angle DA&B is equal to the angle ABC, and the angtle EAB is the half of DAB, cand EIBA the half of _ x-.} ABC; the angle EAB is equal to the angle.EBA': and the side EA (6. 1.) to the side EB. 7n-1 the same manner, it may be demonstrated, that thie straight lines EC, El) are each of them equal to EA, or EB; therefore the four stiaight lines LA, EB, EC, ED are equal to one another; and tlhe circle described from the centre E, at the distance of one of the n must )pass through the extremnities of the other three, and be described ahouat the, square AB..:D. "'t]Yicl \iias to be dnop, To descgibe an isosceles tricanrgle, hcvisng each of the aznges at tihe bast' doubtle rf Ihs third angle~ Take any straight iinO AB, and divide ( 1. 2.) it in the point C, so that the rectangle A1, BC may be e-qual to the square ot A.:; and froni the centre A, at the distance AB, describe the circle DBDE, ic Wvhich place (1. 4.) the straight iine BD equal to AC, which is not,eater thari tIhe diametet' of thi circle BDDE; join DA, DC, and about the triangle Ai}C descaibe (5. 4.) the circle ACD,; the trians gle ABD is such as is required, fliat is, each of the angles ABD, ADI) is double of the angie lA D. Becau.use the rectangle B.BIC e eqtlI to te square of AC. and LAC equal to B D, the rectangle AB.B G is equal to le scquare of't I-D; and because from the point r - P B vithout the circle A'D two straight lines BCA, tD ar e draw n to the crcumference, one ( Wvlhich cuts, and the other meets " tlhe circle, and the rectanr'e AP. / \_ -BC'ontatined by the whole ot th. Cutting line, and the pa:t of it L Without the circle, is equal to the, square bf.PD, which meets it; the straight line B D touches (37o' 43., the cirle AC. O, And bccausd g D touches the circle, and 1)C i$ drawn from the point o~ contct D, tihe angRle BDC is equa~l (32; 3.) to the angle BDGC in the alternate segment of the eircleg to;each of thes-e add the angte CDA: therefore the whole angle BDA is equal to the two angles CDA, DAC; but the exterior'angle BCD is eqtial (32-. I.: to he angles CDA, DAC; therctbre also BDA is q'a. to BC(D hbut BDA is equal (5. 1.) to CDB, because the side AD is equal to the side AB; thereu'ore CBD, ob IDBA is equal to BCD; and consequently the three angles BI)DA DBA, BC,'D, are qiual to oe atother. A id because the atgle DBC is equal to the angleBCGD the side BD is equ al (,. I.) to thie side DC - hil BDwas madel equal to CA; tii'efo e also CA a.S; equal to CD, and the angle CDA equa (5. 1.) to tothe angle D AC;'therefore the -Mak.le.s Ca DA, DAC to. Mgtherk are, dulbe ou tAheO angle DAC; but, BCD is equal to the an-gles CDA), DA:C (32. 1.); th-erefore fe.lso BCD is double of DACG aBUt BCD is equal to'each of the angeles BDA, ID 1, anld therefore each of:the angles BDA, DBA, is double-'of the angle DAB; wherei'oe'an isosceies triangle ABD is described, havirl:g each of the ean oges at the base double oft the irf angle, Whick was to be doen o oe...ite angle BAD is tile fift1 par of o we i'gilyt 8:4at!,';e ~ For since each of the angles A fD a"nd ADB i - equal to twice the ~! angle BAD, they ale togrether etqual to toI.ur ties BID, nd tS Ahe t fore all the hrec- antg le's ABB! AD B, tIAl, tkn together, are " equal to five tiines the anlo I BAD. ut the three aidlt Is ABD,'; ADB, BAiD are equal.to two ti'nt artgle.s theret1i re five titles tfh"6 angle BAD is es the fifth part of 6 two right angleS." - Cor. 2. Beca use B AD is the fifth p art of two, or the te nth part': of four right antghes, all the angles about the centr e tare together "~ equal to ten times the angle BA D, anod anay therefore be dividedc; irnto ten parts each equal to B D. And as th3ese ten equal angles at' the centre, oiust stand on ten equal arches. therefore the arch BD;G is one tenth of the circumf'trence; and toe straight litl.e 113D that is; 5' AC, is thereffore equal to the side of ah euilate:tas d lncnsa' iansrib. C ed in the cirCle.HBDE OP.XL lO To:inscr Te a er d q'lateral qnd equianaon.ib iz a gzivei n circa'e; Let ABCD:E be the given circle, it is requre to inscribe an equilateral and equianglair p entagon ia the circle ABCDIEo Describe (10. 4.) an isosceles triangle FGP I, having ach of the ane gles at G, I, doubite of the angle at F; aond in the circle AB( lT>E inscribe (z. 4.) the triangle ACD equiangular to thI tirisanle FGH, so tha the angle CAD he equal t be equal to the ang at and each of the anglet ACD, CDA equal to the angle _ at G or H; wherefore each of the angles ACD, CDA is don- ble of the angle CAD. Bisect by the straight lines CY1, t)B -B and join AB; BC, DE, E.. / / I / ABuDE Is the pentagon re - / quired.' -i ~ Becausoe the angles ACP -_CDA are each of them douible of CAD, and are bibected by th"e st'raiff-t linel es' i, E D3 tBhIe five -ngls' iDA C,; - ECDA. I BDA are. equal to e notler bu equal angles stand upon eqUoal (6. 3.) arches t;, etie five arches AB, BC, CD9, }1E, EA tie eqtua to oi t aro P,;" i. aned equal ar-chies are subtended by equal (Oct..;4~) straic,!,:is; ghi th'ereto.rc.- th1e five straight lines AJ, BC, CD, 1D E, ia', equal to one alnothie' Where'ore the pentagon A. t1; i's equilateral. It aIso sequiangalar; because the arch AB is equal to tie aich' DE; if to eacIl be, added BCD, the whole ABCD is equall to the whole ED,' and the angle.AED st e.nds on. 1,a1e archl A-.CD.,a te' an gIe B.'.i, oA} th Ih ag rc :~ 02e ILEfAdE"I-'I> ED)CB; therefore the angle IKAE is equal (2'7. 3.) to the angie' L " for the same reason, each of the angles ABC, BCD, (DE is equal to the angle BALF or AED: therefore the pentagon AB{CDE is equiangular: and it has been shown that it is equilateral. Whereftre, n the given circle, an equilateral arnd equiangular pontagaon has been inscribed. -Which was to be done. Otherwise. 1" Divide the radius of thc given circle, so that thle rectangie coni, "C tained by the whole and one,fhe part ma be qual to the part be quare "' of the other (11. 2.) Apply in the circle, on each side of a given' point, a line equal to the greater of these parts; then (2. Core 1O0 4.o) "' each of the arches cut off will be one-tenth of the circuemference, and' therefore the arch made up of both will be one-fifth of the circuYm f' erence; and if the straight line subtending this arch be drawn, it'; will be the side of an equilateral pentagon inscribed in the circlc." PROP~ X1.o PROB. To describe az equilateral and equiangulaer ena-goo, aebout a giveab c circle. Let ABCDE be the given circle, it is required to describe an equila= teral and equianglular pentagon about the cirele.ABt DE. Let the angles of a peutagon, il-scribed in the circle, by the last pro= position, be in the points A, B, C, D, E, so tehat the arches AB, BCd CD, DE, E are equal (I I. 4.) and throtugh- the points A, B, C, D, Edraw G, HK, K, LM, L VMG, touching (I7. 3. the circle; take the centre F, and j.in FB, F, FC, FL, FD. Arnd because the straight line KL touches the circle ABCY E in the point C, to which F: is drawn from the centre F, FC is Cierpetldicular (18. 3.) to KL; therefore each of the angles at C is a righit aigle; for the samne reason, the angOles at the points B, D are right angles; and because FCK is a right angle, the square of FK is equal (47. 1.) to the squares of FC, CK. For the same reason, the squaroe of FK is equal to thie squares of?B9 BK: therefore the squares of FC, CK are equal to the square, of FB, BK, of which the square of FC is equal to the square of FB; the remaining square of'C is therefore equal to the -'emaining square of BK, and thle straight line CK equal to BK; and because FB is equal to FtC, and F' common to the triangles BFlK, CFK, the two BF, FK are equal to the two CF, F; and the base BK is equal to the base iC: therefore the angle BFW is equal (8S 1.) to the angle KFC, and the angle BKF to FKC; wherefore the angle BFC is double of the arngle >I1.C, and K1(C double of FKC ~ for the same reason, the angle'FD is double of the angle CL, and CLD double of CLF: and because the arch BC is equal to the arch ID, tllhe angle BFC is equal (27. 3.) to the anile CFD; and BFC is double of the angle 4(F1-C andl CF.T ) do!tble of CFT tl-teref,. tile th nle N.i F P i OF GEOMEAYlRA BOOK l. 1o equal to the angle CFL; now the right angle FCK is equal to the right angle FCL; and therefore in the Atwo triangles FKC, FLC, there are two angles of one equal to two angles of the other, each to each, and U the side FC, which is adjacent to the equal angles in each, is common to both; therefore the other sides B are equal (26. 1.) to the other sides, and the third angle to the third angle; therefore the straight line KC is C ( L equal to CL, and the angle FKC to the angle FLC: and because KC is equal to CL, KL is double of KC; in the same manner, it may be shown that HK is double of BK; and because BK is equal to KC, as was demonstrated, and KL is double of KC, and HK double dofBK, HK is equal to KL; in like manner, it may be shown that GH, GM, ML, are each of them equal to HK or KL: therefore the pentagon GHKLM is equilateral. It is also equiangular; for, since the angle FKC is equal to the angle FLC, and the angle HKL double of the angle FKC and K-LM double of FL,', as was before demonstrated, the angle HKL is equal to KLM; and in like manner it may be shown, that each of the angles KHG, HGM, GML is equal to the angle HIKL or KLM; thereforethe five angles GHK, HKL, KL, KLM, LMG, MGH being equal to one another,: the pentagon GHKLM is equiangular: and it is equilateral as was-demonstrated; -and it is described about the circle ABCDE, Which was to be done. PROP. XIIo. PROB. Sob inscribe a circle inl a given equilateral and equiangular pentagon., Let ABCDE be the given equilateral and equiangular pentagon; it is required to inscribe a circle in the pentagon ABCDE. Bisect (9. 1.) the angles BCD, CDE by the straight lines CF, DF, and from the point F, in which they meet, draw the straight lines FB, FA, FE; therefore, since BC is equal to CD, and CF common to the triangles BCF, DCF, the two sides BC, CF are equal to the two DC, CF; and the angle BCF is equal to the angle DCF: therefore the base BF is equal (4. 1.) to the base FD, and the other angles to the other angles, to which the equal sides are opposite: therefore the angle CBF is equal to the angle CDF; and because the angle CDE is double of:CDF, and CDE equal to CBA, and CDF to CBF; CBA is also double of the angle C1lF; theelobro the! angle ABF is equal to the angl.c CBF; wherefore the angle ABC is - bisected by the straight line BF: In the same manner, it may be demon- / strated that the anvdles BAE. AEDl, ) V 4are bisected by the strai.1lt lines AF,EF: firom n e Ioint F draw (12. I.) FG, FH, FK, FL, FM perpendiculars to the straight lines A B, BC, CD, DE, EA: and because the / angle HCF is equal to KCF, and the C' K D right angle Fl'C, equal tn the right angne FKC; in the triangles F11HC'FKC there are two angles of one equal to two angles of the other, and the side FC, which is opposite to one of the equal angles in each, is comnmon to both; tlherefore, the other sides shall be equal (26. 1.), each to each; wherefore the perpendicular FEH is equal to the perpendi[. cular FK; in the sa;l, manner it may be demonstrated, that FL, FMI FG are each of them equal to FH or Fi'K; therefbre the five straight lines FG, FH, FK, FL, FMA are equal to one another; wherefore the circle described from the centre F, at the distance of one of' these five, will pass through the extremities of the other four, and touch the straight lines AB, BC, CD, DE, EA, because that the angles at the points G, H, K, L, MI are right angles, and that a straight line drawn from the xtrenity of the diatnmeter of a circle at right angles to it, touches (Cor. 16, 3.) the circle: therefore each of the straight lines AB, BC, CD, DE, EA touches the circle; wherefore the circle is i~n scribed in the pentagon ABCDiE. Which was to be done. PROP. XiV. F'R0 1OB. bo describe a circTe trbout a irven ecuil tberal anad eqlulgui(r Let ABCDE be the given equilateral and equiancgular pentagon;;t is required to describe a circle doabout it. Bisect (9. 1. the angles BC(D, CDE by the straight lines CF, FI) and from the point F, in which they neet, draw the straight lines 14B, FA,.FE to the points B, A, E. It may be iemonstriate, inthe same manner as in the precedino iproposhtion, that the an-. gles CBA, BEE, AED are bisected tv y the straight litles F! 9F Ai, - FE: and betause that the angle BCD is equalito the angle CDE, and that FCD is thi I/.half of the angle BCD, and C'DF tho -..?aefO31 (' C])_E - tr',.'.':i::F CDi e us e ua~ s ~'?" Tr~~~~~~~~~~~'~[ to FD)U; w1herefbre tihe side CF is equal (6. I.) to the sidp FD: In like manner it may be demonstrated, that FB, FA, FE are each them equat to FC, or FD): theretore the five straight lines FA, FB, PC, FD, FE are equal to one another; and the circle described from the centre F, at tie distance of one of them, will pass through the ~extremities of the other four, and be described about the equilatera) and equiangular pentagon ABCDE. 1qWhich was to be done. PROP. XV. P1RO.'o inscribe an eudzasate:ra a equianl e lar exegon in a griven ciz'ciCe Let ABCDEF be the given circle; it is required to insscrsibe s?,equilateraland equiangular hexagon in it. Find the centre G of the circle A1~CDEF,:and draw the diametrc AGD: and firom D, as a centre, at the distance DG, describe the ciri cle EGCH, join EG, CG, and produce them to the points B, F; and join AB, BC D,DE, GD,, FA; the hexagon AB.CDEF is eqilai'aurl'a and equianfrular. Because G is the centre of the circle ABCDEF, GE is equal.to CD:and because D is l4he centre of the circle PGCIi,'G- 3DE is equal to DG; wherefore GE is equal to ED, and the t rinUlgie EGD is eqoui lateral; and there.-ore its thrice angles EGGD, GDE, DE G are equal to one another (Cor. 5. 1.); and the three angles of a triangle are equal (32. 1.) to two right apgles; thereforse the angle EGD is tfh third part of two rifght angles In the A same manner it may be demonstrated that tihe angle DGC is also the third part of'two right angles: and because the straight F line GO makes with EB the adjacent angles EGC, CGB equal (t31'' to two right angles: t-le. renainilng angle CGB is the tlird part of two right angles: there- 7 i tore tiht angles IG D, DGC, CGB, are equal to one another: anll also thle angles ve'rtical to them, BGA,.AGV, FGi G (1.5. 1.);:G therefore the six angles EG-I), 1)C, CG, ~BGA, SAGF, FG@E- are equal to one ano- k ~ ther. But eqail angles at the centre' / stand upon equal (26. 3.) arches; there-.... fore the six arcelis A 13 BC, CD, DE, EF, FA are equal to ae' another and equal arches ar e subtendeid l.yr equal (2'39. 3.) straight lines; thierefore the six straig;ht lines aret equlal to one anothier, and the lexagon ABCDI EI is equilateral. It is also equiatllgslar; for, sitnce the irch A-F is equal to D,1', o eacil f these add the arch ABCD;.thefrore thi! whole arch F ABCP sbhall h))o equal to the whole EDCBA': a'r the angle FgP. stands upon ttie 9.rch FABC ), alnd thle Pang! AE, upvon EIDCOA: thlereco the "ie ~ngl k 0ou- ]ELEMENTS, &c. AFE is equal to FED-: in the same manner it may be demonsira:ied:e that the other angles of the hexagon ABCDEF are each of them: equal to the angle AFE or FED; therefore the hexagon is equiangu? lar; it is also equilateral, as was shown; and it is inscribed in the given circle ABCDEF. Which-was to be done. Con. From this it is manifest, that the side of the hexagon is equalto the straight line from the centre, th-at is, to the radius of the circle. And if through the points A, B, C, D, E, F, there be drawn straight lines touching the circle, an equilateral and equiangular hexagon shall be described about it, which may be demonstrated from what has been said of the pentagon; and likewise a circle may be inscribed in a given equilateral and equiangular hexagon, and circumscribed about it, by t: method like to that used for the pertagon. PROP. XVI. PROB. To inscribe an equilateral and equiangular quindecagon in a givencircle. Let ABED be the given circle; it is required to inscribe an equis lateral and equiangular quindecagon in the circle ABCD. Let AC be the side of an equilateral triangle inscribed (2. 4.) it the circle, and AB the side of an equi. lateral and equiangular pentagon inscrib- A ed (il1. 4.) in- the same; therefore, of such equal parts as the whole circumference ABCDF contains fifteen, the / arch ABC, being the third- part of the whole, contains five; and the arch AB, E which is the fifth part of the whole, contains three; therefore B C their dif- c ference contains two of the same parts: bisect (30. 3.) BC in E; therefore BE, EC are, each of them, the fifteenth part of the whole circumference AB D: therefore if the straight lines BE, EC be drawn, and straightt lines equal to them be placed (I. 4.) around in the whole circle, an equilateral and equiangular quindecagon will be inscribed in it. Which was to. be done. And in the same manner as was done in the pentagon, if through the points of division made by inscribing the qtu;ndecagon, straight lines be drawn touching the circle, an equilateral and equiangular quindecagon may be described about it: And likewise, as in the penta? gon, a circle mnay be inscribed in a given equilateral and. equiangular quindecagon, and circumscribed about it. GAF rN the demonstrations of this bookl there are certain s signs or:harcacters which it has been found conveni;nt to em ploy. "' i. The letters A, B, C, &c. are used to denote magnitudes of any 6k Inid. The letters ir n, p9 q, are used to denote numbers only.'C2. The sign + (plus), written between two letters, that denote "' magnitudes or numbers, signifies tile sum of those magnitudes or "nurmters. Thus, A + B is the sum of the two magnitudes denoted "' by the letters A and B; nr+n is the Sum of the numbers denoted -t' by tit and n. "3. The sign - (minus), written between tvo letters, signifies the excess of the magnitude detoted by the first of these letters, which "' is Stpposed the greatest, above that which is denoted by the other.' Thus,, A B signifies the excess of the magnittde A above the "' magnitude B. "4. When a nunmbe-r, -or a letter denoting a number, is written clo se to another letter denoting a magnitude of any kind, it sig"nifies that the magnitude is mlltipihed by the number. Thus,'3A signifies three times A; nB%, nm irnes B, or a multiple of B by " in. When the number is intended to multiply two or more mag. "nitudes that follow, it is written rthus, m(A+B), which signifies' the sum of A and B taken in times; rn(A — B) is 7m7 times the ex-' cess of. A ab)ove B.' Also, wh!en two letters that denote numbers are written close to " one anothler, they denote the product of those numbers, when mul"-'tiplied into one another. Thus, w7n ijs the product of In into n; a nd:' mnA is A multiplied by the product of in into a2.' 5. The sign = signifies the equcality of the magnitudes denoted ' by tihe letteri that stand bn the opposite sides of' it; A- sig.n'' that A is equal to B; A+ B=C — D signifies that the sum of A and "i B is equal to the excess of C above D. 6"8. The sign 7 is used to signify that the magnitudes between', which it is placed are unequal, and that the magnitude to which the "' opening of the lines is turned is grbater than the othier Thus A 7' 7 B signifies that A is greater than B; and A Z B signifies that A; i;s e.s.s than B;" DEFINITIONS, A less magnitude is said to be a part of a greater magnitude, Whleh the less measures the greater, that is, when the less is contained a 66iain nuiaber of times, exactly, in the greater. A dlreater magnitude is said to be a mnultiple of a less,; Wheih the greater is measured by the less, that is, when the gfeater'contains the less 3 certain number of times exactly. III. Ratio is a mutual relation of two magnitudes, of the same kind, to camr another, in res;pect of quantity; Iv. Si'agnitudes are said to be of the same kind, when the less can be multiplied so as to exceed the greater; and it is only such magnitudes that are said to have a ratio to one another/ V. f, there be four magnitudes, and if any equimultiples whatsoever 6ie taken of the first and third, and any equimultiples whatsoever of the second and fourth, and if; according as the multiple of the first is greater thai'the multiple of the second, equal to it, or less, the multiple of the third is als6 greater than the multiple 6t the fourth, equal to it, or less; then the first of the Inagnitudes is said to have to the second the same ratio tlhat the third has to the fourm:th:Magnitudes are said/ to be proportionals, when the first has tfie satie: ratio to the second that t'he third has to the fourth; and the third to theo fourth the same ratio which the fifth hhs to the sixth, and so on whatever be their number..' When four magnitudes, A., B, C, ) are plroprtifonals, it is usual to say that A is to B as C to D, and to write t ihen thu-s, A; A B ~' 4C -: XD, or''thusl A --, C I' *D7" VII. When of the equimultiples of four magnitudes, taken as in the fiftih definition, the multiple of the first is greater than that of the second, but the multiple of the third is not greater than the multiple of the fourth: then the first is said to have to the second a greater ratio than the third magnitude has to the fourth: and, on the contrary, the tihird is said to have to the fourth a less ratio than the first has to the second. When there is any numbiler of magnitudes greater than two, of whichI the first has to thee second the same ratio that the'second has to the third, and the second to the third the same ratio which the third has to the fourtl, and so on, the'magnitudes are said to be contis nual proportionals. IX. When three magnitudes are continual proportionals, the second is said to be a mean proportional between the other two. X. When there is any number of magnitudes of the same kind, the first is said to have to the last the ratio compounded of the ratio which the first has to the second, and of the ratio which the second has to the third, and of the ratio which the third has to the fourth, and so oni unto the last mragnitude. For example, if A, B, C, -D, be four magnitudes of tlihe same kind, the first A is said to have to the last D, the ratio compounded of the ratio of A to B, and of the ratio of B to C, and of the ratio of C to D; or, the ratio of A to D is said to be compounded of the ratios of A to B, Bto C,and C to D. And'ifA:B -: E: F; andB: C':: G: Itand C: D:: I: L, thern, since by this definition A has to D the ratio compounded of the rat~ios of A to B, B to C, C to D; A may also be said to have to D the ratio compounded of the ratios which are the same with the ratios of E to F, G to EH, and K to L. in like manner, th&'lsame things being supposed, if M has to N the same ratio whichlA ha to P, then, for shortness' sake, M is said to have to N a ratio compounded of the same ratios, which compound the ratio of A to D; that is, a rhati6 compounded of the ratios'of E t.o F, G to II, and K to L. If three magnitudes are continual'proportionals, the ratio of the first to the third is said to be duplicate of the ratio of the first to the second. T' h'us, if A be to B as B to C, the ratio of AX to C is said to he duplio "cate of the.ratio of A to B. HIence,-since by the last definition, ~' the ratio of A to C is compounded of the ratios of A to B3, and l: iii E mEEN'S'toI C, a ratio, which is compounded of two equia ratios, is dupii3 "cate of either of these ratios." XII, If four magnitudes are continual proportionals, the ratio of the first to the fourth is said to be triplicate of the ratio of the first to the second, or of the ratio of the second to the third, &c. "So also, if there are five continual proportionals; the ratio of the first " to the fifth is called quadruplicate of the ratio of the first to the "second; and so on, according to the number of ratios. Hence, a " ratio comlpounded of three equal ratios, is triplicate of any one of'' those ratios; a ratio compounded -of four equal ratios quadruplii" cate," &c. XIII, In proportionals, the antecedent terms are called homologous to onanother, as also the consequents to one another. Geometers make use of the following technical words to signify certain ways of chanaing either the order or magnitude of proportionals, so as that they continue still to be proportionals. XIV. Pernmutando, or alternando, by permutation, or alternately; this word is used when there are four proportionals, and it is inferred, that the first has tile same ratio to the third which the second has to thoi fourth; or that the first is to the third as the second to the fourth:'See Prop. 16. of this Book. XV. Invertendo, by inversion: When there are four proportionals, and it is inferred, that the second is to the first, as the fourth to the third. Prop. A. Book 5. -.YI:o tComponendo, by composition: When there are four proportionals, and it is inferred, that the first, together with tiihe second, is to the second'as the third, together with the fourth, is to trhe fourth.'18th Prop; Book 5. iDividendo, by division: when there are four proportionals, and it is inferred, that the excess of the first above the second, is to the second, as the excess of' the third above the fourth, is to the fourth. 17th Prop. Book 5. Convertendo, by conversion; when there are four proportionals, and it is infierredi that the first is to its excess above the second, as the third to its excess above the fourth. Prop, D. Book 5. OiF GEOMETRY. BOOK.' i 1 XIX.'x waquali (sc. distantia), or ex maquo, from equality of distance; whent there is any number of magnitudes more than two, and as many others, so that they are proportiona!ls when taken two and two of each rank, and it is inferred, that the first is to the last of the first rank of magnitudes, as the first is to the last of the others; Of' this there are the two following kinds which arise from the different order in which the magnitudes are taken two and two. XX. Ex equali, from equality; this term is used simply by itself, when the first magnitude is to the second of' the first rank, as the first to the second of the pther rank; and as the second is to the third of the first rank, so is the second to the third of the other; and so on in order, and the inference is as mentioned in the preceding definition; whence this is called ordinate proportion. It is demonstrated in the 22d Prop. Book 5. XXI. Zx mquali, in proportiono perturbata, seu inordinata: from equality, in perturbate, or disorderly proportion; this-term is used when the first magnitude is to fthe second of the first rank, as the last but one is to the last of the second rank; and as the second is to the third of the first rank, so is the last but two to the last bhlt one of the second rank; and as the third is to the fourth of the first rank, so is the third from the last, to the last but two, of the second rank; andr so on in a cross, or inverse, order; and the inference is as in the 19th definition. It is demonstrated in the 23d Prop. of Book 5. AXIOMSo I-QUIMULTIPLES of the same, or of equal magnitudes, are equal to onu another. II. Those magnitudes of which the same, or equal mnagnitudes, are equio multiples, are equal to one another. III. A multiple of a greater magnitude is greater than the same multiple of g less. lY. That magnitude of which a multiple is greater than the same muiltiple of another, is greater than that other magnitude. la3, 12'PiiELEMiiENTS PROP. 1. THEOR. If any tumber of magnitudes be equimulltiples of as mnany others, each oj pach, what multiple soevcr any one of thefrst is of its part, libe same multiple is the sum of all the first of the sum of all the rest, Let any number of magnitudes A, B, and C be equimultiples of as many others, D1, E, and F, each of each, A+B+C is the same mulh tiple of D-E- F, that A is of D. Let A contain D, B contain E, and C contain F, each the same ruqrn hber of times, as, for instance, three times. Then, because A contains D three timnies A=D D +. For the rame reason, B=E E+E; And also, C =F F+ F. Therefore, adding equals to equals (Ax. 2. 1.), A+B+C is equal to D+E+F, taken three times. In the same manner, if A, B, and C were each any other equirnultiple of D, E, and F, it would be shown that A+B+C was the same multiple of D-E+-F. Therefore, &c. Q. E. D. Coa. Hence, if m- be any number, mD-mEgq-nr.F=mr(Dl) -I F). For mD, mE, and mF are multiples of D, E, and F by in, therefore.their.sum,is also a multiple of D-+E+-F by in. PROP. Ilo THEOR. If to a multiple of a magnitude by any nurrtber, a multiple of the same "miagnitude by any number be added, the stum ill be the samne mutilti. ple ofthat- magnitude that the sum of the two numnbers is of unity Let A=mC, and BEnC; A+B=(m+-n)C. For, since A=mC, A=C+C4-C+ &c. C being repeated mtinmes, IFor the same reason, B-C+C+ &c. C beingo repeated i times. Therefore, adding equals to equals, A+B is equal to C taken mn +- times; that is) A+B-(m +- n)C. Therefore A+B contains C as oft as there are units in m+n. Q. E. D. CoR. I. In the same way, if there be any number of, multiples whatsoever, as A=mE, B=nE, C-pE, it is shown, that A+B+-C=: (m+n+p)E. CoR. 2. Hence also, since A+B- C (n —+-+np)E, and since A =mE, B=nE, and C=pE, mE+-nE+pE =(in —i+p)E. PROP. III. THEOR. If thefirst of three magnitudes contain the second as oft as there are tnilts in a certain number, and if the second contain the third also, as often.as there are units in a.certain number, the-first will contain the third as oft as there are units in the product of these two nu. rbr ~et A=mB, and B=AIC. then A==mn-C., OF GEOME-TIRY. B001K Y. Sinee B=S niC,mBnCC-tlC-+&c. repeated mi- times. But nC+nC &c. repeated m times is equal to C (2. Cor. 2. 5.), multiplied by n+i- — &c. n being added to itself m times; but rt added to itself in times, is it multiplied by 7n, or rin. Therefore ntC - nC + &C. repeated n times z-nmnC; whence also mBInnC, and by hypothesis -=mB, therefore-A=mnC. rTerefore, &c. Q. E. D. PROP. IV. THEOR. tf the first of four magnitcudes hos tlbe same ratio to the'ccornd which the third has to thefouarth, and i'any equimstiples whateiver be taken of ez.efirst and third, and any ehatever of the second and fourth; the mnvltiple rf the firstshall have th.e same ratio to thie multiple of the second, that the m?ultiple of the third has to the multiplid of the fourthi Let A: B:: C: D, and let m and n be any two numbers; mlA B: i:IC:rD. Take of mA and nzC equimultiples by any iiumber p, and of nB and nsD equimultiples by any number q. Then the equi:nultiples of nrA, and mC by p, are equimultiples also of A and C, for they contain A and C as oft as there are unitf in pam (S..), and are equal to pinmA and pmC. IFor the same reason the multiples of nB and nrD by q, ate qnB, qniD. Since, therefore, A: B:: C: D, and of A and C there are taken any equimultiples, vii. pmiA and pmC, and of B3 and D, any equimuiltiples qnB, qnD, if pnmA be greater than qnB, p7wC must be greater than qnD (d.f: 5. 5.); if equal, equal; and if less, less. BuS pmA, pnimC are also equimtlltiples of vmA and nC, and qazB, qnD gre equimultiples of nB and nD, the'r;efore (def: 5 5..), rnA: ~iB::- mC nD. Therefore, &c. Q. E. D. COR. In the same manner it may be demonstrated, that if A: B:. C: 1D, and of A and C eqCuinmultiples be taken by any number mz, viz.?MA and PiC, mA B:: mC:. TI'his may also be considered as in?.:cluded in-the proposition, and as being the case Tvhen -Il PROP. V. TH E OR tf.one,magnitudce. e the samne nmZltiple of anotier,,/iich a mtagnitudP. tiken frorn thefirslt is of a mnagnlitde taken fromn te other; -the re' nmainderis the same?nultle of the renmarindr', that the zhole is-of the zvh6"c. Let "Ai:A and tnB be aRy equirnuitiplos of the t~wo namagnitudes A and B, of which A. is greater than B; rlaA-mB is the same multiple of A- B that rtA is of A, that i',.A — mB —=-vt(A-B). Let D be the excess of A abov'e B, then A — BD, and adding B toboth,A=DqI —B. - Therefore (. 5.)-nIt.. nD — +nB; tale,rB from. bot,ll and?a A —- tnB-=, D b If DA- B i thelfre nr.& m mB(A I.. Tr lvefore, &SC..E DE ~.11.4 bELEMENT6 PROP. VI. THEOR. (ifjrom a multiple of a magnitude by any number a multiple of the same magnitude by a less number be taken away, the remainder will be the samne multiple of that magnitude that the difference of the numbers is of unity. Let *nA and nA be multiples of the magnitude A, by the numbers M and n, and let in be greater than n; mA —nA contains A as oft as s —n contains unity, or mA-nA=(m —n)A. Let rn-n=q; then m=n+q. Therefore (2. 5.) mrA=nnA+qA; take nA from both, and mA-nA=q&. Therefore mA- nA contains A as oft as there are units in q, that is, in m- n, or mA- nA=(m - n) A. Therefore, &c. Q. E. D. -Co. When the difference of the two numbers is equal to unity, or -n —i, then tnA-nA=A. PROP. A. THEOR. iYffour magnitudes be p.roportionals, they are proportionals also whev~ taken inversely. If A B:: C: D, then also B: A::D: C. Let MA and mC be any equimultiples of A and CB; nB and nD any equimultiples of B and D. Then, because A: B:: C: D, if m-A be less than nB, mC will be less than nD (def. 5. 5.), that is, if nB be greater than mA, nD will be greater than inC. For the same reason, if-nBn —A, nD=mC, and if nB / mA, nD L mC. But nB, nD are any equimultiples of B and D, and mA, inC any equimultiples of A and C, therefore (def. 5. 5.), B: A:: D: C Therefore, &c,. Q. E. D. PROP. B. TIHEOR,.f. the first be the sa:te mnultiple of the second, or the same part of i2t that the third is of thefourth; the first is to the second as the third to tle fourth. First, if mA, v. B hbe etuimultiples of the magnitiudes A and B, mA A:: mB: B. Take of rnm annd 0mB equirlltiples-by any- number n; and of A and B equimultiples by any numbei p; these wil' be nm.A (3. 5.), p.A nmnB (3. 5.), pB. Now, if nmA be greater than pA, nm is also greater than p; and if nm is greater 1tan p, nnmB is greater than pB, therefore, when nmA is greater than pA, nmB is greater than pBo In the same manner, if nmA=pA, nmB= pB, and if nmA/LpA, 91n,. Z/pB.'Now, ntmA, nnB are ann eqaximultiples of mA and mB anO OF GEOMETRY. BOOKV'' I. A pB are any equimultiples of A and B, therefore mA A A: mB B (def. 5. 5.). Next, Let C be the same part of A that D is of B; then A is the same multiple of C that B is of D, and therefore, as has been demonstrated, A: C: B: D, and inversely (A. 5.) C A: e 1) ~ B. Therefore, &c. Q. E. D. PROP. C. THEOR. If the first be to the second as the third to thefourth; and if the first be a multiple or a part of the second, the third is the same multiple or the same part of thefourth. Let A: B:: C: D, and first, let A be a multiple of B, C is the same multiple of D, that is, if A-m=iB, C=mD. Take of A and C equimultiples by anly number as 2, viz. 2A and 2C; and of B and D, take equimultiples by the number 2m, viz. 2rmB, 2#mD (3. 5.); then, because A —rnB, e2A-*2mB; and since A: B: G C: D, and since 2A=2mtB, therefore 2C =2mrD (def. 5. 5.), and C=miD, that is, C contains D m times, or as often as A contains B. Next, Let A be a part of B, C is the same part of D. For, since A:B::C:D, inversely (A. 5.), B:A:: 1):C. ButAbeing a part of B, B is a multiple of A-; and therefore, as is shewn above, D is the same multiple of C, and therefore C is the same part of D that A. is of B. Therefore, &c. Q. E1D. PROP. VII. THEOR. Equal ntagnitutdes have the same ratio to the samne mragnitude; and dthe same has the same ratio to equal magnitudes. Let A and B be equal magnitudes, and C any other; A: C: B: C, Let mA, moB, be any equimultiples of A and B; and nC any multiple of C. Because A-B, mA-=mB (Ax. 1..5.); wherefore, if mA be greater than:nC, rmB is greater than nIC; and if trA=,nC, mB=nC; or, if,mzA2nC mB /nC. But mA and mB are any equimultiples of Aanl B, land nC is any multiple of C, therefore (def. 5. 5.) A: C:: B: C. Again, if A- B, C: A:: C: B; for, as has been proved, A: C::B: C, and inversly (A. 5.), C: A:: C: B. Therefore, &c, Q. E. D. PROP. VIII. THEOR. -Of -unequal magnitudes, the greater has a greater ratio to the slame than'the less has; and the samn ite itude has a greater ratio to the less'than it has -to the greater. Let A+B be a magnitude greater than A, and C a third magnitudeb RcLE NPIlNTS +B hia's to C a grieater patio than A has to C; anid C has a greatei ratio to A than it has to A+B. Let mn be such a number that m7A and ImB are each of them greate' than C; and let nC be the least multiple of C that exceeds mA +mBn theni n - C, that is, (n - I)C (1. 5.) will be less than mA+inB, or mA+rmB, that is, m(A+B) is greater than (n - 1.)C. But because 4iC iS greater than mA+tB; and C less thaln mB, nC'- C is greater than mA, or mA is less than nC —C, that is, than (n-1.)C. There: tore the multiple of A+B by m exceeds the multiple of C by n —1 but the multiple of A by m does not exceed the multiple of C by n —; therefore A+B has a greater ratio to C that A has to C (def. 7. 5.). Again, because the multiple of C by n-l, exceeds the multiple of A by im, but does not exceed the multiple of A+B by n, C has a greator ratio to A than it has to A+B'(def. 7. 5.). Therefire, &c. Q. E, D PROP. IX. THEOILt Ijagnzicudes whihlch hfave the same ratio to the same magnitude are eq ua to one another.; and those to which the sarne ragnitude has the same -ratio are equal to on0e another~ If A: C:: B C, A=B. a For, if not, let A be greater than B; then, becaiuse A is greater that D, two numibers, m and n, may be found, as in the last proposition, such that mA shall exceed aLC, while vmB does not exceed nC. But because A C:: B C and if mA exceed nC, mnB must also exceed aC ldef. 5: 5.); and it is also shewn that niB does not exceed nC, which is imrpossible. Themefore A is not greater than B; and in the:'ame way it is demonstrated that B is not greater. than A; therefore A is equal to B. Next, let C: A:: C: B, AB. — For by inversion (A. 5.) A C:: B ~ C; rnd therefore by the first case, A —-B PROP. X. THEOR.'That mnagnittud-; which has a greater ratio tikan another has to the samne magnitude, is the greatest of the tw~,o A dnd that megnitude, to which'the same has a greater ratio thaOt it has to an other magnitude, is the least of the two. If the ratio of A, to C be greater than thdat of B to -CA is greater %han B. Because A: (C 7B: C, tWvo numbers mn and n may be found, such -that mA7nC, and mnBLZnC (def. 7. 5.). Therefore also mnA7mB, and A7B (Ax.:4. 5.). Again, let C: B7 C: A; BLZA. For two numbers, m and n mny be found, such that vrC 7nB, and mCZnA (def. 7. 5.) Therefore,'since nB is less, and nA greater than the same magnitude inC, nBZ nA, (and threfo're'B A. Therefor:e, &c. QE. F). PROP. XI. THEOR. f11atios that are equoal to the same ratio are equal to one another. If A: B:: C::; and also C: D: E: F; thenA:B:: E.: F, Take mA, mnC, mE, any equimultiples of A,, and E; and nB, nDl, n:F any equimultiples of B, D, and F. Because A: B:: C D, if mA 7 nB,- mrC 7 nD (def. 5.5 ); but if &C 7 nD,?mE 7 nF (def. 5. 5.), because C: D: E: F; therefore if mA 7nB, mE7 nF. In the same manner, ifmA-InB, mE =nF; and if mAznB, nE /_ nF. Now, mA, mE are )any equimultiples whatever of A and E; and nB, nF any whatever of B and FF; therefore A: B E: F (def. 5. 5.) Therefore, &c. Q. E. D' PROP. XI1. THEOR. Iff any number'of magniztudes be proportionals, as one of the antecedentb is to its consequent, so are all the antecedents, taken together, to al:'the consequents, IfA: B:: C D, and C: D: E: F; then also, A: B:: A+~C +E; B+ D +F. Take mA, mtC, mE any equimultiples of A, C, and E; and nB, nD~ nF, any equimultiples of B, D, and F. Then, because A: B:: C D, if mA7mB, nC7nD (def. 5. 5.); and when mC7nD, mE 7nF9'because C D': E: F. Therefore, if nA7nB, mA-IC —nCmE7nB +nD+nF: In the same manner, if mA=nB, mtA+miC+mE-nB + hDA —nF; and ifmA z nB, mAiA+mCrr m+E Z nB+?- D+nF. Now, mA +mC+-mE —R-(A, C+E) (Cor. 1. 5.), so that,Ar and m-A+mC+mE are any equimultiples of A, and ofA+C+E. And for'the same reason niB, and nB+nD+nF are any equimultiples of B, and of Ba D+F; therefore (def. 5. 5.) A: B:: A+- C +E: B +- D+F Therefore, &c. E, D. PROP. XIlI. THIEOR. If the first have to the second the same ratio which the third has to ti'e fourth, but the third to thefourth a greater ratio than the fifth has to the sixth; the first has also to the second a greater ratio than thefifth'has to the sixth. IfA: B:: C: D; but C: D7E: F; then also, A: B7E: F. Because C: D 7E: F, there are two numbers m and n, such that inC7-nD, but mELnF (def. 7. 5.). Now, ifmC7nD, mA7nB, beo cause A: B: C: D. Therefore mA 7 B, and mnELnF, wherefore, A: B7: F (def. 7, 5.). Therefore, &c. Q. E.o D. i 18 E LEMENTS PROP. XIV. THEOR. f the first have to the second the same ratio which the third has to iihe fourth, and if thefirst be greater thanl the third, the second shall be greater than thefourth; if equal, equal.; and if less, less. IfA:B::C: D; then if A7C, B7DD;if A=C, B=D; and if A LC, B/_ D. First, let A7C; then A: B7C: B (8. 5.), but A: B:: C:D therefore C: D'7C: B (13. 5.), and therefore B 7 D (10. 5.) In the same manner, it is proved, that if A —C B ID; and if AL C, BL D. Therefore, &c. QE. D. PROP.' XV. THEOR. -Magnitudes have the same ratio to ome another iwhich their equimultilles have. If A and B be two magnitudes, and m any number, A. B:: mA qL Because A:B::A:B (7. 5.); A:B-:: A-dA:B*+B (12..), or A: B:: 2A: 2B. And in the;same manner since A: B:: A: 2B, A:- B:: A+2A: B+2B (12. 5.), or A: B:: 3A:f3B; and so cn, for all the equimultiples of A and B. Therefore, &c. E. Da PROP. XVI. THEOR. iffour magnitudes of the same kind be proportionals, they will'also be proportionals when taken alternately. If A: B:: C: D, then alternately, A: C:: B: D. Take mA, imB any equimultiples of A and B, and nC, nD any equimultiples of C and D. Then (15. 5.) A: B:: mA: mB; now A: B:: C: D, therefore (I1. 5.) C: D::rnA: mB. But C: D:?iC: nD (15. 5.); therefore mA: mnB::: nD (I 1.- 5 ): where. fore if uA7nC, nB 7nD (14. 5.); if mAnC. mB —nD, or if i A aC, m.BL/nD; therefore (def. 5. 5.) A: C:: B: D. Therefore, &c. Q. E. D. PROP. XVII. THEOR. If magnitudes, taken jointly, be proportionals, they witl also be proper-. tionals wahen taken separately; that is, if the first, together with the second, have to the second the same ratio vwhich the third, toether with thefourth, has to thefourth, the first will have to the second tfte same ratio which the third has to the fourth. If A+B B:: c+D:D, then by division A i B: C: Do OF GEOMETRY. BOOK V. 119 Take mnA and nB any multiples of A and B, by the numbers mn and 2-; and first let mA 7nB: to each of themn add 1nB, then mA+mB 7 nmB+nB. But mA+mnB=m(A+B) (Cor. 1. 5.), and,nB+nB= (md-n)B (2. Cor. 2. 5.), therefore m(A —B) 7(m+n)B. And because A+: B:: C+D: D, if mn(A+B)7(,t;+ — n)B, m(C+JD)7(m+-n)D, or mC-+mD 7mD+nD, that is, taking mD from both, mC 7nD. Therefore, when mA is greater than nB, InC is greater than nD. In like manner, it is demonstrated, that if mnA=nB, mC =nD, and if rnA LZnB, that inD/_ nD; therefore A B:: C: D (def. 6. b.). Therefore, &c. Q. E. D. PROP. XVIT_ TIHEOR. If magnitudes, taken separately, be proportionals, they will also be pro. portionals when taken jointly, that is, if thefirst be to the second as the third to thefourth, the first and second together will be to the se-. cond as the third and fourth together to the fourth. If A: B:: C: D, then, by composition, A+B: B: C+D: D. Take m(A+B), and nB any multiples whatever ofA+-B and B: and first, let m be greater than n. Then, because A+B is also greater than B, m(A+B) 7nB. For the same reason, m(C+D) 7nD. In this case, therefore, that is, when m 7n, mn(A- B) is greater than nB, and m(C+D) is greater than nD. And in the same manner it may be proved, that when m=n, m(A-+B) is greater than nB, and m(C+D) greater than nD. Next, let mn L n, or.n 7rm, then m (A+B) may be greater than nB, or may be equal to it, or may be less; first, let m(A+B) be greater than snB; then also, mA+-mB 7 nB; take rnB, which is less than nB, from both, andrnA 7 nB-mB, or nA7 (n - m)B (6. 5.). B:t ifmA 7 (n — n) B,m nC7(n-tn)D, becatse A: B:: C: D. Now, (n —m)D=nD — mD (6. 5.), therefore, tnC 7nD mnD, and adding mD to both, mnC-+m, D7nD, that is (!. 5.), n(C+D) 7nD. If, therefore,?n(A+B) 7nB,.n(C —+D) 7nD. In the same manner -it will be proved, that if m(A+B)=nB,m(CD) —nD; and if m(A+-B) / nB,rn(C+D)L nD; therefore (def. 5. 5.)? A+B: B:: C+D: D. Therefore, &c. Q. E. D. PROP. XIX. THEOR. Ifa whole magnitude be to a whole, as a magnitude taken from thefirs t is to a magnitude taken from the other; the remainder will be to ithe remainder as the whole to the;hole. IfA: B:: C: D, and ifC beless than A, A-C:B-D:: B.,Because A: B:: C: D,.alternately (16. 5.), A: C ~B: D; and therefore by division (17. 5.) A-C: C B —D) D. Wherefore, againalternately, A-C B] ~ C D; but A: B:: C D. theref -} i ~U t&ELEiM E N i S fore (11. 5.) A-C B-D:: A: B. Therefore, &c. EQ. E. i CoR. A —C: B-D: C- D3. PROP. 3. THEOR. f/foeqr magnitudes be proportionals, they are also proportionals by con,, version, that is, thefirst is to its excess above the second, as the thirc to its excess above the fourth? If A: B:: C: D, by conversion, A: A-B:: C: C-. For, since A: B: C: D, by division (17. 5.), A-B: B: C-l-i D, and inversely (A. 5.), B; A-B:: D: C — D1; therefore, by comr position (18. 5.), A: A-B:: C: C-D. Therefore, &c. Q. E. D. Con. In the same way, it may be proved that A: A+B:: C C C+D. iPROP. XX. THEORo Yf there be th'ree inagnitudes, and other three, ewhich tak.n two and tuzt, have the same ratio; if the first be greater than the third, the fourth is greater than the sixth; if equal, equal; and if less, less. If there be three magnitudes, A, B, and C, and other three D, lE and F; and if A: B: D: E; and also B: C, B C,: E:F, then if A7 C, D7F; if A=C, E 3' D=F: and if A L C, DZFo 1. First, let A7C; thenA:B7C: B (8. 5.). But A:B:: E therefore alsQ D: E7C: B (18. 5.). 1Now B: C:: E: F, and inverseIy (A.5.), C: B:: F E; and it has been shewn that D: E7C: B5 therefore D: E7F: E (13. 5.), and consequently D?7F (10. 5.) Next, let A=C; then A: B: C: B (7. 5.), but A: B:: D E: therefore, C: B:: E, but C B: F E E, therefore, D3: E': F E (11. 5.), and D= —F (9. 5.). Lastly, let A/C. Then C7A, and because, as was already shewn, C B:F:B:, and B A:: E D; therefore, by the first case, ifC7A, F7D3 that is, if AZC. D, LF. Therefore, &c. QE. D. PROP. X1I. TH'EOh. If there be three r,agszitudes, and othe' tIhree, which have:/the san3e -,: tio taken t"wo and two, but in a cross order; if thefirstnmaatgitude be greater than the third, the fourth is greater than the sixth; ifjequzco. equal; and f less, less. If there be three magnitudes, A, B, C, and other tiree, D, E, and F, such that A: B::: F, arnd B: C,: D-:;if A7C, D714 if A=C, D-=F. and if A /, D F, OF GEOMETRY. BOO)K VY I2 First, let A7C. Then A: BC: B (8. 5.), I A, B, C, but A:B:: E:F, therefore F: F7C: B D, E, F.; (13. 5). Now, B: C:: D: E, and inversely, C: B:: E: D; therefore,E:F7E: D (13. 5.), wherefore, 7:F (10. 5.). Next, letA=C. Then(7. 5.) A: B C: B;but A: B: E F, therefore, C: B:: E:F (11. 5.); butB: C D: E, and inverse, ly, C: B E: D, therefore (11. 5. ), E: F: E: D, and, consequently, D-F (9. 5.). Lastly, let A/ C. Then C 7 A, and, as was already proved, C;: E: D; and B: A, F E, therefore, by the first case, since T7A% F 7 D, that is, D_ F. Therefore, &c. Q. E. D..PROP. XXII. THEORP9 lf there be any number of mnagnitudes, and as many others, which, takey two and two in order, have the same ratio; the first will have to the last of thefirst magnitudes, the sapae ratio which. thefirst of the ot.e0, has to the last,* First, let there be three magnitudes, A, B, C, and other three, D, E. F, which, taken two and two, in order, have the saime ratio, yvg. A j;:: D: E, and B: C:: E: F; then A: C:: D: F. Take of A and D any equimultiples whatever, mA, mI); and of B and D any whatever, nB, nE: and of C and F any whatever, qC, qF~:Because A: B: D:E, mA: nB:: D:nE A, B, C(4. 5.); and for the same reason, nB: qC:: E: I U, E, F, qF. Therefore(20. 5.), according as A isgreat- I7n, nB, Q or than qC, equal to it, or less, mD is greater than mD, riE, qF. qF, equal to it or less; butm nA, wrD are any equimultiples of A and D; and qC, qF are any equimultiples of C and F' therefore (def. 5. 5.), A: C:: D: F: Again, let there be four magnitudes, apd othler four which, tagken two and two in order, have the same ratio, viz. A B:: E: F; B C::F:G; C:D:: G: H, then A D:: E: H. -- For, since A, B, C are threc magnitudes, and E, F, G other three, which, taken two iA B, C, D, and two, have the same ratio, by the forego- E, F, G, Iti_ ing case, A: C::E: G. And because al- - so C: D:: G: H, by that samecase, A: D: E: H Inthe am mnanner is the demonstration extended to any number of magnitudes9, Therefore, &c. Q. E. Do: N B? This Tproposition is usually cited by the words' erx aquali," or "e( alo qu.~ X 22 ELEMEINTS PROP. XXIII. TH3EOR. If there be any number of magnitudes, and as many others, which, takue two and tteo, in a cross order, have the same r,,tio; the first wilk have to the last of thefirst magnitudes the same ratio which the firs4 ofthe others has to the last.e - First, let there be three magnitudes, A, B, C, and other three, D~ E, and F, which, taken two and two in a cross order, have the sanme ratio, viz. A: B:: E: F, and B: C:: D: E, then A: C:: D: F, Take of A, B, and D, any equimultiples mA, rmB, mD; and of Ct, E, F any equimultiples nC, nE nF. Because A: B:: E: F, and because also A: B:: mA: 1mB (15. 5), andE: F:: nE: nF; therefore, snA: mB:: nE: nF (11. 5.). Again, because:-: C: E, mB nC':: mD nE A, B, C, (4. 5.); and it has been just shewn that nmA:E, F, nB:: nE: nF; therefore, if' mA 7nC, mD 7nF nmA, nB, nC, (21. 5 ); if mA=nC, D=-nF; and if ntA/_nC, rnD, nB, nF. { mDZ- nF. Now, nmA and mnD are any equimultiples of A and- Dl, and nC, nF any equimultiples of C and F; there. fore, A: C:: D: F (def. 5. 5.). Next, Let there be four magnitudes, A, B, C, and D, and other four, E, F, G, and EI, which taken two and two, in a cross order, have the same ratio, viz. A: B:: G: H; B: C:: F: G, and C:D: E:F, thenA: D E: AB, - D, Hi.For, since A, B, C, are three magnitudes, E, F, G., i and F, G, H: other three, which taken two and i two, in a cross order, have the same ratio, by the first case, A: C:: Fr: H. But C: D:: E: F, therefore, again, by the first case, A D: E: H. In the same manner may the demonstration be extended to any number of magnitudes. Therefore, &c. Q. E. D. PROP. XXIV. THEOR. f the first has to the second the same ratio which the third has to thefourth; antd the fifth to the second, the same ratio which the sixth has to the fourth;l th first and fifth together, shall have to the second, the same ratio which the third and sixth together, have to thefourth. Let A: B:: D, and also E: B:: F: D, thenA+-E:B::C — F: D. Because E: B:: F: D, by-inversion, B1: E:: D: F. But by hypothesis, A: B:: C: D, therefore, ex xquali (22. 5.), A: E:: C F; and by composition (18. 5.), A-E: E:: C- +F: F. And again by hypothesis, E: B::- F: D, therefore, ex-oequali (22. 5.), A+E B:: C -.F: D. Therefore, &c. Q. E. D. * N. B. This proposition is usually cited by the words'" ex eqtuali in pronportione pe':: V tiarbata o r. "es Tqeguo inverse.X!?r PF GEOM1ETRY. BOOK V i PROP. E. THEOR. Jf four magnitudes be proportionals, the sunm of the first is to their dif/erence as the sum of the other two to their dfference. Let A: B: C: D; then if A7B, A+B: A —:: C-D: C-D; or if A/ZB A+B B -A:: C+D: D -C. Vor, if A 7B, then because A: B:: C - D, by division (17. 65. A -- B: B:: C - D: D, and by inversion (A. 5.), B: A-B:: D: C-D. But, by composition (18. 5., A+B: B:: C+-D: D, therefore, ex mquali (22. 5o') A+B: A-B:: C~D: C —D. iT the same manner, if B 7 A, it is proved, that A+B: B-A:: C+D D-C. Therefore, &;e. E. D. PROP. F. THEOR. Ratios which are compounded of equal ratios, are equal lo one another. Let the ratios of A to B, and of B to C, which compound the ratio'of A to C, be equal, each to each, to the ratios of D to E, and E to F, which compound the ratio of D to F, A;. C': D: F. For, first, if the ratio of A to B be equal to A A, B, B C that of D to E, and the ratio of B to C'equato I), E, F. that of E to F, ex oequali (22. 5.), A: C: D: F. And next, if the ratio of A to B be equal to,that of E to F, and'the ratio of B to C equal to that of D to E, ex- equali inversely (23. 5.), A: C:: D: F. In the same manner may the proposition: be demonstrated, whatever be the number of ratios. Therefires &: X, F Pu OF OOK E T Y NBOOK Siin SihLAseR rectilineal figures -.. lre those which have their leveral angles equal, each to bach, and the sides about thetqua! atgles proportionalss'wo sides of one figure are said to be reciprocally proportional to two -sides of another, when onie of the sides of the first is to one of the sides of the second, as the remaining side of the second is to the remaining side of the firsts II. A straight linb is said to be cut in extreme and mean ratio, when the whole is to the greater segment, as the greater segment is to the less& - 1 iThe altitude of any figure is the straight line drawn from its vertex perpendicular to the PROP. L TIIEOR..v...'rangl"gs andpaiallelog- mns, of the same altit tde, are one to anotke'as their bases. Let the triangles ABC, ACD, and the parallelogram'us EC, CF have the same /ltitude, viz. the perpendicular drawn from the point A M BD: T-hon, as the -base BC, is to the base CD),.so is the triangle ABS to the triangle ACD, and theparallelogranm EC to the parallelogram CF. Produce BD both ways to the points H, L, and take any numbceiofe straight lines BG, GiH, each equal to the base BC; and DK, KL, any number of them, each equal to the base CD; and join AG, AH, AK, AL. Then, because CB, BG, GH are all equal, the triangles AHG, AGB, ABC are all equal (38. 1.); Therefore, whatever multiple the base HC is -of the base BC, the same multiple is the triangle AHC of the triangle ABC. For the same reason, whatever the base LC is of the base CD, the same multiple is the tri- EA A F angle ALC of the triangle ra _ADC. But if the base J-IC be equal to the base CL, the triangle AHC is also equal to thle triangle ALC (38. 1.): and if the base IIC be greater - than the base CL, like- C D L wise the triangle AHC is greater than the triangle ALC; and if less, less. Therefore, sinco there are four magnitudes, viz. the two bases BC, CD, and the two triangles ABC, ACD; and of the base B.C and the triangle ABC, the first and third, any equimultiples whatever have been taken, viz. the base HC, and the triangle AHC; and of the base CD and triangle ACD, the second and fourth, have been taken any equimultiples whatever, viz. the base CL and triangle ALC; and since it has been shown, that if the base HC be greater than the base CL., the triangle AHUC is greator than the triangle ALC; and if equal, equal; and if less, less; Therefore (def. 5. 5.), as the base BC is to the base CD, so is the triangle ABC to the triangle AUD. And because the parallelogram CE is double of the triangle ABC (41. 1.), and the parallelogram CF double of the triangle Ai DA, and because magnitudes have the same ratio which their equimultiples have (15. 5.); as the triangle ABC is to the triangle ACD, so. is the parallelogram EC to the parallelogram CF. And'because it has been shown, that, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD; and as the triangle ABC to the triangle AC)D, so is the parallelogram EiC to the parallelogran- CF; therefore, as the base BC is to the base CD, so is (11. 5.) the parallelogram EC to the parallelogram CF. Wherefore triangles, &c. Q. E. D, Con. From this it is plain, that triangles and parallelograms that have equal altitudes, are to one another as their bases. Let the figures be placed so as to have their bases in the same straight line; and having drawn perpendiculars from the vertices of thetriangles to the bases, the straight line which joins the vertices is parallel to that in which their bases are (33. 1.), because the per~n-diculars are both equal and parallel to one another. Thien, if'th same construction be made as in the propositiona the demonstratiotn will be the same. PROP. II. THEOR. Yfua straight line be drawrn parallel to one of the sides of a triangle, iet will cut the other sides, or the other sides produced, proportionally. AZnd if the sides, or the sides produced, be cut proportionally, the straight line which joins the points of section will be parallel to the remaining side of the triangle. Let DE be drawn parallel to BC, one of the sides of the triangkl; ABC: BD is to DA as CE to EA. Join BE, CD; then the triangle IBDE is equal to the triangle CDE.(37. 1.), because they are on the -same base DE and between the same parallels DE, BC: but ADE is another triangle. and equal -magnitudes have, to the same, the same ratio (7. 5.); therefore, as the triangle BDE to the triangle ADE, so is ihe triangle CDE to the triangle ADE; but as the triangle BOE to the triangle ADE, so is (1. 6 ) BD to DA, because having the same altitude, viz.- the perpen. dicular drawn from the point E to AB, they are to one another as their bases; and for the same reason, as the triangle CDE to the triangle ADE, so is CE to EA. Therefore, as BD to DA, so is CE to EA (l. 5.) Next, let the sides AB, AC of the triangle ABC, or these sidez A.. I,, T produced, be cut proportionally in thle points D, E, that is, so thag AD be-to DA, as CE to EA, and join DE; DE is parallel to BC. The same construction being made, because as BD to DA, so is CE to EA; and as BD to DA, so is the triangle BDE to the triangle AR1E (1. 6.): and as CE to EA, so is the triangle CDE to the triangle ADE; therefore the triangle BDE, is to the triangle ADE, as the triangle CDE to tile triangle ADE; that is, the triangles BDE, CDE have the same ratio to the triangle ADE; and therefore (9. 5.) the triangle BDE is equal to the triangle CDE:I And they are on the same base DE' 7; hut equal trianglec o'n the sam.e base are between thfi OF GEOMETRY. BOOK Vl. i27 samne parallels (39. 1.); therefore DE is parallel to BC. Wherefore, f a straight line, &c. Q. E. D.. PROP. Ill. THEOR. if the angle of a triangle be bisected by a straight line which also cuts the base; the segrnents of the base shall have the sanme ratio which the other. sides of the triangle have to one another; And if the segments of the base have the same ratio which the other sides of the triangle have to one another, the straight line drawnfrom the vertex to the point of section, bisects the vertical angle..Let the angle BAC, of any triangle ABC, be divided into two equal.ngles, by the straight line AD; BD is to DC as BA to AC. Through the point C draw CE parallel (31. 1.) to DA, and let BA produced meet CE in E. Because the straight line AC meets the parallels AD, EC, the angle ACE is equal to the alternate angle CAD (29. 1.): But-CAD, by the hypothesis, is equal to the angle BAD; wherefore BAD is equal to the angle AGE. Again, because the straight line BAE meets the parallels AD, EC, the exterior angle BAD is equal to the interior and opposite angle AEC: But the angle AiCE has been proved equal to the / angle BAD; therefore also ACE is equal to the angle AEC, and consequently the side AE is equal to the / side (6. I.j AC. And because AD is drwvn parallel to one of the sides / of the- triangle BCE, viz. to EC, BD is to DC, as BA to AE (2. 6.); D D C but AE is equal to AC; therefore, as BD to DC, so is BA to ACX (7. 5.)NeSt, let BD be to DC, as BA to AC, and join AD; the angle BAC is divided into two equal angles, by the straight line AD. Tthe same construction being made; because, as BD to DC, so j BA to AC; and as BD to DC, so - is BA to AE (2. 6.), because AD is parallel to EC; therorefo AB is to AC, as AB to AE (ii. 5.): Consequently AC is equal to AE (9. 5.), and the angle ALC is.therefore equal to the angle ACE (5. 1.) But the angle AEC is equal'to- the exterior and opposite angle BAD; and the angle ACE is equal to the alternate an- ~D C gie CAD (29. 1.: Wherefore also the angle BAD is equal to tlhe an: 12,,.3.SEL t4EMEiNTS gle CAD: Therefore the angle BAC is cut into two equail angles by the straight line AD.'Therefore, if the angle, &c. Q. E. D. PROP. A. THEOR. fJ the exterior angle of a triangle be bisected by a straight line'which also cuts the base produced; the segments between the bisecting line and the extremities of the base have the same ratio which the other sides of the triangles have to one another; A.nd if the segomets of the base produced have the same ratio Which the other sides of the triangle have, the straight line, drawn from the vertex to the point of section, bisect the exterior angle of the triangle. Let the exterior angle CAE, of any triangle ABC, be bisected by the straight line AD which meets the base produced in D; BD is to DC, as' BA to AC. rhlrough C draw CF parallel to AD (31. 1.): and because the straight line AC meets the parallels AD, FC, the angle ACF is equal to, the alternate angle CAD (29. 1.): But CAD is equal to the angle DAE (Hyp.): therefore also DAE is equal to the angle ACF. Again, because the-straight line FAE meets the parallels AD, FC the exterior angle DAE is equal to the interior and- opposite angle CFA:; But the angle ACF has been prov- - ed to be equal to the angle DAE; therefore also the angle A ACF is equal to the angle CFA, and consequently the side AF is equal to the side AC (6. 1.); and, because AD is parallel to FC, a side of the triangle BCF, BD is to DC, as BA toAF (2. D 6.); but AF is equal to AC; therefore as BD- is to DC, so is BA to AC. Now let BD be to DC,,as BA to AC, and join AD; the angle CAD is equal to the angle DAE. The same construction being made, because BD is to DC, as BA to AC; and also BD to DC, BA.to AF (2. 6.); therefore BA is to AC, as BA to AF (11. 5.); wherefore AC is equal to AF (9. 5.), and theangle AFC equal (5. 1.) to the angle ACF.: but the angle AFC is equal to the exterior angle EAD, and the angle ACF to the alterhate angle CAD; therefore also EAD) is equal to the angle CAD.'Wherefora, if the exterior, &c. Q: E. D. PROP. IV. THEOR. The sides about the equal angles of eqlltiangular tfrianglcs are proporlionals; and those which are opposite to the equal angles aree honmolo.gous sides, that is, are the antecedents or consequents of the ratios. Let ABC, T bee equiangtular triangles, having the angle ABC3( OF GEOaMET:I:.. BOOK VA. eqtual:o the a-ngle DCE, and the — angle ACB to the angle DEC, and,consequently (32. 1.) the angle BAC equal to the angle CDE. The sides about the equal angles of the triangles ABC, DCE are proportionals; and those are the homologous sides'which are opposite to th equal angles. Let the triangle DCE be placed, so- that its side CE may be Cognati guous to -BC, and in the same straight line with it: And, because the angles ABC, ACB are together less than two right angles (17. 1.), ABC and DEC, which is equal to ACB, are al- G so less than two right angles: wherefore B3A, ED produced shall meet (Cor. 29. 1.); ilet them be produced and meet in the A - -)oint F; and because the angle ABC is equal to.the angle DCE, BF is parallel: (2-8. 1.) to CD. Again, because -the angle ACB is equal to the angle DEC. AC is.pa'ralel to FE (2.3, 1.);-'Therefore Fre CD is a parallelogrnam; and consequently AYF B is equal to CD, and AC to FD (34. 1.) And-because AC is parallel to FE, one of the sides o.f the triangle i'BE, BA:.AF BC: CE (S. 6.): but AFJ is equal to CD; therei iore (7. 6,), BA: CD:: BC: CE; and alternately,' BA: BC C CE'( 16. 5.): Again, because CD is parallel to BF, BC: CE: t Y:D: DE (2. 6.); but FD is equal to AC; therefore BC: CE AG: 3DE; and alternaltely, BC: CA" CE: ED. Therefore, because it has been'proved that AB: BC:: DC: CE; and BC: CA:: CE', ED, ex eouali, BA: AU:: CD: DEe Therefore the sides, &ec Q. E. D. PROPo V. TIHEOR. Jf t'.e sides of -o Ujongles, a out each of their angles, be prope rtioi -als, the triangles shall be euiandar, and haclve their equal..angrjl oppoSite to the honmologous sides. Let the triangles ABC, -DEF lave their sides propoltionals, so thsla AB is to BC. as DE to EL; and BC to CA, as ElF to -FD.; and caQm. sequently, ex' equali, BA to AC, as ED to DEF; the triangle AB:C -is equiangular to the triangle DEF, and their equal angles are opposite to the homologous sides, viz. the angle ABC being equal to the RaPg!i't)EF, and BCA to EFD, and also BAC to EDFR At the points E, F, in the jstraight line LEF, make (23, 1.) the angle FEG equal to the angle ABC, and the angle EFG -equal to BCA, wherefore the remaining-angle BAC is equal to E the remaining angle EGF (32. 1.),and the triangle ABC is therefore equiangular to the triangle GEF; and consequently they LB C & have their sides opposite to. the equal angles proportionals (4. 6.). Wherefore, AB: BC: GE: EF; but by supposition, AB BC: DE: EF, therefore, DE: EF:; GE: EF. Therefore (11. 5.) DE and GL have the same ratio to EF, and consequently are equal (9. 5.). For: the same reason, DF is equal to FYG: And because, in the triangles:DEF, GEF, DE is equal to EG, and EF common, and also the base DF equal- to the base GF; therefore the angle DEF is equal (8. 1.). to the angle GEF, and the other angles to the other angles, which are subtended by the equal sides (4. 1,) Wherefore the angle DFE is equal to the angle GFE, and EDF to EGF: and because the angle DEF is equal to the angle G.EF, and GEF to the angle ABC; therefore the angle ABC is equal to the angle DEF: For the same reason, the angle ACB is equal to the angle DFE, and the angle at A to the angle at D. Therefore the triangle ABC is equiangular to the triangle DEF. Wherefore, if the sides, &c. Q. E, D. PROP. VI. THEOR. If two triangles have one. angle of the one equal to one angle of the other, and the sides about the equal- angles proportionals, the triangles shalt be equiangular, and shall have those angles equal'ehich are opposite to the homologous sides. Let the triangles ABC, DEF have the angle BAC in the one equal: to the angle EDF in the other, and the sides about those angles pro~ portionals; that is, BA to AC, as ED to DF; the triangles ABC, DEF are equiangular, and have the angle ABC equal to the angle DEF, and ACB to DFE.'At the points D, F, is the straight line DF, make (23. 1.) the angle F DG equal to either of the angles _D BAC, EDF'; and the angle DFE equal to the anglo / ACB; wherefore the remaining angle at B is equal / to the remaining one at G - (32. 1.), and cunsequently7 OF GEl0~1 leIT R. BOO VI',.'he triangle ABC is equiangular to the triangle DGF; and therefore BA: AC:: D (4. 6 ): DF. But by hypothesis, BA: AC:: ED: DF; and therefore ED: DF:: GDL: ( i.5. 5.DF; wherefore ED is equal (9. 5.) to 3DG: and DF is common to the two triangles EDV, GDF; therefore the two sides ED, DFare equal to the two sides GD, DF; but the angle EDF is also equa! to the angle GDF; wherefture thle base EF is.equal to the base FG (4. 1.), and the triangle EDF to the tri' angle GDF, and the renmaining angles to the remaining angles, each to each, which are subtended by the equal sides.: Thereifore the angle DFG is equal to the angle DFE, and the angle at G to the angle at E: But the angle DFG is equal to the angle ACB; therefore the angle ACB is equal the angle DFE, and the angle BAC is equal to the angleEDF (Hyp.); wherefore also the remaining angle at B -is eq al; to the remaining angle at E. Therefore the triangle ABC is equian gular to the triangle DEF. Wherefore, if two triangles, c. Q. E. Do PROP. VIo. THEOR. rf two triangles have one angle of the one equal to one angle of the other, and the sides about two other algles proportionals, then, if each of the remaining angles be either less, or not less, than a righi agle, the trri_ -aigles shall be equiaingular, and have those angles equal about wh.ch:the sides are prorortionals. Let the two triangles ABC, DEF have one angle in the one equal to one angle in the other, viza the angle BAC to the anglie EDF, and the sides about two other angles ABiC, )DEF proporlionals, so that AB is to BC, as DE to EF; and, in the first case, let each of the remailning angles at C, F be less than a right angle. The triangle ABC is equiangular to the triangle DEF, that is, the angle AB_ isequal to the an gle DEF, and the rernaining angle at C to the remaininlg angle at F. For, if the angles ABC, DEF be n:,t equal, one of them is greater than the other: Let ABC be the greator, and at the point B, in the straight line AB, make the angle ABG equal to the angle (23. i.) Ij-, DIEF: and because the angleat A -. is equal to the angle at D, and the/! f) -angle ABG to the angle DEF; / the remaining angle AGB is equal (32. 1.) to the renmaining angle DFE: Therefore the e tiangle ABG is equiangtular to the triangle DEF; wherefore (4. C.), AB: BG:: DE EF; but, by hypothesis, DE: Eli': AB: BC, tlereforse ~ B, ABR B EG: (11; 56.) arid becatuse AB has the same ratio to each of the liihes BC, BG; B C s equal (9. 5.) to BG, ani tlherefore the angle BGC is equal:to the angle BCG (6. 1.): But the angle BCG is, by hypothesis, less than a" right angle; therefore also the angle BGC is less than a right angle, a&nd the adjacent angle A.GB must be greater than a right angle (13. LE. B]ut it was proved that the angle AGB is equal to the angle at F; therefore th, anglee at F is greater than a riaht angle: But- by the hypothesis, it is less than a right angle; which is absurd. Therefore the angles ABC, DEF are not unequal, that is, they are equal: And the anigle at A: is equal to the angle at D; whlerefore the remaining angle at C is equl t the remaining angle at F.: Therefore the triangle ABC is equiangular to the triangle DEF. Next, let each of the angles at C, F be not less than a right angle; bhe triangle ABC is also, in this case, equiengular to the triangle DEFo The sane construction being made, it may be proved; in like man-.ner, that BC is equal to BG, and the angle at C equal to ihe angle BGC: But the angle at C is not less than a right angle; ihee'fore the angle / BGC is not less than a right... angle: Wherefore, two an- E] gles of the triangle BGC are together not less than two right angles, which is impossible (17. l. )-:and therefore the triangle ABEC may be proved to be equiangular to the triangle DEF, as in the first case. PROP. V1Lo TIHEOR. in a right angled triangle if a perpendicitlar be draw nfrom the right angle to the base; the triangles on each. s de of it are similar to the'hole triangle, and to one another. Let ABC be a right angled triangle, having the right angle BAC; and from the point A let AD be dirawn perpendicular to the base BC the triangies ABD, ADC aare similar to the -Whole triangle ABC, and to one another. Because the angle BAC is equal to the angle ADB, each of them being a right angle, and the -angle at B common to the two triangles ABC, ABD; thoe remnaining angle ACB is equal to the remaininganL gle BAD (32. i.): therefore the trigngle ABC is cequiangular to the'triangle A BD, and the sides abol ut Wh~ir e'qual angles are proportion-?t, (4. 6.); wherefore the triangles are similar (def. I. 6.). In the,ii manner, it mayv b demon strated, that t:he trianlle ADC is equiana .OF GEOIOiPETL. $BOO&K ~L!: gular and similar to the triangle ABC,: and the triangles AMID, ADC, being each equiangular,and similar to ABC,-and equiangular and simi. larto one another. Therefore, in a right angled, &c. Q. E. D. COR. Fromn this it is manifest, that the perpendicular, drawn fro m the-right angle of a right angled triangle, to the base, is a mean proportional between the segments of the base; and ilso that each of the sides is a mnean proportional between the base, and its segment adjacent to that side. For in the triangles BDA, ADC, BD: DA: DA: DC (4. 6.) and in the triangles ABC, BDA, BC BA:: BA: BD (4. 6.);; and in the {riaigles ABC, ACD, BC: CA CA: CD (4. 6.. PROP. IX. PROB. From a given straight line cut off any part required, that is, a pa.rt Which shall be contained in it a given number of times. Let AB be the given straight line; it is required to cut off from AB, a part which shall be contained in it a given number of times. From the point A draw a straight line AC making any angle witklAB;;and in AC take any point D, and take AC such that it shall contain AD, as oft as AB.is to contain the part, which is to be cut off froin it; join BCj anddraw DE parallel to it: then AE is the part required to be cut off. Because ED is parallel to one of the sides of the triangle ABC, viz. to BC, CD: DA:: DE: EA. (2. 6.); and by composition (18. 5.), CA: AD::BA: AE; But GA is a multiple of AD; therefore (C..5.) BA is the samrne multiple of AE, or contains AE the same num- B c ber of times that AC contains AD; and thereFore, whatever part AD is of AC, AE, is the same of AB; wherefore, from the straight line AB the part required is cut off. Whlich was to be done. PROP. X.,PROB. To divide a givelZ straight line similarly to a given. divided straight linle that is, into parts that shall have the same ratios to one another vWhicl the parts of the divided given straight line have. Let AB be the straight line given to be divided, and AC the divided line: it is required to divide AB similarly to AC.'Let AC be divided in the points D, E: and let AB1, AC be placed so as to contain any angle, and join 13C, and through the points D, E,:draw (31. 1.) DF, EG, parallel to BC; and through D draw DIIK, parallel to AB; therefore each of the figures FHI, HB. is a parallelof ';'34, - Lf g il'g E TM E I. gram: wherelore DH is equal (34. 1.) to A FG, and 1HK to GD; and because HE is parallel to KC, one of the sides of the triangle DKC,CE: ED:: (2. 6.) KIH: ID; But KFH=BG, and HI-)=GF; therefore CE: ED: BG ~ GF: Again, because FD is parallel to EG, one of the sides of 4" the triangle AGE, ED - DA:' GF -: AiBut it has been proved that CE *- ED:: L_ BG: GF; therefore the given strai-.ht line AB is divided similarly to ACG. hich ia - to be done. PROP. XI. PROBo To find a third propotional to to o given straight lineso Let AB, AC be the two given straight lines, and let them be placqed so as to contain any angle; it is requir._x. ed to find a third proportional to AB, AC Produce AB, AC to the points D, E and make BD equal to AC and havin B joined BC, through D draw DE paralKel to it (31. 1.). Because BC is parallel to DE, a side of the triangle AiDE, AB: (2. 60.) BD: -: AC: CE; but BE —AC: therefore AB' AC:: AC: CE. Wherefore to the two D given straight lines AB, AC a third pro. portional, CE is found. Which was to be done. PROP. XII. PROBo To find afoburth proportional to three give'n straight lines. Let A, B, C be the three given straight lines; it is required to find:afourth proportional to A, B, C. Take two straight lines DE, DF, containing any angle EDF; and upon these make DG equal to A, GE equal to B,3 and DH1 equal to C;:and having joined GH, draw EF parallel (31. 1.) to it through the D ~//x.\ f. -- / \, /. k, /.' x. OcV GEOMETI~Y. BOOK VI.'' point E. And because GH is parallel to EF, one of the sides of the triangle 1)EF, DG: GE:: D: HF (2. 6.); but DG=A, GE-=B and DH=C; and therefore A: B:: C: HF. Wherefore to the three: given straight lines, A, B, C, a fourth proportional Hi' is found.'Which was to be done. PROP. XIIT. PROB. To find n mean proportional between two given straight lines. Let AB, BC be the two given straight lines; it is required to find a nmean proportional between them. Place AB, BC in a straight line, and upon AC describe the semieir cle ADC, and from the point B (11. I.)draw BD at right angles to AC, and join AD, DC. Because the angle ADC in a sernicircle is a right angle (31. 3.) and because in the right angled triangle ADC, DB is drawn from the right angle, perpendicular to A the base, DB is a mean proportional between AB, BC, the segments of the base (Car. 8. 6,); therefore between the two given straight lines AB, BC, a mean proportion? al DB is found. Which was to be done. PROP, IIt. PRO Bo Equal parallelograms which have one angle of the one equca. to one angLf of the other, have their sides about the equal angles reclprocally proportional;: nd parallelograms which have one angle of~ the one equal to one angle of the other, and their sides about the equal angles reci., procally proportional, are equal to one, another. Let AB, BC be equal parallel- F ograms, which have the angles at B equal, and let the sides DB, BE be-placed in the same straight line; wherefore also FB, JG ID 113 are in one straight line (14. 1.): the sides of the parallelogratms AB, BC, about the equal angle]s,' are reciprocally proportional; that is, DB is to BE, as GB to & C BF. Complete the parallelogram FE; and because the parallelograms, AB, BC are equal, and FE is another parallelogram. A. *.: E (7, 5r,) ) L LMEAN TS but because the parallelograms AB, FE have the same altitude, AB: FE:: DB: BE (1. 6.), also, B.: FE:: GB: BF (1. 6.); therefore D]B: BE:: G1B; B1F (11. 5.). Wherefore, the sides of the parallelograms AB, BC about their equal angles are reciprocally proportional. But, let the sides about the equal angles be reciprocally proportion. al, viz. as DB to BE, so GB to BF; the parallelogram.AB is equal to the parallelogram BC. Because, DB: BE:: GB: BF, and DB BE:: AB: FE, and GB: BF:: BC EF, therefore, AB: FE:;BC FE (11. 5.) Wherefore the parallelogram AB is equal (9. 5.) to the parallelogranm;BC. Thereftre equal parallelograms, &c. Q. E. Do PROP. XV. THEOR,.Equal triangles which have one angle of the onle equ al to one angle. nf-the other have their sides about the equal tangles reciprocally proportional: Jlnd triangles which have one angle in the one equul to one anlgle in the other, and theirq sides about the equal alngles reciprocally pro. portional, are equal to one anothexr Let ABC, ADE be equal triangles, which have the a:ngle BAC equal to the angl. DAE: the sides about the.,".. equal angles of the triangles J are reciprocally proportion-' -_ _ zll; that is, CA is to A s, as EA to AB. Let the triangles be placed so that their sides CAx, AD be in one straight line; wherefore also EA and AB are in one straight line (14. E 1.); join BD. Because the triangie AB is equal to the triangle ADE, and ABD is another trianleo; therefore, triangle CAB: trian-.gle BAD:: trianle EAD: triangle BA); but CAB: BAD: CA AD and EAD: BAD:: EA: A1B; therefore CA: AD:: EA: AB (1 t 5.), wherefore the sides of the triaregles ABC, ADE about the:equal angles are reciprocally:proportional. But let the sides of the trikngles ABC, ADE, about the equal angles be reciprocally proportiotal, viz. CA to AD, as EA to AB; the triangle ABC is equal to the. tianIle ADE. Having joined B Di) as bctore; because CA: AD: E, iA 3; and since CA: AD:: triangle ABC: trianlle BAD3 (1. 6); and also A-. AF t triangl e L ADl: a c:-'AT BAD (11. 5.); therefore, trian)~ AB,. triagle BAI)D. r t,;aler A - tY iaB:Tgle!:3 that. -: tDm triangles ABC, EAD lave' the samn ratio to the triangle BAD;wherefore the triangle ABC is equal (9. 5.) to the triangle EAD,h Therefore equal triangles, &c. Q. Eo Do PROP. XV.T THiEOR four straitghit li@nes be p;oportionals, the rectangle contained by the exs gremes is equal to the rectanagle contained by the tnehns; Jlnd if the rectangle contained by the extremes be equal to the rectangle contsai ed by the means, the jour straight lines are proportionalso Let the'four straight lines, AB, CD, E, F be proportionals, Vio a~ AB to CD, so E to F; the rectangle contained by AB, F is equal to the rectangle containe by CD, E. From the points A, C draw ( 1. 1o) AG, CIl at right angles to AB, CD; and make AG equal to iF, and CH equal to E, and complete the parallelograms BG, DHo Because AB CD D: E: F; and since A4=CH, and F=AG, AB: CD (7. 5.):-CH: AG; therefore the sides of the parallelograms BG, DH about the equal angles are.recirocally proportional; but parallelograms which have their sides about equal angles reciprocally proportional, are equal to one another (14, 6.); therefore the parallelogram BG is equal to the parallelograrm DIH: and the parallelogram BG is - contained by the straight lines A B, F; because AG is equal to F; and. the parallelogram Di Es contained by CD and E. because CH is equa! 1 to E: therefore the rectangle con- tained by the straight lines AB, F i is equal to that which is contained by CD and E. And if the rectangle contained 2A - by the straight lines AB, F be equal to that which is contained by CD, F; these four lines are proportionals, viz. AB is to CD, as E to F, The same construction being made, because the rectangle contain, ed by the straight lines AB, F is equal to that which is contained by CD, E, and the rectangle BG is contained by AB, F, because AG-is equal to F; and the rectangle DH, by CD, E, because CH is equa' to E; therefore the parallelogram BG is equal to the parallelogramn DH, and they are equiangular: but the sides about the equal angles ofequal parallelograms are reciproally proportional (14. 6.): whereo fore AB CD:: CH: AG; but CH=E,,andAG= F, therefore AB 4 )CD: E. F. Whleref're; if fer, &c> Q. E, DP ti' lrh~'e;s'Oat'i Oiesie oioS Ocpl"ne Is, the rectarngle eotlaii'ned by the exd, trenmes is equal to the sqzuare o:lhe,rneanl: An.d if the,rectangle con, tained b1y the extremies he eqva to the squar's q' the mnean, the three. araight ines are pTi0Ortionals. Let the three straight ines, A, C h e proportionals, viz. as A to 3, so B to C; the- rectangle contained by A, C is equal to the square of B. Take D eqiual to 3:' and beauGe as A to B, so B to C, and that 3: is equal to D; A is (7. 57) to B, a's D to C but if four straight lines be proportionals, the rectangle contained by the extremes is equal to that which is contained by the means (16. 67); A - - therefore the rectangle A.C. = the rectangle B B.D; but the rectangle B.D is equal to the square of B, betca; Se B-D; therefore the rectdngle C A. is eqtual to wte: square of B. And if the rectangle contained by A, C be equal to the square of B; AB:"B::B C. The same conslruction being made, because the rectangle containl ed by A, C is equal -to the square of B, and the square of B is equal to the rectan:ale contai ned by B, I, because B is equal to D; there-,-bre the r ctangl aontainedl by A, C is equal to tbint contained by B, ~ u; but if;he Iectangte contained by the extremes be equal to that onttained by the metans, the four straight lines are proportionals (Ib'. 6.): therefore A: B: B": C, but B=D; wherefore A: B:: B: C Therefore, if three strai.ght lines, (&c. Q. l. Do'P-9,O'r iYll. P1RO1. T?'on~ a cgivle St';'gh't iitne to describe a rectilineealfgTure simiila r, a mite~ sXiilmaly situated to a given rectilineal figure. Let ASDI I> e. l"1Cs_1 stagt line, and CD. EF the given rectilineal:tqgure of four s;uer.i i is requi'red upon the giYen straight line AB to deseriboe a rctillincil fi-ure-imIilr andd smimilarly situated to CDFEF. Join ia F, n at theVO polinti s A, B in the straighlt line AB, mnake (23. l ) ~t}he isar; a.;-I..r.. etla! to the an.Igle at C, and the angale B AG equal -o the angli C.D1 F tlereaore, the rtialing a.Igle C Ei) is equal to tlhe #ren ma;ial r PIa~l' J - A ( i. ) wherefore the triangle IFCD is equia angular to tlthe triaangle GA Again, at the points G, B in the straighti line GCB, arn.e (03. 1.) the angle BGIi equal to the angle DFE, and the, an nle GB H equnal to FID)E; therefore the renmainig angle FED.is equ0al to the, remnaining angle C-1TB, and the triangle FDE equiangular to the triangjle GiB-1t: thlen because thes aigle AGB is equal to the: ang-le CFDl and BSG-H-t Dto r. th4e whole0 angle AGH is equal to th., v'l ie CFE: tbr the same reason, the angle ABH is equal to th gangle CDE; also the angle at A is equal to the ange at C,, and the anife H H? -to F ED Therefoire the rectilineal fi ure ABHR G is eqularigufar to CDEF but likewise these figures have their sides abo{ut thle equal angles proportionals: for the trian3gles GAB, F 2D.. beirlng equian:gular, BA A AG" DC.CF..) fo r thie same reas,'o.) AG; GB" CF;'F and because of' the t4eC.u angular triangles BGH, D'FE, GB:. "- D ~ F - th' E. h exr quali (22. 5.) AG- GIH _: - i F.:In the samem-nanner, it may'be proved, that AB: BBI:: C: DE. Alo (4. 6.), GH B: HB FE EDJ. Wherefobre, becaus the tlre tilineal figures ABHG, CDE.F are equiangular, and Ihave Uleir svide, about the equal angles proportionals, they are simnilar to aAo~.~er (cle. 1. 6.). Next, Let it be required to describe upon a given.straight line AB a reetilineal figure similar, and similarly situated to the rectiineal figure CDKEF. Join DE, and upon the given straight line AB descibe the rectiili. neal figure ABHG similar, and similarly situated to the quadrilateral -figure CD EF,by the former case; and at the points B, H- ii the st'raiaht line BH-I make the angle HB1L equal to the angole EDK, ancd the an glo BHIL equal to the angle DEX(; therefor: the rainingn anrgle a t g is -equal to the remaining angle at L; and becanuse the.gures ABHG, CDEF are similar. the angle GHIB is equal to thle anm-le, FED, arnd BHL is equal to DE K; wherefore the whole ansle GCITL is equal to the whole angle, FEK; for the same reason the angle AtBL equal'to the angle CDK: therefore the five-sided figures AGl LhIL CFE; )D are equiangllar;'and because the figures EMl1, C1i'ED a. mi...lare GlHisto B asEtoED; ands RB to L; and as, so is EYD to P,[K (4.6'j therefore, ex zaquali (22. 5.), GH is to HL, as FE to EK; for this same reason, A is to 3BL, as CD to D:' arid BL is to LI, as (i 6.) I)DKto KtE, because the triangles B LH, DKIE are equiangular: tkLerc.'re, because the five-sided figures AGRILB, Ci ElKD are equianulla', and have Their sides about the equal t.ngles proportionais, they alie similar to one another: and in the same manner a rectilineal figure of six, or more, sides may be described upon a given straight line similar to -one given, and so on. -Which was to be done. PROP. XIX. THEOR. iznimlar triangles are to one oanother in the duplicate r atio of the homen logous s"eso Let ABC, DF be similar triangles, having the angle B equal to the angle E,:and let AB be to BC, fas DE to EF, so that the side BC is homologous to EF (def. 13. 5.): the triangle ABC has to the triangle DEF, the duplicate ratio of- that which BC has to B C B Take BG a third proportional to B(C and EF (i 1. 6.), or such that BC: EF EF: BG, and join GA. Then because AB: BC DE: EF, alternately (16. 5.), AB: DE BC:EF; but:BC: EF;; EF: BG; therefore (11. 5.) AB: DE EF: G: wherefore the sides of the friangles ABG, DEF, which are about the equal angles, are reciproeally proportional: but triangles, which have the sides about two tqu a angles reciprocally propor- A tional, are equal to one another (i5. 6.): therefore the triangle ABG is equal / to the triangle DEF; and because that BC is to EF, as EF to BG; aind that if three etraight lines be proportionals, the first has to - the third'the duplicate rato of that which it has to the second; BC therefore has to BG the dui plicate ratio of that which BC has to EF. But as BC to BG, so is (1. 6.) the triangle ABC to-the triangle ABG: therefore the triangle ABC has to the triangle ABG the duplicate ratio of that which BC has to EF: -and the triangle ABG is equal to the triangle DEF; wherec ore also the triangle ABC has to the triangle DEF the duplicate ratio of that which BC has to EF. Therefore, similar triangles, &Gc fl E.E D t. Cep"o F-rea t:is it is manifest, that if three. straeight lines be p*,, OF GEOMIVETRY. B00K~ V- iie prfionals, as the first is to the third, so is any triangle upon the first to a similar, and similarly described triangle upon the second. PROP. XX. THEOR. S.imilar polygonls may be -divided into the same number. of similar tri. aingtgs, having the same ratio-to one another ha4t the polygons have and the.polygons, have to -one another the dutplicate ratio -of that -uhich their homologous sides hate. Let ABCDE, FGHKL be similar polygons, and let AB be the ho-. mologous side to FG: the polygons ABCDE, FGHKL may be divided into the same'number of similar triangles, whereof each has to each:the same ratio which the polygons have; and the polygon ABCDE has to the polygon FGHEKL a ratio duplicate of that which the side AB has -to the side FG. Join BE, EC, GL, MLH: and b1ecause the polygon ABCDE is siniif ar to the polygon FGHK-L, the angle BAE is equal to the angle GFL (def. 1. 6.), and BA: AE:: GF: FL (def. 1. 6.); wherefore, because the triangles ABE, FGL have an angle in one equal to an an*.gle in the other, and their sides about these equal angles proportionals, the triangle ABE is equianglav (6. 6.), and therefore similar, to the -triangle FGL (4. -6.): wherefore the angle ABE is equal to the -angle FGL: and, because the polygons are similar, the whole angle ABC i4 equal (def. 1. 6.) to the whole angle FGH; therefore the remaining angle EBC is equal to the remaining angle LGfH: now'because the triangles ABE, FGL are similar, EB: BA LG ~b: GF; and also because the polygons are similar, AB: BC: FG: GH (def. 1. 6.); there. fore, ex equali (22. 5.); EB: BC LG: GII; that is, the sides about the equal angles EBC, LGH are proportionals; therefor-e (6. ~6.) the triangle EBC is equiangular to the ritangle LGHI, and sinilar to it (4. 6.). For the same reason, the triangle ECD is like-:vise similar to the triangle LHK; therefore the similar polygons ABCDE, FGHKL'are divided into the same number of similar tarangles, Also these trisanlgis have, each to each, the same ratio which Ithc poiygons have to one another, the antecedents being ABE, BiO23C ECD, and the consequents FG4L, LGH, LHK: and the polygon A BCDE has to the polygon: FerGrxL the duplicate ratio of that whichi the side AB has to the homologous side FG. Because the triangle APBE is similar to the triangle FGL, ABE has to FGL the duplicate ratio (t19 6.) of that which the side BE has to the side GL: for the same reason, the triangle BEC has-to -GLH the.'duplicate-ratio of that which BE has to GL: therefore, as the triano gle ABE to the t'iangle FGL, so (41. 5.) is the triangle BEGJ to theo triangle GEH. Again, because the triangle EBC is similar to the triangle LCGHI, BC has to LGH the duplicate ratio of thlat wvhich ihel side EC has to the side LI1: fo) the same reason,: the triangle LECD has to the triangle LHIK, the duplicate ratio of that which EG has to LH: therefore, as the triangle EBC to the triangle LGH, so is (11. 5.) the triangle: ECDD to the triangle LH-K: but it has been proved, that the triangle EBC is likewise to the triangle LLGH1, as the trianngle ABE to the triangle FGL. -Thereifore, as the triangle ABE is to the teri angle FGL, so is the triangle tgBC to the triangle LGH, and the triangle- ECD to the triangle LHK-: and therefore, as one of' the antecedents to one of the consequeorts, so are all the antecedents to all=AB.CD: bt AD. CB + )DB.AC=AD.AC +- DB.ACg because CB.AC.- Therefore AD. AC+DB.AC, that is (1. 2.), (AD+ / DB) AC=AB.CD. And because the saides of equal rectangles are recipro-' -. cally proportional (4.6.), AD~DB'.... DC A:: B: AC. Wherefore, &c. QE, D. PROP. F. THEOR..f two points be taken in the diameter of a circle, such that the, reiaangle contained by the segments inztercepted betireen them.l and the centre of the circle be equal to the square of the ritlds:-and iJ fromn these points two straight lines be drawn to a.jy point rwhatsoever in the circumference of the circle, the ratio of these lines a;ill be the sanie with the ratio of the segments intercepted betwveen the Twojfirst mentioned points ana.d the circumference (of the circle. Let ABC be a circle, of which the centre is D, and in DA produced, let the points E and F be such that the rectantgle ED, )DF is equal to the square of AD; fiom E and F to any point B in the circumference; let EB, FeD be drawn; FB: BE::FA AE. Join BD, and because the rectangle PD, DE is equal to the' square of AD, that is, of DB, FD: DB:: DB: DE (17. 6.). The two triangles, FIDB, BIDE have therefore Ithe, sides proportionlal that are about the common angle D; thlerefore theSr anr eqliangui. lar (6. 6.), the angle DE:B being equal to the angle DBF, and DBE to DPFRB. Now since the sid.es abouit the.se equal angles are al.so nrropr;, tI 60 E -' 1 ELL~iEN IS tional (4. 6.), PFB BD:: BE E), and alternately (16. 5.), FB BE::BD: ED, or PB:BE::AD:DE. But because PD.: DA: DA: DE, by division (17. 5.), FA: DA: AE ED, and alternately (11. 5.) FA: AE:: DA: ED. Now it has been shewn that B: BE:: AD: DE, therefore FB: BE': FA: AE. Therefore, &c. Q. E. D. CORd If AB be drawn, because FB: BE:: FA: AE, the angle FBE is bisected (3. 6.) by AB. Also, since FD: DC:: DC: DE, by composition (18. 5.), FC: DC: CE: ED- and since it has been shewn that FA: AD (DC):: AE: ED, therefore, ex amquo, FA AE.::FC:CE. But FB:BE::FA:AE, therefore, FB:BE: FC: CE (11. 5.); so that if FB be produced to G, and if BC be drawn, the angle EBG is bisected by the line BC (A 6.) PROP. G. THEOR. f/from the extremity of the diameter of a circle a straight line hbedraw in the circle, and if either within the circle or produced wzithout it, it meet a line perpendicular to the same diameter, the rectangle containo,ed by the straight line drawn in the circle, and the segment of it initer" cepted between the extremity of the diameter and the perpendicula,; is equal to the rectangle contained by the diameter, and the segmen2 of it cut of by the perpendicular. Let ABC be a circle, of which AC is a diameter, let DE be perE B \ a' AD a~~~~j —— PLhJ K'~ OF GEOMETRY. BOOK Xl. 15;7 pendicular to the diameter AC, and let AB meet DE in F; the rectangle BA.AF is equal'to the rectangle CA.AD. Join BC, and because ABC is an angle in a semicircle, it is a -right angle (31. 3.): Now, the angle ADF is also a right angle (Hyp.); and the anglo BAC is either the same with DAF, or vertical to it; therefore the triangles ABC, ADF are equiangular, and BA: AC: AD: AF (4.6.); therefore also the rectangle BA.AF, contained by the extremes, is equal to the rectangle AC.AD contained by the means (16. 6.). If therefore, &c. Q. E.D D. PROP. H. THEOR. The perpendiculars drawn from the three angles of any triangle to the opposite sides intersect one another in the same point. Let ABC be a triangle, BD and CE two perpendiculars intersecting one another in F: Let AF be joined, and produced if necessary, -let it meet BU in G, AG is perpendicular to BC. Join DE, and about the triangle AEF let a circle be described, AEF: then,-because AEF is a right angle, the circle described about the triangle AEF will harve AF for its diameter (31. 3.). In the _A same manner, the circle described about the triangle ADF has AF for its diameter; therefore ithe points A, E, F and D are in the circumference of the same circle. But because the angle EFB is equal to the angle DFC (15. 1.), and also the angle BEF to the angle CDF, being both right angles, the triangles BEF, and _CDF-are equiangular, and therefore BF:'EF:: CF:FD (4. 6.), B or alternately (16. 5.) BF: FC C.: EF: FD. Since, then, the sides about the equal angles BFC, EFD are proportionals, the triangles BFC, EFD are also equiangular (6. 6.); wherefore the angle FCB is equal to the angle EDF. But EDF is equal to EAF, because they are angles in the same segment (21. 3.); therefore the angle EAF is equal to the angle FCG: Now, the angles AFE, CFG are also equal, because they are vertical angles; therefore the remaining angles A FEF, FGC are also equal (32. 1.): But AIF is a right angle, therefow-:FGC is a right angle, and AG is perplndicular to BC. QE. D. Con. The triangle ADE is similar to the triangle ABC. For the two triangles BAD, CAE having the angles at D and E right angles,'and-the angle at A common, are equiangular, and therefore BA: AD:: CA: AE, and alternately BA: CA AD: AE; therefore the two.triangles BAC, DAE, have the angle at A common, and the sides about, ELEMENTS, &c. that angle proportionals, therefore they are equiangular (6. 6.) and similar. Hence the rectangles BA.AE, CA.AD are equal. PROP. K. I. THEOR. If fromn any angle of a triangle.a perpendizclar be drawn to the opposite side or base: the rectangle contained by the sumn and difference of the other two sides, is equal to the rectangle contained by the sumiand'dfference of the segments, into which-the base is divided by the perpendicular. Let ABC be a triangle, AD a perpendicular drawn from the angle A on the base BC, so that BD, DC are the segments of the base: (AC+AB) (AC - AB)=(CD+DB) (CD- DB).:' From A as a centre with the radius AC, the greater of the two sides, describe the circle CFG: produce AB to meet the circumference in E and F, and CB to meet it in G. Then because AF=AC, BF=AB+ AC, the sum of the sides; and since AE=AC, BE=AC, -AB= the difference of the sides. Also, because AD drawn from the centre cuts GC at right angles, it bisects it; therefore, when the perpendicular falls within the triangle, BG=DG —DB=)DC - DBthe difference of the segments of the base, and BC —-BD+DC=- the sum of the segments. But when AD falls without the triangle, BG= DG+ —DBCD+DB= the sum of the segments of the base, and BC -CD- DB= the difference of the segments of the base. Now, in both cases, because B is the intersection of the two lines FE, GC, drawn in the circle, FB.BEz-CIC:5,RG; that is, as has been shewn, (AC+-AB) (AC-AB)=(CD+J-i]) (CD-P.-]B). Therefore? &c. SUPPLEMENT TO TtEIID ELEMENTS G E 0 IM-E TRY. (G E 0 MA E T- vat SUPPLEMENT, 1BOOK I. )OF THE QUADRATURE OF THE CIRCI4oX DEFINITIOTNS. A CHoRD of an arch of a circle is the straight line joining the extremities of the arch; or the straight line which subtends the arch. [Io The perimeter of any figure is the length of the line or lines, by which it is bounded. HI. The area of any figure is the space contained within ito AXIO0L The least line that can be drawn between two points, is a straight lit.e: and if two figures have the same straight line for their base, that which is contained within the other, if its bounding line or lines be -not any where convex toward the base, has the least perimeter. C'oRn:.. Hence the perimeter of any polygon inscribed in a circle is less than the circumference of the circle. - Cnr. 2. If fiom a point two straight lines be drawn touching a cirs cle, these two lines are together greater than the- arch intercepted be-:tween-them; and hence the perimeter of any polygon described aboit a eircle is greater than the circumference of the Mdircle~: SUPPLEMENT TO THE ELEMENTS PROP. 1. THEOR. iffrom the greater of two unequal magnitudes there be taken away- its. half, and fobtn the retrvminder its half; and so on; There will at length remain a magnitude less than the least of the proposed magf nitudeso Let AB and C be two unequal magnitudes, of which AB is the greater. If from AB there be taken away its half, and from the remainder its half, and so on; there Ad shall at length remain a magnitude less than C. For C may be multipled so as, at length, to become greater than AB. Let DE, therefore, be a multiple of C, which is greater than AB, and let it contain the parts DF, FG, GE, each equal to C. From AB take BH equal to its half, and from:. the remainder AH, take HK equal to its half, and so on, until there be as many divisions in AB as there are in DE; And let the divisions in AB be AK, KH, HB. And because DE is greater than AB, and EG taken from DE is not greater than its half, but BH taken from AB is- equal to its half; therefore -the remainder GD is greater than J3 C E the remainder HA. Again, because GD is greater than HA, and GF is not greater than the half of GD, but HK is equal to the half of HA; therefore, the remainder FD is greater than the remainder AK. And FD is equal to C, therefore C is greater than AK that is, AK is less than C QE. D. PROP. II. THEOR. Equilateral polygons, of the same number of sides, inscribed in circles% are similar, and are to one another as the squares of the diameters of the circles, ~Lt ABCDEF and GHIKLIX be two equilateral polygons of the same- number of sides' inscribed in the circles ABD, and GHK ABCDEF and GHIKLM are similar, and are to one another as the squares: of the diameters of the circles ABD, GI(K. Find N and O the centres of the circles, join AN and BN, as also GO and HO, and produce AN and GO till they meet the circumfer' onces in P and- Ko OF GEOMETRY. BOOK 1i 3 Because the straight lines AB, BC, CD, DE, EV, FA, are all equal, (the arches AB, BC, CD, DE, EF, FA are also equal (28. 3.). For the same reason, the arches GH, HI, 1IK, KL, LM, MG are all equal, and they'are equal in number to the others; therefore, whatever part the arch AB is of the whole circumference, ABD, the same is the arch GH of the circumference GHK. But the angle ANB is the same part of four right angles, that the arch AB is of the circumfer.. once ABD (33. 6.); and the angle GOH is the same part of four right angles that the arch GH is of the circumference GHK (33. 6,) therefore the angles ANB, GOH are each of them the same part of four right angles, and therefore they are equal to one another. The isosceles triangles ANB, GOH are therefore equiangular (6. 6.), and the angle ABN equal to the angle GHO; in the same manner, by joining NC, 0I, it may be proved that the angles NBC,, QHI are AN 1/ equal to one another, and to the angle ABN. Thdrefore the %whole angle ABC is equal to the whole GHI; Iand the same may be proved of the angles BCD, HIK, and of the rest. Therefore, the polygons ABCDEF and GHIKLM are equiangular to one another -and since they are equilateral, the sides about the equal angles are proportion. als; the polygon ABCD is therefore simi lar to the polygon GHIKLM (def. 1. 6.). And because similar polypois are as th~e squares of :64 -SUPPLEMENT TO THE ELEMENTS their homologous sides (-20. 6.), the polygon ABCDEF is to the polygon GHIKLM as the square of AB to the square of GHi; but because the triangles ANB, G Oil are equiangular, the square of AB is to the square of GH as the square of AN to the square of GO (4. 6.), or as four times the square of AN to four times the- square (15. 5.) of GO, that is, as the square of AD to the square of GK (2. Cor. 8. 2.). Therefore also, the polygon ABCDEF is to the polygon GHIKLMI as the square of AD to the square of GK; and they have also been shewn to be similar. Therefore, &c. Q. E. D. Cop. Every equilateral polygon inscribed in a circle is also equiangular: For the isosceles triangles, which have their common verte: in the centre, are all equal and similar; therefore, the angles at their bases are all equal, and the angles of the polygon are therefore also equal, PROP. Ill. THEOR,:Th side of any equilateral polygon inscribed in a circle bei.g given,,c find_ the side of a polygon of the same number of sides described aborts the circlet Let ABCDEF be an equilateral polygon inscribed in the circle ABD it is required to find the side of an equilateral polygon of the same.number of sides described about the circle. Find G the centre ofthe circle; join GA, GB, bisect the arch Al.in; I; and through H draw KHL touching the circle in H, and meets Ing GA and GB produced in K and L; KL is the side of the polygon reqiuired. Produce GF to N, so that GN may be equal to GL; join KN, and from G draw GM at right angles to KN, join also HG. Because the arch AB is bisected in H, the angle AGHI is equal to the angle BGH (27. 3.); and because KL touches thie circle in H,the angles LIIG,K HG are l tight angles ( 16. 3.); therefore, there are two angles of the tri-;angle HGK, equal to two anAles of the triangie HIGL, each to each. But the zide GH i I common to these tiianales; therefore they are equalt-(2,,. 1.), and GL is equal to KE. V/ Again, in the triangles Ki-L,:KGN, because GN is equal to GL; and GKcomrnon, and also the angle LGK equal to the an -. e` KGN; therefore the lease IL is equal to the base iN (4. 1.). Dot because the triangle KGN is i.oscells the angle GKN is egu OF G'orMETR. BOO0K L a -to the angle GNK, and the angles GMK, GTMN are both right angles by construction; wherefore, the triangles GMK, GMN have two angles of the one equal to two angles of the other, and they have also the side GM common, therefore they are equal (26. 1.), and the side KM is equal to the side MN, so that KN is bisected in M. But KN is equal to KL, and therefore their halves KM and KH are also equal, WYherefore, in the triangles GIKH, GKM, the two sides GK and KR are equal to the two GK and KM, each to each; and the angles GKH, GKM, are also equal, therefore GM is equal to GHI (4. i.); wherefore, the point M is in the circumference of the circle; and because KMG is a right angle, KM touches the circle. And in the same man. ner, by joining the centre and the other angular points of the inscribed polygon, aln equilateral polygn may be described about the circle, the sides of which will each be equal to KL, and will be equal in numbev to the sides of the inscribed polygon. Therefore, KL is the side of an equilateral polygon, described about the circle, of the same num. ber of sides with the inscribed polygon ABCDEF; which was to be found. CoR. 1. Because GL, GK, GN, and the other straight lines drawn from the centre G to the angular points of the polygon described about the circle ABD are all equal; if a circle be described from the centre X, with the distance G1K, the polygon will be inscribed in that circle; and therefore, it is similar to the polygon ABCDEF (2. 1.). CoR. 20. It is evident that AB, a side of the inscribed polygon is to.KL, a side of the circumscribed, as the perpendicular from G upon AB, to the perpendicular from G upon KL, that is, to the radius of the circle; therefore also, because magnitudes have the same ratio withI their equimultiples (15. 5.), the perimeter of the inscribed polygon is to the perimeter of the circumscribed, as the perpendicular from the centre, on a side of the inscribed polygon;/to the radius of the circle. PROP. IV. THEOR. A circle being given, two similar polygons may be found, the one de~ scribed about the circle, and the other inscribed in it, which shall digf fe rfrom one another by a space less than any given space. Let ABC be the given circle, and the square of D any given space; ~a polygon may be inscribed in the circle ABC, and a similar polygon described about it, so that the difference between them shall be less than the square of D. In the circle ABC apply the straight line AE equal to D, and let AB be a fourth part of the circumference of the circle. From the cir. tumference AB take away its half, and from the remainder its half, and so on till the circumference AF is found less than the circumference AE (1. 1. Sup.). Find the centre G; draw the diameter AC, as also lie straight lines AF and FG and having bisected the circumference 166 SUPPLEMENT TO THE ELEMENTS AF in K, join KG, and draw HL touching the circle in K, and meets ing GA and GF produced in it and L; join CF. Because the isoscles triangles HGL and AGF have the common angle AGF, they are equiangular (6. 6.), and the angles GHK, GAF are therefore equal to one another. But the angle GKH, CFA are also equal, for they are right angles; therefore the triangles HGK, ACF, are likewise equiangular (32. 1.) And because the arch AF was found by taking from the arch AB its half, and firom that remainder its half, and so on, AF will be contained a certain number of times, exactly, in the arch AB, and therefore it will also be contained a certain number of times, exactly, in the whole circumference, ABC; and the straight line AF is therefore the side of an equilateral polygon inscribed in the circle ABC. Wherefore also, HL is the side of an equilateral polygon, of the same number of sides,'described about ABC (3. 1. Sup.). Let the polygon described about the circle be called M, and the polygon inscribed be called N; then, B L T because these polygons are similar (Cor. 3. 1.) they are as the squateu of the homologous sides HL and AF (Sup. 3. Cor. 20. 6.), that is, because the triangles HLG, AFG are similar, as the square of HG to the square 6f AG, that is of GK. But the triangles HGK, ACF have been proved to be similar, and therefore the square of AC is to the square of CF as the polygon M to the polygon N; and, by conversion, the square of AC is to its excess above the squares of CF, that is, to the square of AF (47. 1 ), as the polygon M to its excess above the polygon N. But the square of AC, that is, the square described about the circle ABC is greater than the equilateral polygon of eight sides described about the circle, because it contains that polygon; and, for the same reason, the polygon of eight sides is greater than the polygon of sixteen, and so on; therefore, the square of AC is greater than any polygon described about the circle by the continual bisection of the arch AB; it is therefore greater than the polygon Me. Now, it OF'GEOM:ETRY. BOOK tL l6has been demonstrated, that the square of AC is to the square of AF as the polygon M to the difference of the polygons; therefore, since the square of AC is greater than M, the square of AF is greater than the difference of the polygons (14. 5.). The difference of the polygons is therefore less than the square of AF; but AF is less than D X therefore, the difference of the polygons is less than the square of D: that is, than the given space. Therefore, &c. Q. E. D. CoiL. 1. Because: the polygons Mt 2an N.n.- -Ur, one- another more thaniehither of tnem divters from the circle, the difference between each of them and the circle is less than the given space, viz, the square of D. And therefore, however small any given space may be, a polygon may be inscribed in the circle, and another described about it, each of which shall differ from the circle by a space less than the given spuceA CoR. 2. The space B which is greater than any polygon that can be inscribed in- the circle A, and less than any polygon that can be described about it, is equal to the circle A. If not, let them be unes qual;- and first, let B exceed A by the space C. Then, because the polygons described about the circle A are all greater than B, by hypothesis; and because B is greater than A by the space C, therefore no polygon can be described about the circle A, but what must exceed it by a space greater than C, which is absurd. In the same manner, if B be less than A by the space C, it is shewn that no polygon can be inscribed in the circle A, but what is less than A by a space greater than C, which is also absurd. Therefore, A and B are not unequal, that is, they are equal to one another. PROP. V. THEOR, 2"he area of any circle is equal to the rectangle contained by the semin diameter, and a straight line equal to half the circumference. Let ABC be a circle of which the centre is D, and the diameter AC; if in AC produced there be taken AH equal to half the circum. ference, the area of the circle is equal to the rectangle contained by DA and AHl SUPPLFEMENT -TO THE ELEMENTS Let AB be the side of any equilateral polygon inscribed'in the eirele ABC; bisect the cilcumference AB in G, and through G draw EUF touching the circle, and leeting DA produced in E, and DM produced in F; EF will be the side of an equilateral polygon de. scribed about the circle ABU (3.1. Sup.). In AC produced tak. AK equal to half the perimetei of the polygon whose side is AB and AL equal to half the perimeter of the polygon whose side is EF~ Then AK will be less, and AL greater than the straight line AH (Ax. 1. Sup.). Now, because in the triangle EDF, DG is drawn perpendicular to the base, the triangle EDF is equal to the rectangle contained by DG and the half of EF (41. 1); and as the same is true of all the other equal triangles having their vertices in D, which make up the polygon described about the circle; therefore, the whole poly? gon is equal to the rectangle contained by DG and AL, half the perimeter of the polygon (I. 2.), or by DA and AL. But AL is greater than AH, therefore the rectangle DA.AL is greater than the rectangle DA.AH; the rectangle DA.AII is therefore less than the rectangle DA.AL, that is, than any polygon described about the circle ABC. Again, the triangle ADB is equal to the rectangle contained by DM the perpendicular, and one half of the base AB, and it is therefore kess than the rectangle contained by DG, or DA, and the half of AB& iF" i~~~~~_ OF GEOMETRY. BOOIK-. 1t And as the same is true of all- the other triangles having their verti-:es in-D, which make up the inscribed polygon, therefore the whole of the inscribed polygon is less than the rectangle contained by DA, and AK half the perimeter of the polygon. Now, the rectangle DA.AK is less than DA.AH; much more, therefore, is the polygon whose side is AB less than DA.AH; and the rectangle DA.AHt is thereforegreater than any polygon inscribed in the circle ABC. But the same rectangle DA.AH has been proved to be less than any polytgon described about' the circle ABC; therefore, the rectangle DA.AHI is equal to the circle ABC (2. Cor. 4. 1. Sup.) Now, DA is the semidiameter of the circle ABC, and All the half of its circumferenceo Therefore, &c. Q. E. D. CoR. 1. Because DA: ALt-:: DAY: DX.AH (I. 6.), and because By this proposition, DA.A!H=the area of the circle, of which DA i, the radius: therefore, as the radius of any circle to the semicircumference, or as the diameter to the whole circumference, so is tih sqtare of the radius to the area of the circle. CoR. 2. Hence a polygon may be described about a circle, the e. rimeter of which shall exceed the circtumference of the circle by v line that is less than any given line. Let NO be the given line. Take in NO the part NP less than its half, and less also than AD, and leta polygon be described about the circle ABC, so that its excess above ABC may be less than; the square of NP (I. Cor. 4. 1. Sup.). Let.hie side of this polygon be EF. And since, as has been proved, the circle is equal to the rectangle DA.AH, and the polygon to the recte:angle DA.AL, the excess of the polygon above the circle is equal to the rectangle DA.HL; therefore the rectangle DA.HL is les tha~ the square of NP; and therefore, since DA is greater than NP, Hl. is less than NP, and twice HU[ less than twice NP, wherefore, much noreis twice HL':'less than NO. But i-L is the differenpe betweeqr hlalf the perimeter of the polygon whose side is EF, and half the cire Cuniference of the circle; therefore, twice iL is the difference be,tween the whole perimeter of the polygon and the whole circumfeir; erce of the circle (5. 5.). The difference, therefore, between tha perimenter of the polygon and the circumference of the circle is less ihan the given line NO. Con. 3. Hence also, a polygon may be inscribed in a circle, such that the excess of the circumference above the perimeter of the polyp gon may be less than any given line. This is proved like the pro, ceding-. PROP. VI. THIEO:R T7e aoreas of circlfs are to o:ne another in the duplicate ratio, r.o as thd s~,uares of their diamneters. laet ABD and GIII be two circles. of which tihe diameters are 4p and GL'; the circle IABD is to tihe circle G HL as the square: of- AD.to tbe wsuare of GL,!r 1T70 SUiPPLEMENT TOi THE ELEMENTS Let ABCDEF and G IKLMN be two equilateral polygons f~tihsame number of sides inscribed in the circles ABD, GHIL; and let Q / J / be such a space that the square of AD is to the square of GL as the circle ABD to the space Q. Because the polygons ABCDEF and OGHKI,MN are equilateral and of the same number of sides, they are similar (2. 1. Sup.), and their areas are as the squares of the diameters of the circles in which they are inscribed. Therefore AD2' GL2:: polygon AB(CDEF: polygon GHKLMN; but AD2 GL2: circle ABD: Q; and therefore, ABCDEF: GHKLM:: circle ABD. Q. Now, circle ABD 7 ABC DEF; therefore Q 7 GUKLMN (14..), that is, Q is greater than any polygon inscribed in the circle:GHL. in the same manner it is demonstrated, that Q is less than any poly. gon described about the circle GH-L; wherefore the space Q is equal to the circle GHL (2. Cor. 4. 1. Sup.). Now, by hypothesis. the circle ABD is to the space Q as the square of AD to the square of GL; therefore the circle ABI) is to the circle GHIL as the square of AD to the square of GL. Therefore, &c. Q. E. D. CoR. 1. Hence the circumferences f cir'cles are to one anotlher as their diameters. OF GEOMIETRY. BOOK i. 1i7 Let the straight line X be equal to half the circumference of the:circle ABD, and the straight line Y to half the circumference of the X:cile GHL: And because the rectangles AO.X and GP Y are equal to the circles ABD and GUL (5. 1. Sup.); therefore AO.X: (B:P.Y e AD2 GL2: AO2: GP2; and alternately, AO.X AOW" GP.Y. GP2; whence, because rectangles that have equal altitudes are as their bases (1. 6.), X: AO 1: GP, and again alternately, X: Y: AO: GP; wherefore, takina the doubles of each, the circuinfer~. ence ABD is to the circumference GHL as the diameter AL) to the diameter GL. COR. 2,- The circle that is described upon the side of a right angled triangle opposite to the right angle, is equal to the two circles described on the other two sides. For the-circle described upon SR is to the circle described upon RT as the square of' SR to the square of LRT; and the circle described upon TS is to the circle described upon Rr as the square of ST to the square of RT. Wherefore, the circles described on SR and on ST are to the circle described on RT as the squares of SR and of sfr to the square of RT (24. 5.,). But the squarets of RS and of S'F are equal to the square of RT (47. 1.); therefore the circles described on RS and ST are equal to the circle described on RT. PROP. VII. THEOR. Equiangular parallelograms -are to one another as the products of the, nunmbers proportional to their sides. Let AC and DF be two equiangular parallelograms, and let M, N, P and Q be four numbers, such that AB: BC:: M: N; AB: D-E: M:: P; and AB: EF:: M, and therefore ex mequali, BC: EF: N::Q. The parallelogram AC is to the parallelogram DF as MN to PQ. Let NP be the product of N into P, and the ratio of MN to PQ will be compounded of the ratios (def. 10. 5.) of MN to NP, and of.P to PQ.:Bt the ratio of MN to NP is the same with that of M !A1 tjSUPPLE MENT TO'THE ELEMiENT - B -..... f A B D E *o P (15. 5.), because MN and NP are equimultiples of MP and F iaid for. the samhe reason, the ratio of NP'to PQ is the same with that of N to Q; therefore the ratio of IMN to PQ' is compounded of the ratios of M to P, and of N to Q. Now, the ratio of MI to P is the apjne withthat of the side AB to the side DE (by Hyp.); and the *atio of N to Q the same with that of the side BC to the- side EF. Theref'ore, the ratio of MN to PQ i comipounded of the ratios of AB to DE, and of BC to EF. And the ratio of the parallelogram AC to the parallelogram DF is compounded of the same ratios (23. 6.) iherefore, the: par'llelogram AC is to the parallelogram DF as MNi the' product 6f ihe fiumbers M and N, to PQ, the product of the numrbers P and Q. Therefore, &c. Q. E. D..Coi. i. Hence, if GHI be to KL as the number M to the number F; thie sq'uare desciibed on GH evill be to the square described on G i - H K KL as MM, the square of the nunlter M to NN, the square of the number N. ConR 2. If A, B, C, D, &c, are any lines, aid m, n, i, s, &c. riumin bers proportional to them; viz. A: B: rn: n, A: C:: m: r, A: D:: r n s, &c.; and if' the rectangle contained by any two of the lines be equal to the square of a third line, the product of the nunmberi, proportional to the first two, will be equal to the square of the numbers proportional to the third; that is, if A.C=B2, nmXr —-Xii, or — na. For by this Prop. A.C: B2:: mrXr: n2; but A.C-=B2, therefore nXr=n2. Nearly in the same way it may be demonstrated, that "whatever is the relation between the rectangles contained by these lines, there is the same between the products of the numbers propor4 tional to them. So also conversely if mn and r be numbers proportional to the lines Aii nd C; if also A.C=B2 and if a number a be found such,'that rn -mr, then A: B::n:n. or let A: B::t:' q, then since, mw,q,'r are proportional to -A, B, and C, -and A.C=B- B therefore, as has just been proved, q'1m X r; but a.=q X r, by hypothsis, -therefore - =q2, and n=g; wherefore A: B:. "CHOLIUM. ii ordei to have numbers proportional to any set of magnitudes of -he same kind, suppose one of them to be divided into any number, vi -E equal parts, and let H h, one of thooae paxtes z et H be found b OF LGEOMETRY. BOOlK i, tL tiraes in the magnitude B, r times in C s times in D, &c., then it is evident that the numbers m, n, r, s are proportional-to the magnitudes A, B) C and D. When therefore it is said in any of the following propositions, that a line as A = a number in, it is understood that A-nm X H, or that A is equal- to the given magnitude EH multipled by mn, and the same is understood of the other magnitudes, B, C, D, and their proportional numbers, H being the common measure of all the magnitudes. This common measure is omitted for the sake of brevity in the arithmetical expression; but is always implied, when a line, or other geometrical magnitude, is said to be equal to a number. Also, when there are fractions in the number to which the magnitude is called equali it is meant that the common measure H is farther subdivided into such parts as the numerical fraction indicates. Thus, if A=360.375, it is meant that there is a certain magnitude H, such that 375 A —-360XIH+00 X H, or that A is equal to 360 times H, together with 375 of the thousandth parts of H. And the same is true in all other cases,' where numbers are used to express the relations of geof/ttrical magnitudes. PROP. VIII. THEORh Tlhe perpendicular drawn from the centre of a circle on the chord of any arch is a mean proportional between half the radius and the line made up of the radius and the perpendicular drawn from the centre on the chord of double that arch: AnJd the chord of the arch is a mean profportional between the diameter and a line which is the difference- between the radius and the aforesaid perpendicular from the centrei Let ADB be a circle, of which the centre is C; DBE any arch,, and-DB the half of it; let the chords DE, DB be drawn; as also CF and CG at right angles to DE and DB: if CF be produced it will meet the circumference in B: let it meet it again in A, and let AC be -biseotod in H; CG is a mean proportional between Al and AF; arnd,, [ /A]) 1El ad4 SUPPLEMENT'ATO'tHE ELEIMVENTS BD a mean- proportional between AB, and BF, the excess of the ran dius above CF. Join AD; and because ADB is a right angle, being an angle in a semicircle; and because CGB is also a right angle, the triangles ABD, CBG are equiangular, and, AB: AD:: BC: CG (4. 6.), or alternately. AB: BC:AD:: CG; and therefore, because AB is double of BC, AD is double of CG, and the square of AD therefore equal to four tihmes the square of CG. But, because AD B is a right angled tria ngle, and DE a perpendicular on AB, AD is a mean proporti:nal between AB and AF (8. 6.)-, and AID2=AB.AF (17. 6.), or since AB is=-4AH, AD2=4AH. AF. Therefore also, -because 4CG"=AD2, 4CG-2=4AL.HAF, and CG2-' AH.AF; wherefore CG is a mean proportional between /IH and AF9 that is, between half the radius and the line made up of the radius, and the perpendicular on the chord of twice the arch BD. Again, it is evident, that BD is a mean proportional between AB and BF (8. 6.), that is, between the diameter and the excess of the radius above the perpendicular, on the chord of twice the arch DIIB Therefore, &c. Q. E. D. PROP. IX. THEOR.* The circumference of a circle exceeds three times the diameter, by a line'& less than ten of the parts, of which the diameter contains seventy, but greater than ten of the parts whereof the diameter contains seventy-one. Let ABD be a circle, of which the centre is C, and the diameter AB; the circumference is greater than three times AB, by a line less 10 1 10 than 7, or -, of AC, but greater than;- of AC. 70' 7' "1 In the circle ABD apply the straight line BD equal to the radir. / I* n this proposition, the character -.. placed after a numnber, signifies that something is to be added to it; and the character -.on tbheother hand, signifies that something is to bh taken away froin it. OF GEOMlkET.RY. BOOK I. l'7 BC: Draw DF perpendicular to BC, and let it meet the circumfer. ence again in E; draw also CG perpendicular to BD: produce BC to A, bisect AC in H, and join CD. It is evident, that the arches BD, BE are each of them one-sixth of the circumference (Cor. 15. 4.), and that therefore the arch DBE is one third of the circumference. Wherefore,'the line (8. 1. Sup.) CG is a mean proportional between All, half the radius, and the line AF. Now because the sides BD, DC, of the triangle BDC are equal the angles DCF, DBF are also.equal; and the angles DFC, DFi being equal, and the side DF common to the triangles DBF, DCF, the base.BF is equal to the base CF, and BC is bisected in F. Therefore, if AC or BC=1000, AH"-500, CF=500, AF=1500, and CG being a mean proportional between AH and AF, CG2=(17. 6.) AH.AF=-500 X 1500-=750000; wherefore CG-=866.0254+, be* cause (866.0254) is less than 750000. Hence also, AC+-CG-1866. 0254+. Now, as CG is the perpendicular drawn from the centre C, on the chord of one-sixth of the circumference, if P = the perpendicular from C on the chord of one-twelfth of the circumference, P will be a mean proportional between AllH (8. 1. Sup.) and AC+ CG, and P2a AH (AC + CG)=500 X(1866.0254+)933012.7 +-. Therefore, P=965.9258-, because (965.9258)2 isless than 933012.7. Hence also, AC+P-19f65.9258 —. Again, if Q = the perpendicular drawn from C on the chord of one:twenty-fourth of the circumference, Q will be a mean proportional between AH and AC~-P, and Q= —AH (AC+P)=500( 1965.9258+) =982962.9 +-; and therefore Q-991.4449+, because (991.14449)2 is less than 982962,9. Therefore also AC+Q-1991.4449-. In like manner, if S be the perpendicular from C on the chord of one forty-eighth of the circumference, S2 = AH (AC + Q) = 500 (1991.4449+)-=995722.45 -+; and S-997.8589-, because (997. 8589)2 is less than 995722.45. Hence also, AC-S —1997.8589+. Lastly, if T be the perpendicular from C on the chord of one nine~ ty-sixth of the circumference, T2= AH (AC+S)=500 (1997.8589 +) = 998929.45 +, and T - 999.46 45 E — Thus T, the perpendicular on the chord of one ninety-sixth of the circumference, is greater than 999.46458 of those parts of which the radius contains 1000. But by the last proposition, the chord of one ninety-sixth part of the circumference is a mean proportional between the diameter and the excess of the radius above S, the perpendicular from -the centre on the chord of one forty-eighth of the circumference. Therefore, the square of the chord of one ninety-sixth of the circumference = AB (AC - S) = 2000 X (2.1411 —.) - 4282.-.2; and therefore the chord itself =65.4386 -, because (65.4386)2 is greater than 4282.2. Now, the chord of one ninty-sixth of the circumfercnce, or the side of-an equilateral polvygon of ninety-six sides inscribhed in the circle, of 43 I/. SUPP'LE1:MENT TO THE ELEMLENTS being 65.4386 —, the perimeter of that polygon will be - (65,43186 _)96 - 6282.1056-. Let the perimeter of the circumscribed polygon of the same numrn ber of sides, be M, then (2 Cor. 2. 1. Sup.) T: AC:: 6282.1056 —:, that: is, (since T- 999.46458-, as already shewn), 999.46458 -: 1000 6282.1056 -: I; if then N be such, that 999.4645S-8 1000 e 62S2.1056-: N; ex tequo perturb. 999.46458+: 999.46458: N: M; and, since the first is greater than the second, the third is greater than the fourth, or N is fgreater than M. Now, if a fourth proportional be found to 999.46458, 1000 and 6282.1056 viz. 6285.461-, then, because, 999.46458: 1000: 6282.1056: 6285.461-, and as before, 999.46458: 1000:: 6282.1056-: N; therefore, 6282.1056 ~ 6282.1056-:6285.461 -N, and as the fir of these proportionals is greater than the second, the third, viz. 62854 1 7 461 —-s greater than N, the fourth, But N was proved to be greater than M; much more, therefore, is 6285.461 greater than M, the perimeter of a polygon of ninety-six sides circumscribed about the circle-; that is, the perimeter of that polygon is less than 6285.461; now, the circumference of the circle is less than the perimeter of the polygon; much more, therefore, is it less than 6285.461; wherefbre the circumference of a circle is less than 6285.461 of those part of which the radius contains 1000. The circumference, therefore, has to the diameter a less ratio (8. 5.) than 6285.461 has to 2000, or than 3i42.7305 has to 1000: but the ratio of 22 to 7 is greater than the ratio of 3142.7305 to 1000, therefore the circumference has a less ratio to the diameter than 22 has to 7, qr the circumifere e'W!ess than 22 of the parts of w'hich the diameter contains 7. — F GEOM~ETRY. BOO-K 1. 177 It remains to demonstrate, that the part by which the circumference exceeds the diameter is greater than 1 of the diameter. It was before shewn, that CG2= 750000; wherefore CG = 868. 02545-, because (866.02545)2 is greater than 750000; therefore AC4 CG=1366.02545 —. Now, P being, as before, the perpendicular from the centre on the chord of one twelfth of the circumference, P2 = AH (AC+CG) =500X (1866.02545) —- 933012.73-; and P=965.92585-, because (965.92585)2 is greater than 633012.73. Hence also, AC+P = 1965.92585-. Next, as Q=the perpendicular drawn from the centre on the chord of one twenty-fourth of the circumference, Q2 - All (AC+P)500 X (1965.92585 -)982962.93 —; and Q = 991.44495 —, because (991.44495)2 is greater than 982962.93. Hence also, AC-+ 1 991.44495-. In like manner, as S is the perpendicular from C on the chord of one forty-eighth of the circumference, S2-Al (AC+Q) =500(1~)9 1 o 44495 —) 995722.475-, and S = (997.85895- ) because (997. 85895)2 is greater than 995722.475. But the square of the chord of the ninety-sixth part of the circumference = AB (AC- S) = 2000 (2.14105+) - 4282.1+, and the chord itself - 65.4377 - because (65.4377)2 is less than 4282. f Now the chord of one ninety-sixth part of the circumference being -65.4377+, the perimeter of a polygon of ninety-six sides inseribed.in the circle= (65.4377+) 9t=66e82.019+ — But the circumference of the circle is greater than the perimeter of the inscribed polygon; therefore the circumference is greater than 6282.019, of those parts of which the radius contains 1000; or than 3141.009 of the parts of which the radius contains 500, or the diameter contains 1000. Now, 3141.009 has to 1000 a greater ratio than 3+ 71 to 1; therefore the circumference of the circle has a greater ratio to the diameter than 3+ 7 has to I; that is, the excess of the circumference above three times the diameter is greater than ten of those parts of which the diameter contains 71; and it has already been shewn to be less than ten of those of which the diameter contains 70. Therefore, &c. QE. D. Con. 1. Hence the diameter of a circle being given, the circumrfr. ence may be found nearly, by making as 7 to 22, so the given diametqr to a fourth proportional, which will be greater than the circumfier 10 once. And if as 1 to 3+ 1, or as 71 to 223, so the given diameter to a fonrt proportional, this will be nearly equal to the circumference. butf will be less than ito 1-?8 SUPPLEMENT TO THE ELEMEiNTS I 10 I CoR. 2. Because the difference between - and- is - theres, 7 71 497 fore the lines found by these proportions differ by - of the diameter. Therefore the difference of either of them from the circumnference must be less than the 497th part of the diameter. CoR. 3. As 7 to 22, so the square of the radius to the area of the circle nearly. For it has been shewn, that (I. Cot. 5. 1. Sup.) the diameter of a circle is to its circumference as the square of the radius to the area of the circle; bhut the diameter is to the circutnference nearly as 7 to 22, therefore the square of the radius is to the area of the circle nearly in that same ratio. SC HOLIUM. It is evident that the method employed in this proposition, for finds. ing the limits of the ratio of the circumference of the diameter, may be-carried to a greater degree of exactness, by finding the perimeter of an inscribed and of a circumscribed polygon of a greater number of sides than 96. The manner in which the perimeters of such po~ iygons approach nearer to one another, as the number of their sides increases, may be seen from the following rTable, which is constructo ed oan the principles explained in the foregoing Proposition, and in which the~ radius is supposed- o o NO. of Sides Perimeter of' the Perimeter of the of the Poly- inscribed Poly- ~ circumscribed gon. gon. Polygon. ___ _6...000000 6.822033- i 1 2d 6.211 ~657+ 60.430781L- -094 6.265257+- 6.31932048 6.278700 i.2921q 7396 6.282063+i 6.285430-. 192 6.2829044+ 6.2 8374^7- 384 6.283115+ 6.283327 —.. 768 6.283167+ 6.28322211 536 6.283180+4- 6.283195- - 3072 6.283184+ 6.283188-.6144 6.283185 — 6.283186The part that is wanting in the numbers of the second column 9 tW make up the entire perimeter of any of the inscribed polygons, is less than unit in the sixth decimal place; and in like manner, the part by which the numbers in the last column exceed the perimeter of any of the eircumscribed polygonsA is lens' a.. nit n the sirah.~~~~~~~~~~~ta,.: h isin aes OF G-EOMETR~ Y BOOK 1.E?'c3 decimal place, that is than l of the radius. Also, as the numn'1000000 bers in the second column are less than the perimeters of the in. scribed polygons, they are each of them less than the circumference, of the circle; and for the same reason, each of those in the third column is greater than the circumference. But when the arch of 6 of the circumference is bisected ten times, the number of sides inr the polygon is 6144, and the numbers in the Table differ from one another only by i part of the radius and therefore the- peri. a000000 meters of the polygons-diffWer by less than that quantity; and conseu quently the circumference of the circle, which is greater than the least, and less than the greatest of these numbers, is determined with. in less than the millioneth part of the radius. Hence also, if R be the radius of any circle, the circumference is greater than R X6.283185, or than 2R X 3.141592, but less than 2R X 3. 141593; and these numbers differ from one another only by a nillioneth part of the radius. So also R2+-3.141592 is less, and RB X 3.141593 greater than the area of the circle; and these numbers differ from one another only by a millioneth part of the square of the radius. In this way, also, the circumference and the area of the circle may be found still nearer to the truth; but neither by this, nor by any other method yet known to geometers, can they be exactly determined, though the errors of both may be reduced to a less quantity than any that can be assigned. ELEMV1ENTS GEOME T R Y SUPPLEMENT. BOOK II. 01? TIE INTERSECTION OF PLANE;S.'DEFINITIONS. A sTRAIGHT line is perpendicular or at right angles to a plane, wherti it makes right angles with every straight line which it meets in that plane. II. A plane is perpendicular to a plane, wvhen the straight lines drawn in one of the planes perpendicular to the common section of the two. planes, are perpendicular to the other plane. The inclination of a straight line to a plane is the acute angle contain. ed by that straight line, and another drawn from the point in whichi the first line meets the plane, to the point in which a perpendicular to the plane, drawn fromi any point of the first line, meets the same plane, IV. The angle made by two planes whicli cut one another, is the angle contained by two straight lines drawn from any, the same point in the line of their common section, at right angles to that line, the one, in the one plane, and the other, in the other. Of the two adjacent angles made by two lines drawn in this nmanner, that which.s acute is also called the inclination of the planes to one another. SUPPLEMENT 0 TO THE ELEENTS, &c, iA' V. Two planes are said to have the same, or a like inclination to one another, which two other planes have, when the angles of inclination above defined are equal to one another. ~I. A straight line is said to, be parallel to a pIlne, when it does not meet the plane, though produced ever so far. VII. Planes are said to be parallelto one another, which do not meet, though produced ever so far. VIII. A solid angle is an angle made by the meeting of more than two plane angles, which are not in the same plane in one point. PROP. I. THIEOR. One part of a straight line cannot be in a plane and another part about it, If it be possible, let AB, part of the straight line ABC, be in the plane, and the part BC above it: and since the straight line AB is in the plane, it can be produced in that plane (2. Post. 1.); let it be produced to D: Then ABC and ABD are two straight lines, and they have the common segment A B AB, which is'impossible (Cor. def. 3. 1.). Therefore ABC is not a straight line. Wherefore one part, &c. Q. E. D. PROP. II. THEOR. lAny three straightt lines which meet one another, not in the same point, are in one plane. Let the three straight lines AB, CD, CB meet one another in the points B, C and E; AB, CD, CB are in one plane. AD Let any plane pass through the straight line EB, and let the plane be turned about EB, produced, if necessary, until it pass through the point C: Then, because the points E, C are in this plane, the straight line EC is in it (def. 5. 1.): for the same reason, the straight line BC is in the same; and, by the hypothesis, EB is in it; therefore t th Ithree straight lines EC, CB, BE are in one plane: but the whole of the lines DO, 8 SUPPLEI1AIENT TO THE ELEMENTS AB, and BC produced, are in the same plane with the parts of theno EC, EB, BC (1. 2. Sup.). Therefore AB, CD, CB, are all in one plane. Wherefore, &c. Q. E. D. Con. It is manifest, that any two straight lines which cut one another are in one plane: Also, that any three points whatever are in one plane. PROP. III. THEOR. If twb planes cut one another, their common section is a straight linej Let two planes AB, BC cut one another, and let B and D be two points in the line of their common section. From B to.D draw the straight line BD; and because the points B and D are in the ilane AB, the straight'line BD is in that plane (def. 5. 1.): for the same reason it is in the plane CB; the straight line BD is there- A fore common to the planes AB and BC, or it is the common section of these planes. Therefore, &c. Q. E. D. PROP. IV. THEOR. If a straight line stand at right angles to each.of two straight lines zia the point of their intersection, it will also be at right angles to the plane in which these lines are. Let the straight line AB stand at right angles to each of the straight lines EF, CD in A, the point of their intersection: AB is also at right angles to the plane passing through EF, CD. Through A draw any line AG in the plane in which are EF and CD; let G be any point in that line; draw GHI parallel to AD; and make HF=HA, join FG; and when produced let it meet CA in D; join BD, BG, BF. Because Gil is parallel to AD, and FH=HA: C \ therefore FG=GD, so that the line DF is bisected in G. And because BAD is a right angle, BD2=AB2+AD-(47. 1.); and for the same reason, BF2 = AB2+ A AF2, therefore BD2-+ BF2 -- 2AB +AD2+AF"; and because DF is bisected in G (A. 2.), AD2- + AF2=-2AG-2+ E 2GF', therefore BD- + BF -= 2AB2 + 2AG 2+ 2GF2. But BD- + BF2 (. 2.) 2IBGS + 2GF2' therefore 2BG' + W24FW 2AB' + OF GEOMETRY. BOOK Ilo 180 4AG2 + 2GF2; and taken 2GF2 from both, 2BG2 - 2AB- + 2AG", or BG2=AB2+AG2; whence BAG (48. 1.) is a right angle, Now AG is aniy straight line drawn in the plane of the lines AD, AF; and when a straight line is at right angles to any straight line which it meets with in a plane, it is at right angles to the plane itself (def. 1. 2. Sup.). AB is therefore at right angles to the plane of the lines AF, AD. Therefore, &G.. Q. E. D. PROP. V. TIHEOR. f three straight lines meet all in one point, and a straight line stand at right angles to each of them in that poiat these three straight lines are in one and the same plane. Let the straight line AB stand at right angles to each of the straight lines BC, BD, BE, in B, the point where they meet; BC, BD, BE are in one and the same plane. If not, let BD and BE, if possible, be in one plane, and BC be above it; and let a plane pass through AB, BC, the common section of which with the plane, in which BD and BE are, shall be a straight (3. 2. Sup.) line; let this be BF: therefore the three straight lines AB, BC, BF are all in one plane, viz. that which passes through AD, BC; and because AB stands at right angles to each of the straight lines BD, BE, it is also at right angles (4. 2. Sup.) to the plane passing through them; and therefore makes right angles with every straight line meeting it in Al that plane; but BF which is in that plane meets it; therefore the angle ABF is a right angle; but the angle ABC. by the hypothesis is also a right angle; therefore the an-. gle ABF is equal tothe angle ABC, and they are both in the same plane, which is impossible: therefore the straightline BC is not above the plane in which are BD and BE: J.3 Wherefore the three straight lines BC, BD, BE are in one and the same plane. Therefore, if three straight lines, &c. Q. E. D. PROP. VI. THEOR. Two straight lines which are at right angles to the same plane, are parallel to one another. Lot the straight lines AB, CD be at right angles to the same plane BDE; AB is parallel to CD. h 1 384 SUPPLEMIENT TO THE ELEMENTS Let them-meet the plane in the points B, D. Draw DE at right angles to DB, in the plane BDE, and let E be any point in it; Join AE, AD, EB. Because ABE is a right angle, AB2 + BE2 = (47. 1.) AE2, and because BDE is a right angle, BE -- BDP-+DE2; therefore AB- + B0D + DE =2 AE2; now, AB2 B D-92=AD, because ABD is a right angle, therefore AD2 +DE2=AE2, and ADE is therefore a (48. 1.) right angle. Thero-.3 D fore ED is perpendicular to the three lines BD9 DA, DC, whence these lines are in one plane (5. 2. Sup.). But AB is in the plane in which are BD, DA, because any three straight lines, which meet one another, are in one plane (2. 2. Sup.): Therefore AB, BD, DC are in one plane; and each of the angles ABD, BDC is a-right angle; therefore AB is parallel (28. 1.) to CD. Wherefore, if two straight lines, &c. Q. E. D. PROP. VII. THEOR. if two straight lines be parallel, and one of them at right angles to a plane; the other is also at right angles to the same plane. Let AB, CD be two parallel straight lines, and let one of them AB be at right angles to a plane; the A C other CD is at right angles to the i same plane. For, if Cl) be not perpendicular to the plane to which AB is perpendicular, let DG be perpendicu- i har to it. Then (6. 2. Sup.) DG is I parallel to AB: DG and DU there-.-_TJ__ Iore are both parallel to AB, and are drawn through the same point - h D, which is impossible (11. Ax. i.) \ Therefore, &c. Q. E. D. PROP. VIII. THEOlR. Two straight lines whiich are eachn of them parallel to the same straigha inie, though not both in the same plane zwith it, are parallel to one another. Let AB, CD be each of them parallel to EF, and not in the same plane with it; AB shall be parallel to CD. In EF take any point G, from which draw, in the plane passing through EF, AB, the straight line GHI at right angles to EF; and if! the plane passing through EF, CD, draw GK at right angles to the OF GEOMETRY. BOOK IH. 685 same EF. And because EF is perpendicular both to-GH and GK, it is perpendicular (4. 2. Sup.) to the plane HGIK passing through them: and EF is parallel.to- AB; therefore AB is at right anglte (7. 2. Sup.) to th" plane H(G;K. For the same reason, CD is A B likewise at right angles to the plane HGK. Therefore AB, CD are each of them at right angles to the plane IIGK. But 0if two straight lines are at right angles to the same plane, they / are parallel -(6. 2. Sup.) to one C D another. Therefore AB'is parallel td.CDR. Wherefore two straight lines, &c. Q.E.E D. PROP. IX. THEOIRP If two straight lines meeting one another.be parallel to two otbers tha,' meet one another, though not in the same plare.with the first two; the,frst two and the other two shall conlain eq.wal angles. Let the two straight lines Al, BC which meet one another he parallel'to the two straight lines DE, EF that meet one another, and are not in the same plane with AB, -BC. he angle ABC is equal to the angle DEF. Take BA, BC, ED; EFI all equal to one anothber; and join &1), CFp - BE,.AC, DF: Because BA i's equal,nd paralle'l to ED, there- tA. fore AD is (33. 1.) both equal and parallel to BE. — For"the satne reason, CF is:? eual and parallet to B E. Therefore AT) and CF are each:tthem'equal and parallel to BE. But straight li'nes that are parallel to the same straight lilne, -though not in the same plane with it, are parallel (8 2. Sup.) to one ano — ther. Therefore AD is parallel to CF; and it is equal to it, and AC, DF join them towards, DE the same parts; and therefore (33 1 ) AC is equal and parallel to IWF. And because AB, BC are equal to DE, EF, and the base AC to-the. base DF; the angle ABC is equal (8..) to the angle DEFo Therefore, if two straight lines, &c. Q. E. D. PROP. X. PROQB 2'o draw a straight.line perpendicular to a plane,frotm a. given point above it. Let A be the given point above the plane BH, it is required to draw rorn the point A a straight line perpendicular to the plane BWl s SUPPLE:MENT TO THE ELEMENTS In threplane draw any straight line BC, and from the point A draw. (12;_1.) At) pernwadiculat to -BC. if then AD be. also perpendicular to the plaite BH, the thing required is already done; but if it be not, from the point D draw A (11. 1.), in the plane BH, the _i straight line DE at right angles to BC; and from the point A draw AF perpendicular to DE; and F through' draw (31. 1.) GH parallel to BC: and because BC is at right angles to ED, and DA, 3..Cis at right angles (4. 2. Sup.) to the plane passing through ED, DDAO Ard G.IH is parallel to BC; but if two straight lines be parallei9, bne of which is at right angles to a plane, the other shall be at right (7. 2. Sup.) angles tothe same plane; wherefore GH is at right angles to the plane through ED, I)A, and is perpendicular (def. 1. 2. Sup.) to every straight line meeting it in that plane. But AF, which is in the plane through ED, DA, meets it: Therefore GIH is perpendicuJar to AF, and consequently AF is perpendicular to GH; and AF is also. perpendicular to DE; Therefore AF is perpendicular to each of the' straight lines G'I-, DE. But if a straight line stands at right- an gles to each of two straight lines in the point of their intersection, it is also at right angles to the.-plane passing through them (4. bz. Sup.), And the plane passing through ED, lGH is the plane tBH; therefore AF is perpendicular to the plane BH; so that, from the given point A, above the plane BH, the straight line AF is drawn perpendicular to that plane. Which was to be done. Con. If it be required from a point C in a plane to erect a perpendicular to that plane, take a point A above the plane, and draw AF perpendicular to the plane; then, if from C a line be drawn parallel to AF, it will be.the perpendicular required; for being parallel to AF it will be perpendicular to the same plane to; which AF is perpendicular (7. 2. Sup.). PROP. XL. THEOR. Fromn the same point in a plane, there cannot be two straight lines at ~'ight angles to the plane, upon the same side of it:.land there can be but one perpendicular to a plane from a point above it. For, if it be possible, let the two- straight lines AC, AB be at right angles to a given plane from the same point A in the plane, and upon the same side of it; and let a plane pass through BA, AC; the comr mon section of this plane with the given plane is a straight (3. 2. Sup.) line passing through A: Let IDAE be their common section: Therefore the straight lines AB, AC, 1DAE are in one plane: And because.A. is at right angles to the given plane, it makes right angles with OF GEOviE, TRY. BOOK IL 1: every straight line meeting it in that plane. But DAE, which is in that plane, meets B C CA: therefore CAE is a right angle. For the same reason BAE is a right angle. Wherefore: the angle CAE is equal to the angle BAE; and they are in one plane, which is impossible. Also, from a point above a plane, there can be but one per pendikcular to that plane; for if there could be two, they would be parallel (6. 2.. Sup.)to one another, which is absurd. Therefore, from the same point, &c. Q. E. D, PROP. XII. THEOR. Planes to wohich the soame straight line is perpendicular, are parallel to one another, et:the straight line AB be perpendicular to each of the planes CD, EF: -these planes are parallel to one another. If not, they must meet one another when produced, and their eoram mon section must be a straight line GH, in,lich take any point K, and join AK, BK: Then, because AB is perpendicular to the plane EF, it is perpendicular (def. i. V. Sup.) to the straight line BRK which is in.that plane, c and therefore ABK is a right angle. For the same reason, BAKE is a right angle; where fore the two angles ABK, BAK of the trian- A_ gle ABK are equal to two right angles, which is impossible (17. 1.): Therefore the planes CD, EF, though produced, do not meet one another; that is, they are parallel (de. 7. 2.,ap.).'Therefore planes, &c. Q. E. D. PROP. XIII. THEOR. If two straight lines meeting one another, be parallel to two straight lines'which'also meet one another, but are not in ti.e same plane with the first twao: the plane which passes through the first two is parallel to the plane passing through the others. Let AB, BC, two straight lines meeting one another, be parallbl to DE, EF that meet one another, but are not in the same platie:with AB, BC: The planes through AB, BC, and- DYE, EF shall not rmee't, though produced. From the point B draw BG perpendicular (10.?. Sup.) to the plane which passes through DE, EF, and let it meet- that plane in G; iand through G draw GH parallel to ED (31. 1.)I and GK parallel to EF: f SUPL EIENT TO THE ELEE L E -T And because BG is perpenditcular to the plane through DTE, EF, It mnust make right angles with every straight line meeting it in that plane (1. def. 2. Sup.). But the straight lines _. G-H, G(. in that plane meet it: Therefore each of the K angles BGH, BGAK is a right anele: And because BA iS -parallel (8. 2. Stlp.) to Gil (fr each (of themn is pairal- xD I } lei to D)E), the aiglhes fA,. BGH are together eqtlal (?9. 1 ) to two rigtt angles: Antd BGH is a right angle; therefore also GB A is a right anale~. alnd Gli perp; ndicular to BA: For the sane leason, GB is peripedicular to BC: Sirice, therefore, the straight line GB stands at right angle s to the two straight lines BA, BC, that cut one another in B; GB is perpenrdicular (4. -. Sup.) to the plane ihrough BA, BC: And it is perpendi'cular to thlie plane through DE, E}F; therefsre BG is perpendicular to each of the planes through AB, BC, and DE, EF: But plahes to which the saime straight line is perpendicular, are -parallel (I! 2. Sup.) to one another: Therefore the plane through AB, BC, is parallel to the plane through DE, EVk Wherefore, if two straight lines, &c. Q. E. D. COR. It follows from this demonstration, that if a straight line meet two parallel planes, and be perpendicular to one of them, it must bb' Poerpendicular to the other also,; PROP. XIV.,THUEOIR If two parallel planes be cut by another plane, their comnmon sectiomns w'tth it are parallels. Let the parallel planes AB, CD be cut by the plane EFHG, and let -their common sections with it be EF, GH; EF is paralb lel to GH. - For the straight lines E1 \ and GH are in the same plane, viz EFHIG which cuts the planes AB and C D; and they do not meet though produced; forthe planes in which they are.do not meet; there-,'fore EFand GIH are parallel (4efr 3o. 1.. Q. E. Do 4 F GEO ME.IY. BOOK: OO i Lt;. PROP. XV. THEOR. If two parallel planes be cut by - a, third plane, they have the same incli. nation to that plane. Let AB and CD be twvo parallel planes, and EH a third plane cutting them: The planes AB and D are equally inclined to EH. Let the straight lines EF and Gli be -the common section of the plane E H with the two planes AS and (CD; and from K, any point in EF, draw in the plane EH the stri ight line KM at right angles. to E F, and let it meet GA] in L; draw also KN at right angles to VFi. in the plane NB: and through the strai -ht lines KM, KN, let a plane be made to pass, cutting the plane C D in the line LO. And because EF and GH are the common sections of the plane -EH with the two parallel planes AB and CD, EF is parallel to GH (14. 2. Sup.). But EF is at right angles to the plane that passes through KN and KMl (4. 2. Sup.), because it is aft right angles to the lines KM1 and KN O therefore GH is also at right angles to the same plane (7. 2. Sup.', and it is therefore at right angles to the lines LM, LO which it meets in that plane. Therefore, since L.M and LO are at right angles to LG, the common section of the two planes ( D and EH,`the angle OLM is: the inclination of the pliine CD to the plane ElH (4. def. 2. Sup. ) For the same reason the arngle MKN is the incrlination of the plane AB to the plane EH u. But because KN and LO are parallel, being the common sections of the-parallel planes AB and C D with a third plane, the interior angle NIKM is equal to the exterior angle OLM (29. 1.); that is, the inclination of the plane AB to the plane EH, is equal to the inclination of the plane CD to the same plane EHi, horab fore, &~, q. E, DO SUPPLEMENT TO TIHE ELE MENTS' PROP. XVI. THEOR. If two straight lines be cut by parallel planes, they must be cut in the same ratio. Let thei straight lines AB, CD be cut by the parallel planes GH, KL, MN, in the points A, E, B; C, F, D: As AE is to EB, so is CF to FD. Join AC, BD, AD, and let AD meet the plane KEL in the point X; and join EX, XF: Because the two parallel planes KL, MN are cut by the M plane EBDX, the common sections / iEX, BD, are parallel (14. 2. Sup.). i i For the same reason, because the _ Itwo parallel planes GH, KL are cut -by the plane AXFC, the common sections AC, XF are parallel: And because EX is parallel to BD, a side of the triangle ABD, as AE to EB,'so is(2. 6.) AX to X0.: Again, be-..1cause XF is parallel A, a side ot the triangle ADC,as AX to XD, so is CF to FD: and it was proved that AX is to XD. as AE to EB: Therefore C- - w (11. 5.), as AE to EB,so is CF to FD. Wherefore, if two straight lines, &c /3 - Q. E.D. PROP. XVII. THEOR. If a straight line be at right angles to a plane, every plane which passes through th:t line is at ri.gh.t angles to the first mentioned plane. Let the straight line AB be at right angles to a plane CK; every plane which passes through AB is at right angles to the plane CK. Let any plane D; pass through AB, and let CE be the common section of the planes DF, CK; take any point F in CE, from which'draw F'G in the plane D F at right angles to CE-: And because AlB is perpenIdicular to-the plane CK, therefore- A it is als0 perpendicular to every straight line meeting it in that plane!). def. 2. Sup.); and consenuafntly it is perperdicular to CE: _ Wherefore ABF i's a right angle; Iqut GFB is likewise a right angle; therefore AB is parallel (28. 1.) to FG. And AB is at right angles to the plane CR: therefore FG is 1 3: OF GE OMETRY. BOOK II. 19 also at right angles to the same plane (7. 2. Sup.) But one plane is at right angles to another plane when the straight lines drawn in one of the planes, at right angles to their common section, are also at right angles to the other plane (def. 2. 2.); and any.straight line FG in the plane DE,, which is at right angles to CE, the common section of the planes, has been proved to be perpendicular to the other plane CK; therefore the plane DE is at right angles to the plane CK. In like manner, it may be proved that all the planes which pass through AC are at right angles to the plane CK. Therefore, if a straight line, &c. QE. D. PROP. XVIII. THEOR. if two planesc-utting one another be each of them perpendicular to a third plane, their common section is perpendicular to the same plane. Let:the two planes AB, BC -be each of them perpendicular to a third:plane, and B-D be the common section of the first two; -BD is perpen. dicular to the plane ADC. From D in the plane ADC, draw DE perpendicular to: AD, and DF to:DC.. Because DE is perpendicular to AD, the common section of the planes AB and ADC; and because the plae~ AB is: at right angles to ADC, DE is - atrright angles- to the plane, AB (defY 2. 2.'\ Sup.), and therefore alsoito the straight line BD in that plane (def. 1. 2. Sup.). For the same reason, DF is at right angles to DB. Since BD is therefore at right angles to both the lines DE and DF, it is at right angles to the plane in which DE and DF are, that is, tothe plane ADC (4. 2. Sup.). Wherefore, &c. Q.E.D. PROP. XIX. THEOR..Two straight lines not in the same plane being given in position, to draw a straight line.perpendicuilar to them both. Let AB and CD be the given lines, which are not in the same plane; it is required to draw a straight line which shall be perpendicular both to AB and CD. In ABR take any point E, and through E draw EF parallel-to CD, and let EG be drawn perpendicular to the plane which passes through EB, BF I0, 2. Sup.). Through AB and EG let a plane pass, viz. GJ(K SUPPLE MENT TO tHE ELEMENTS: / im C D And let this plane meet CD in H; from H draw HK perpendicular be AB; and HIK is the line required. Through H, draw HG' paralIe to AB. Then, since HI and GE, which are in the same plane, are both at right angles to the straight line AB, they are parallel to one another. Andtibecause the lines HG, HD are parallel to the lines EB, EF, eachto each, the plane GHD is parallel to t!he plane ( 13. 2 Sup.) BEF; and therefore EG, which is perpendicular to the plane BEF, is perpendi.. cular'also to the plane (Cor.,3. 2. Sup.) GHD. Therefbre. HK, which is parallel to GrE, is also perpendicular toe the- plane GHD (7. 2. Sup.), and it is therefore perpendicular to HD (def. 1. 2. Sup.), which is in that plane, and it is also perpendicular to AB; therefore HIK is drawn perpendicular to the two given lines, AB and CD. Which was to be done. PROP. XX. THEOOR. -If a solid angle be contained by three plane angles, ony.two of these angles are greater than the third. Let the solid angle at A be contained by the three plane angles BACG, CAD' DAB. Any two of them are greater than the third. If the angles B XC, CAD, DAB be all equal, it is evident that any two of them are greater than the third. But if they are not, let BAC be that angle which is not less than either of the other two, and is greater than one of them, DAB; and at the pointfA in the straight line AB, make in the plane which passes through BA, AC, the angle B A E equal (23. 1.) to the angle DAB; and lmake AE equal to AD. arid through A E draw BEC cutting AB, AC in the points B, C, and join DB, DC. And because DA is equal to AE, and AB is common to the B E two triangles ABD, ABE, and also the angle,DAB equal to the angle E&B; therefore the base DB is equal OF GEOMEITRY, BOOK I H 19. Af. 1.) to the base BE. And because BD, DC are greater (20. 1.)than CB, and one of them BD has been proved equal to BE, a part of CB, therefore the other DC is greater than the remaining part EC. And because DA is equal to AE, and AC common, but the base DC greater than the-base EC; therefore the angle DAC is greater (25-. 1.) than the angle EAC; and, by the construction, the angle DAB is equal to the angle BAE; wherefore the angles DAB, DAC are together:great, or than BAE, EAC, that is, than the angle BAC. But BAC is. not less than either of the angles DAB, DAtU; therefore BAC, with either of-t-hem, is. greater than the other. ]Wtherefore, if a solid angle, &c5, Q. E.D. PROP. XXI. THEORe lThe plane aGngles which contain any solid angle are together less tha. four right angles. Let A bea solid angle contained by any number of plane angles BACCAD, DAE, EAF, FAB_; these together are less than four right any gles. Let the planes which contain the solid angle at A be cut by another plane, and let the section of them by that plane be the rectilineal figure BCDEF. And because the solid angle at B is contained by three plane angles CBA, ABF, FBC, of which any two are greater (20. 2. Sup.) than the third, the angles CBA, ABF are greater than the angle FBC: For the same reason, the two plane angles at each of the points C, 13, E, F, viz. the angles whichc are at the bases of the triangles Poaving the common vertex A,, aie greatgr than thle third angle.at the same point vwhich is one of the angles of thle figure C BCDEF: therefore all the angles at the bases of the triangles are together reat- -,r than all11 e angles of the figuwre: and because all the angles of tihe triangles are together equal to twice aa many right angles as there are triangles (32. 1 ); that is, as there are sides in the figure BCD EF; and because all the angles of the figure, Ltogether wyith four right angles, zaI likewise equal to twice as many right angles as there are sides in the figure (cor 1. f3. 1.); therefore all the angles of the triangles are equal to all the angles of the rectilineal figure, together M ith fouri righlt angles. But all the angles at the'bases of the triangles are gre ater than all the angles of the rectilineal, as has been proved. Wherefrel tlhe remaining angles of the triangles, viz. those at the vertex, which contain the solid angle at A, arm }ess than four right a'glcs. Thiereosre every solid angale, &c. Q. E, a bH'lr' \r'~~ "s,' 194 SUPPLEM1lENT TO THE ELEMENTS, &ce Otherwise. Let the sum of all the angles at the bases of the triangles- =S; the sum of all the angles of the rectilineal figure BCDEF —;; the sum of the plane angles at A=X, and let R = a right angle. Then, because S+X=twice (32. 1.) as many right angles as there are triangles, or as there are sides of the rectilineal figure BCDEF, and as -+4R is also equal to twice as many right angles as there are sides of the same figure; therefore S+X=2+4R. But because of the three plane angles which contain a solid angle, any two are great. er than the third, S 7:; and therefore X Z. 4R; that is, the sum of the plane angles which contain the solid angle at A is less than fouer right angles. Q. E. Do SCHOLIUM. It is evident, that when any of the angles of the figure BCDEF is exterior, like the angle at D, in h:ie annexed figure, the reasoning A finthe above proposition does not liild, because the solid angles at the base are not all contained by plane angles, of which two belong to the triangular planes, having their common vertex in A, and the third is an interior angle of the rectilineal figure, or base. There-. C fore, it cannot be concluded that S is necessarily greater than'. This proposition, therefore, is sub~ ject to a limitation, which is farther explained in the notes on this book. ELEMENTS OF GE ME T R Y. SUPPLEMENT. BOOK IlI. OF THE COMPARISON OF SOLIDS. DEFINITIONS. A SOLID is that which has length, breadth, and thickness. II. Similar solid figures are such as are contained by the same number of similar planes similarly -situated, and having like inclinations to one another. A pyramid is a solid figure contained by planes that are constituted be. twixt one plane and a point above it in which they meet. IV. A prism is a solid figure contained by plane figures, of which two that are opposite are equal, similar, and parallel to one another; and the others are parallelograms. V. A parallelopiped is a solid figure contained by six quadrilateral figures, whereof every opposite two are parallel. VI. cube is a solid figure-contained by six equal squares. 96 SUPPLEMENT TO THE ELE!IIENTiS VII. A sphere is a solid figure described by the revolution of a semicitcle about a diameter, which remains unmoved. VIII. The axis of a sphete is the fixed straight line about which the semis circle revolves. IX. The centre of a sphere is the same with that of the semicircle. X. t.he diameter of a sphere is any straight line which passes through the centre, and is terminated both ways by the superficies of the spheres XI. A cone is a solid figure described by the revolution of a right angled triangle about one of the sides containing the right angle, which side remains fixed. XII. The axis of a cone is the fixed straight line about which the triangle revolves. XIII. The base of a cone is the circle described by that side, containing the right angle, which revolves. XIV. A cylinder is a solid figure described by the revolution of a right angled parallelogram about one of its sides, which remains'fiXed. XV.'the axis of a cylinder is the fixed straight line about which the pan rallelogram revolves. X~I. The bases of a cyiinder are the circles described by the two reo1lvw ing opposite sides of the parallelogram. XVIL, Simiiar cones and cylinders are those which have their axes, and the diameters of their bases proportionals. OF GEOMtETRY. BOok 11i..' PROP. I. THEOR. f taco solids be contained by the same numbter of equal anod similar planes similarly situated, and if the inclination of any two c,.ntiguous planes in the one solid be the same with the inclination of the two equal, and similarly situated planes in the other, the solids themselves are equal and similar. Let AG and KQ be two solids contained by the same number of equal and similar planes, sinmilarly situated so that the plane AC is similar and equal to the plane KM, the plane AF to the plane KP; BG to LQ, GD to QN, DE to NO, and FH to PR Let also the inclination of the plane AF to the plane A( be the same with that of the plane KP. to the plane KM, and so of the rest; the solid K(Q is equal and similar to the solid AG. Let the solid -Q be applied to the solid AG, so that the bases KM M 0E1;___ i and AC, which are equal and similar, may coincide (8. Ax. i.), the point N coinciding with the point D, K with A, L with B, and so on. And because the plane KM coincides with the plane AC, and, by hypothesis, the inclination of KR to KM is the same with the inclination of AH to AC, the plane KR will be upon the plane AH, and will coincide with it, because they are similar and equal (8. Ax. 1.), and because:their equal sides KN and AD coincide. And in the same manner it is shewn that the other planes of the solid KQ coincide with-the other planes of the solid AG, each with each: wherefore the solids KQ and AG do wholly coincide, and are equal and similar to one arother. Therefore, &c. Q. E. D. PROP. II. THEOR. If a solid be contained by six planes, two and two of,wLhich are parallel, the opposite planes are similar and equal parallelograms. Let the solid CDGII be contained by the parallel planes AC, Gl; 13G, CE; FB, AE: its opposite planes are similar and equal parallelograms. Because the two parallel planes B., CF, are cut by the plane AC 198 SUPPLE MENT TO THE ELEMENTS their common sections AB, CD are parallel (14. 2. Sup.). Again, because the two parallel planes BF, AE are cut by the plane AC, their common sections AD, BC are parallel (14. 2. Sup.): and AB is parallel to CD; therefore iAC is a parallelogram. In like manner, it may be proved that each of the figures i CE, FG, GB, BF, AE is a parallelogram; join All DF; and because AB is parallel A to DC, and BH to CF; the two straight lines AB, BH, which meet one another, are parallel to DC and CF, which meet one another; wherefore, though the first two are not in the same plane with the other two, they contain equal angles (9. 2. E Sup.); the angle A/BH is therefore equal to the angle DCF. And because AB, BR, are equal to DC, CF, and the angle ABH equalito the-angle DCF; therefore the base AH is-equal (4. 1.) to the base DF, and the triangle AB&H to the triangle DCF: For the same reason, the triangle AGH is equal to the triangle DEF: and therefore the parallelogram BG is equal and similar to the parallelogram CE. In the same manner, it may be proved, that the parallelogram AC is equal and similar to the parallelogram GF, and the parallelogram AE to 1BFE Therefore, if a solid, &c. Q. E. D. PROP. II!. THEOR. if a solid parallelopiped be cut by a plane parallel to two of its opposite planes, it will be divided into two solids, which will be to one another as the bases. Let the solid parallelopiped AB"lD be cut by the plane EV, which is parallel to the opposite planes AR, liD, and iivides the whole into the solids ABFV, EGCD; as the base AEFY to the base EHCF, so is the solid ABFV to the solid LEGCD. Produce AHI both ways, and take any number of straight lines HMiVLN, each equal to EH, and any number A K, KL each equal to EA, >and complete the parallelograms LO, KY, HQ, MS, and the solids LP: O; -~ -. C Q. S KR, H1fI, MT; then, because the straight lines LK, KA., AE are all OF GEOM —ETRY. BOOK ilt. 199 equal, and also the straight lines (KO, AY, EF which make equal an. gles with LK, KA, AE, the parallelograms LO, KY, AF are equal and aimilar (36. 1. & def. 1 6.): and likewise the parallelograms KX, KB, AG; as also (2. 3. Sup.) the parallelograms LZ, KP, AR, because they are opposite planes. For the same reason, the parallelograms EC, HQ, MS are equal (36. 1. & def. 1. 6.); and the parallelograms HG, HI, IN, as also (2. 3. Sup,) HD, MU, NT; therefore three planes of the solid LP, are equal and similar to three planes of the solid KR, as also to three planes of the solid AV: but the three planes opposite to these three are equal and similar to them (2. 3. Sup.) in the several solids; therefore the solids LP, KR, AV are contained by equal and similar planes. And because the planes LZ, KP, AR are parallel, and are cut by the plane XV, the inclination of LZ to XP is- equal to that of KP to PB; or of AR to BVY (15. 2. Sup.) and the;amni,~jsiue'of the —other contiguous planes, therefore the solids LP, KR, and. AY, are-equal to one another (1.-3. Sup.). For the same reason, the three sclids& ED-, ERU, MIT are equal to one another; thpere fore what multiple soever the base LF is of the base AF, the:same multiple is the solid LV of the solid AV; for -the same reason, whaff:. ever multiple the base NF is of the base 1HF, the same multiple is the solid NV of the solid ED: And if the base LF be equal to the base NF, the solid LV is equal (1. 3. Sup.) to the solid NV; and if the base LF be greater than the base NPF, the solid LY is greater than the solid NV: and ifless, less. Since then there are four magnitudes, viz, th twolbases AF, FH, and the two solids AV, ED, and of the base AF and solid AV, the base LF and solid LV are any equimultiples whatever; and of the base FL! and solid ED, the base FN and solid NV are any equimultiples whatever; and it has been proved, that if the base LF is greater than the base FN, the solid LY is greater than the solid NV; and if equal, equal; and if less, less: Therefore (def 5. 5.): as the base AF is to the base FEt, so. is the solid AV to 6ie selid ED. Wherefore, if a solid, &c. Q. E. D. Con. Because the parallelogram AF is to the parallelogram FH as YF togC (1. 6.), therefore the solid AV is to the solid ED as YF to PROP. IV. THEOR. Ita alid parallelopiped be cut by a plane passing througgh the diagoznals of tnwo of the opposite plattes, it will be cut into two equal prisms. Let AB be a solid parallelopiped, and DE, CF the (iagonals of the oppositerparallelogramrs Al-, GB, viz. those which are drawn betwixt the equal angles in eacih; and because CD, FE are each of them parallel to GA, though not in the saime plane with it, CD, FE are parallel (8. 2. Sup.); wherefore the diagonals CF, DEJ are in the plane SUPPLEAIE'NT TO. THE ELEMENTS in which the parallels are, and are themselves parallels (14. 2. Sup.),: the plane C B CDEF cuts the solid AB into two equal parts. Because the triungle CGF is equal (34. I.) to the triangle CBF, and the triangle G:DAE to DHE; and since the parallelogram CA is equal (2. 3. Sup.) and similar to the opposite one. BE; and the parallelogram GE D 1 l to -CH: -therefore the planes which contain the prisms CAE, CBE, are equal and similar, each to each; and they are also equally inclined to one another, because the planes A E AC, EB are parallel, as also AF and BD, and they are cut by the plane CE (15. 2. Sup.). Therefore the prism CAE is equal to the prismn CBE (1. 3. Sup.), and the solid AB is cut into two equal prisms by the plane CDEF. QE E. D. N. B. The insisting straight lines of a parallelopiped, mentioned an the following propositions, are the sides of the parallelograms betwixt the base and the plane parallel to it. PROP. V. THEOR, 5dlidpadrallelopipeds-upon the sarme base, and of the same altitude, the ina sisting straight lines of Thich are terminated i~n the same straight lines in the plane opposite to the base, are equal to one another. Let the solid parallelopipeds AlH, AK, be upon the same base AB, and of the same altitude, and let their insisting straight lines AF, AG, LN, LN, be terminated in the same straight line FN, and let the insisting lines CD, CE, BHI, BK be terminated in the same straightline DK; the solid Al is equal to the solid AK. Because CH, CK are paralleloagrams, CB is equal (34. 1.) to each of the opposite sides DII, EK-: wherefore DHI is equal to EK: add, or take away the conmon part HIE; then DE is equal to HK: Wherer fore also the triangle CDE is equal (38. 1.) to the triangle BHK: and the parallelogram DG is equal (36. i ) to the parallelogram HN. For the same reason, the triangle AFG isequal to the triangle LMN, and the parallelogram CF is equal (2. 3. Sup.) to the parallelogram BM, and CG to BIN; for they are opposite. Therefore the planes which I i[ EJ,.A. -OF GEOMETRY. BOOKS Ilt. i o' Wontain the prism DAG are sinmilar and equal to those which contain the prism HLN, each to each; and the contiguous planes are also equally inclined to one another (15. 2. Sup.), because that the parallel planes AD and LHI, as also AE and LK, gre cut by the same plane D N!: therefore the prisms DAG, J:LN are equal (1. 3. Sup.). If therefore the prism LNUI be.taken from the solid, of which the base is the parallelogram AB, and FD.KN the plane opposite to the base; and if from.this same solid there be taken the prism AGD, the remains ing solid, viz. the parallelopiped All is equal to the remaining parakl lelopiped AK. Therefore solid parallelopipeds, &c. Q. E. Do PROP. VI. THIEOR. Solid parallelopipeds upon the same,base, and of (he same altitude, the insisting straight lines of wehich are not terminated in the same straight lines in the plane opposite to the base, are equal to one another. Let the parallelopipeds CM, CN, be upon the same base AB, and of he.same altitude, but their insisting straight lines AF, AG, LM, LN, CD, CE, BH, BK, not terminated in the same straight lines; thesolids CM, CxN are equal to one another. Produce FD, MH, and NG, KE, and let them meet one another in the points O,P, Q, R; and join AO, LP, BQ, CR. Because the planes (def. 6. 3. Sup.), LBHM and ACDF are parallel, and because the/ planoe BHIM is that in which are the parallels LB, MHPQ (def 5. 5. Sup.), and in which also is the figure BLPQ; and because the plane ACDF is that in which are the parallels AC, FDOR, and in which also is the figures CAOR; therefore the figures BLPQ, CAOR, are in parallel planes. In like manner, because the planes ALN G and CBKE are parallel, and the plane ALNG is that in which are the parallels ALS.OPGN, and in which also is the figure ALPO; and the plane CBKE is that in which are the parallels CB, RQEK, and in which also is the C c figure CBQR; therefore the figures ALPO, CBQR are in parallel planes. But the planes ACBL, ORQP are also parallel; therefore the solid CP is a parallelopiped. Now the solid parallelopiped CM is equal (5. 2. Sup.) to the-solid parallelopiped CP; because they are upou tkhe same base, and their insisting straight lines AF, AO, CD, C-R; LM, LP, H-, BQ are termitated in the same straight lines FR, 3]fQ: and the solid CP is equa! (:. 2. Sup) to the solid CP; fobr they are upon,:he same base AGCBL, and their insisting straight lines AO, AG, LP, LN; CR, CE, BQ, BK are terminated in the same straight lines' 3ON, RK; ~Therefore the solid CM is equal to the solid CN. Wherefore solid parallelopipeds, &c. Q. E. D. PROP. VII. THEOR. Solid parallelopipeds wkich are upon equal bases, and of the saame alti. tude, are equal to one another. Let the solid parallelopipeds, AE, CF, be upon equal bases AB, CD, and be of the same altitude; the solid AE is equal to the solid CF. CaseI 1. Let the insisting straight lines be at right angles to the bases,AB, CD, and" let the bases be placed in the same plane, and so as that the sides CL, LB, be in a straight line; therefore the straight line LM, which is at right angles to the plane in which the bases are, in the point L, is commonr (. 2. Sup.) to the two solids AE, CF; let the other insisting lines of the solids be AG, 1lK, BE; DE', OP, CN: and first, let the angle ALB be equal to the angle -`LD; then AL, LD are in a straight line (14. 1.). Produce OD, HB, and let them meet in Q and complete the solid parallelopiped LR, the base of which is the parallelogramn LQ, and ofI which LM1 is one of its insisting straight lines: therefore, because the parallelogram AB is equal to CD, as the base AB is to the base LQ. sBo is (7. 5.) the base CL to the same LQ: and because the solid parallelopiped AR is cut by the plane LMEB. vwhich is parallel to the opposite planes AK, DR; as the base AB is to the P? ML P o! base LQ, so is (3. 3. Sup.) the solid AE to the solid LR: for the same reason because the solid parallelopiped CR is cut by the plane LAIFD, which is parallel to the opposite planes:CP, BR.; as the base CGD to OF GEOMETRY. BOOK III. 208 the base LQ; so is the solid CF to the solid LR; but as the base AB to the base LQ, so the base CD to the base LQ, as has been proved: therefore as the solid AE to the solid LR, so is the solid CF to the solid LR,; and therefore the solid AE is eqlmal (9. 5.) to the solid CF. BUt let the-solid parallelopipeds, SE, CF be upon equal bases SB, CD, and be of the same altitude, and let their insisting straight lines be at right angles to the bases; and place the bases SB, CD in the same plane, so that CL, LB be in a straight line; and let the angles SILB, CLD be unequal; the solid SE is also in this case equal to the- solid CF. Produce DL,'TS until they meet in A, and fror B draw BIparallel to X)A; and let FIB, OD produced meet in Q, and complete the solids AE, LR: therefore the solid AE, of which the base is the parallelogram LE, and AK the plane opposite to it, is equal (5. 3. Sup.) to the solid SE, of which the base is LE, and SX the plane opposite; fbr they are upon the same base LE, and of the same altitude, and their insisting straight lines, viz. LA, LS, BH, BT; MG, MU, EK, EX, are in the same straight lines AT, GX: and because the parallelogram AB is equal (35. 1.) to SB, for they are upon the same base LB, and between the same parallels LB. AT; and because the base SB is equal to the base CD; therefore the base AB is equal to the base CD; but the angle ALB is equal to the angle CLD: therefore, by the first case, the solid AE is equal to the solid CF; but the solid ARE is equal to the solid SE, as was demonstrated; therefore the solid SE is equal to the solid CF. Case 2. If the insisting straight lines AG, lK, BE, LM; CN, RS., 1'7CC- -v. i - DF, OP, be not at right angles to the bases AB, CD; in this case likewise thae solid AE is equal to the solid CF. Because solid paral. lelopipeds on the same base, and of the same altitude, are -equal (6. 3, Sup.), if two solid parallelopipeds be coustituted on the bases AB and CD of the saime altitude with the solids AE and CF, and with their insisting lines perpendicular to their bases, they will Be equal to the solids AE and CF; and, by the first case of this proposition, they will be equal to one another; wherefore, the solids AE and CF are also equal. Wherefore, solid parallelopipeds, &c. Q. E.. D, 04 SUPPLEMELNT T'.O THE' Mi-EMENTS PROP. ViIi. THEOR. Solid parallel opipeds W'hich hAave the same altitude, are to one another as their bases. Let AB, CD be solid parallelopipeds of the same altitude: they are to one another as their bases; that is, as the base AE to the base CF, so is the solid AB to the solid CD. To the straight line FG apply the parallelogram FH equal (Cor. 45. 1.) to AE, so that the angle FGH be equal to the angle LCG XB D-. and complete the solid parallelopiped GK upon the base FH, one of Whose insisting lines is FD, whereby the solids CD, GK must be of the same altitude. Therefore the solid AB is equal (7. 3. Sup.) to the solid GK, because they are upon equal bases AE, FH, and are of the same altitude: and because the solid parallelopiped CK is cut by the plane DG which is parallel to its opposite planes, the base HF is (3. 3. Sup.) to the base FC, as the solid HD to the solid DC: But the base HF is equal to the base AE, and the solid GK to the solid AkB: therefore, as the base AE to the base CF, so is the solid AB to the solid CD. Wherefore solid parallelopipeds, &c. Q.E. D. COR. 1. Fromn this it is manifest, that prisms tUpon triangular bases, and of the same altitude, are to one another as their bases. Let the prisms BNM, DPG, the bases of which are the triangles AEM, CFG, have the same altitude; complete the parallelograms AE, CF- and the solid parallelepipeds AB, CD, in the first of which let AN, and in the other let CP be one of the insisting lines. And because the solid parallelopipeds AB, CD have the same altitude, they are to one another as the base AE is to the base CF; wherefore the prisms, which are their halves (4. S. Sup.) are to one another, as the base AE to ihe base CF; that is, as the triangle AEM to the triangle CFG. CoR. 2. Also a prism and a parallelopiped, wlii-cli hlave the same altitude, aire to'one another as their bases; that is; the prism BNM is to the parallelopiped CD as the triangle AEM to the parallelogramn LG. For by the last Cor. the prism BNM is to the prism DPG as thi triangle AME to the triangle CGF, and therefore the prism BNM is to tWiCe the piism DPG as the triangle AME to twice the triangle COW OFl GEOIMETR-Y. BOOK-Ilf. 2I (4. 5.); that is, the prism BNM is to the pardllelopiped CD as the triangle AME to the parallelogram LG. PROP. IX. THEOR. Solid parallelopipeds arie Co one another in the ratio that is coapoubnd&e of the ratios of thlie areas of their bases, and' of their altitudes. Let AF and GO be two solid parallelopipeds, of which the bases are the parallelograms AC and GK, and the altitudes, the perpendiculars let fall on the planes of these bases from any point in the opposite planes EF and MO; the solid AF is to the solid GO in a ratio compounded of the patios of the base AC to the base GK, and of the perpendicular on AC, to the perpendicular on GK. Case 1. When the insisting lines are perpendicular to the bases AtC and GlK, or when the solids are upright. In GM, one of the insisting lines of the solid GO, take GQ equal to AE, one of the insisting lines of the solid AF, and through Q let a plane pass parallel to the plane GIE, meeting the other insisting lines iii:O of the solid GO in the points R, S and T. It is evident that GS is (u solid parallelopiped (def. 5. 3. Sup.), and that it has the same altitude with AF, viz. GQ or AE. Now the solid AF is to the solid GO in a ratio compounded of the ratios of the solid AF to the solid GS (def. 10. 5.), and of the solid GS to the solid GO; but the ratio of the solid AF:to the solid GS, is the same with that of the base AC to the base GK (8. 3. Sup.), because their altitudes AE and GQ are equal; and the ratio of the solid GS to the solid GO, is thesame with that of GQ to GM (3. 2. Sup.); therefore, the ratio which is compounded of the ratios of the solid AF to the-solid GS, and of the-solid GS to the solid GO, is the same with the ratio which is compounded of the ratios of the base AC to the base GK, and of the altitude AE to the altitude G1i (F. 5.). But the ratio of the solid AF to the solid GO, is that which is compounded of the ratios of AF to GS, and of GS to GO; thercfere 206 SUPPLEAiENT TO THE ELEMENTS the ratio of the solid AF to the solid GO is compounded of the ratios of, the base AC to the base GK, andi of the altitude AE to the altitude GMA-, Case 2. W~hen the insisting lines are not perpendicular to the bases. Let the parallelograms AC and GK be the bases as before, and let AE and GIM be the altitudes of two parallelopipeds Y and Z on these bases. Then, if the upright parallelopipeds AF and GO be constituted on the bases AC and GEK, with the altitudes AEf and GM, they will be equal to the parallelopipeds T and Z (7. 3. Sup.'. - Now, the solids AF and GO, by the first case, are in the ratio compounded of the ratios of the bases- AGC and GK, and of the altitudes A E and G iII; therefore also the solids Y and Z have to one another a ratio that is compounded of the same ratios. Therefore, &c. Q. E. D. Con. 1. Hence, two straight lines may be found having the same ratio with the two parallelopipeds AF and GO. To AB, one of'the sides of the parallelogram. At, apply the parallelogram BY equal to GK, having an angle equal to the angle B AD (44. 1.); and as &E to GM, so let AV be to AX (12.:.), then AD is to AX as the solid AF to the solid GO. For the ratio of AD to AX is compounded of the ratios (def. 10. 5.) of AD to AV, and of AV to AX; but the ratio of AD to AV is the same with that of the parallelogram AC to the parallelogram BV (1. 6.) or GK; and the ratio of AV to AX is the same with that of AE to GM, therefore the ratio oft AD to AX is compounded of the ratios of AC to GK, and of AE to GM (E. 5.). But the ratio of the solid AF to the solid GO is compounded of the same ratios; therefore,. as AD to AX, so is the solid AF to the solid GO. CoR. 2. If AF and GO are two parallelopipeds, and if to AB, to the perpendicular from A upon DC, and to the altitude of' the parallelopiped AF, the numbers L, X1, N be proportional. and if to AB, to Gt-, to the perpendicular from G on LK, and to the altitude of the parailelo. piped GO, the numbers L, 1, in, n be proportional; the solid AF is to the solid GO as LXMXN to XmXmlt For it may be proved, as in thle 7th of the 1st of the Sup. that L X VI-XN is to I/XnrXn in the ratio compounded of the ratio of L XMI to IX7n, and of the ratio of N to n. No w the ratio of' L X IM to i X fn is that of the area of the parallelogram AC to that of the parallelogram GIK; and the ratio of N to n is the ratio of the altitudes of the parallelopipeds, by hypothesis, therefore, the ratio of LX M X N to IXsnXn is compounded of the ratio of the areas of the bases, and of the ratio of the altitudes of the parallelopipeds AF and GO; and the ratioo f the parallelopipeds themselves is shewn, in this proposition, to be compounded of the same ratios; therefore it is the same with that of the product L X M XN to the product IXm Xn. Coy.. 3. Hence all prisms are to one another in the ratio compounded of the ratios of their bases, and of their.' altitudes, For every prism is equal tci a para!lelopiped of the same, altitude withl it, and of an equal base (2, Cor. 3. Su.). OF GEOMVETRY. 3BOOK I0 l 20o PROP. X. - THEOR. Solidparallelopipeds, which have thcir bases and altitudes reciprocally proportional, are equal; avtd parallelopipeds twhich are equal, have their bases and altitudes recip rocally proportional. Let ykG and KQ be two solid parallelopipeds, of which the bases are AC and KM, and the altitudes AE and KO. and let AC be to KM as KO to AE; the solids AG and KQ are equal. As the base AC to the base KM, so let the straight line KO be to the straight line S. Then, since AC is to KM as KO to S, and also by hypothesis, AC to KM as KO to AE, KO has the same ratio to S that it has to AE (I 1. 5.); wherefore AF is equal to S (9. 5.). But the solid AG is to the solid KQ, in the ratio compounded of the ratios of AE to KO, and of AC to KM (9. 3. Sup.), that is, in the ratio compounded of the ratios of AE to KO, and of KO to S. And the ratio of AE to S is also compounded of the same ratios (def. 10. 5.); therefore, the solid AG has to the solid KQ the samne ratio that AE has to S. But AE was proved to be equal to S, therefore AG is equal to KQ. Again, if the solids AG and KQ be equal, the base AC is to the base KM as the altitude KO to the altitude AE.'Take S. so that AC may boe to KM as KO to S, and it will be shewn, as was done above, that the solid AG is to the solid KQ as AE to S; now, the solid AG is, by hypothesis, equal to the solid KQ: therefore, AE is equal to S; but, by construction, AC is to KM, as KO is to 8; therefore, AC is to KM as KO to AE. Therefore, &c. Q. E. D. Con. In the same manner, it may be demonstrated, thlat equal prisms have their bases and altitudes reciprocally proportional, and conversely. PROP. IX. THEOR. Simtilar solid.parallelopipeds are to one another in the triplicate ratio of their homologous sides. Let AG, KQ be two similar parallelopipeds, of which AB and KL are two homologous sides; the ratio of the solid AG to the solid KQ is triplicate of the ratio of' AB to KL. 208 SUPPLEMEiNT TO THE ELEMENTS Because the solids are similar, the parallelograms AF, KB are similar (def. 2. 3. Sup.), as also the parallelograms AH, KR; therefore% DiAB L 77 LA the ratios of AB to KL, of AE to KO, and of AD to EN are all equa4 (def. 1. 6.j. But the ratio of the solid AG to the solid KQ is come pounded of the ratios of AC to EM, and of AE to KO. Now, the ratio of AC to EM, because they are equiangular parallelograms, is compounded (23. 6.) of- the ratios of AB to KL, and of AD to KNo Wherefore, the ratio of AG to KQ is compounded of the three ratios of AB to KL, AD to KN, and AE to KO; and the three ratios have already been proved to be equal; therefore, the ratio that is comra pounded of them, viz. the ratio of the solid AG to the solid KQ, is triplicate of any of them (def. 12. 5.): it is therefore triplicate of the ratio of AB to KL. Therefore, similar solid parallelopipeds, &c. QE.D. Con. 1. If as AB to KL, so EL to m, and as EL to ra, so is rn to l, Then AB is to n as the solid AG to the solid KQ. For the ratio of AB to n is triplicate of the ratio of AB to KL (def. 12. 5.), and is therefore equal to that of the solid AG to the solid KQ. Con. 2. As cubes are similar solids, therefore the cube on AB is to the cube on KL in the triplicate ratio of AB to KL, that is in the same ratio with the solid AG, to the solid EQ. Similar solid parallelopipeds on. And as many degrees, minutes, seconds, &c. as are in any arch, so many degrees, minutes, seconds, &c. are said to be in the angle measured by that arch. Con. 1. Any arch is to the whole circumference of which it is a part, as the.number of degrees. and parts of a degree contained in it is to the number 360. And any angle is to four right angles as the number of degrees and parts of a degree iq thy arch, which is the measure of that angle, is to 360. PLANE TRIGONOMETRiY. 22 3 COn. 2. Hence also, the arches which measure the same angle, whatever be the radii with which they are described, contain the same number of degrees, and parts of a degree. For the number of degrees and parts of a degree contained in each of these arches has the same ratio to the number 360, that the angle which they measure has to four right angles (Cor. Lem. I.). The degrees, minutes, seconds, &c. contained in any arch or angle, are usually written as in this example, 49~. 36'. 24". 42"'; that is, 49 degrees, 36 minutes, 24 seconds, and 42 thirds. III. Two angles, which are together equal to two right angles, or two arches which are together equal to a semicircle, are called the Sup~ plements of one another. I~. A straight line CD drawn through C, one of the extremities of the arch AC, perpendicular to the diameter passing through the other K extremity A, is called the Sine of the arch AC, or of the angle ABC, of which AC is the measure. COR. 1. The sine of a quadrant, or of a right angle, is equal to the ra- F dins. ~ Con. 2. The sine of an arch is half the chord of twice that arch: this is evident by producing the sine I of any arch till it cut the circumference. V. The segment DA of the diameter passing through A, one extremity of the arch AC, between the sine CD and the point A, is called the Thersed sine of the arch AC, or of the angle ABC. VI. A straight line AE touching the circle at A, one extremity of the arch AC, and meeting the diameter BC, which passes through C the other extremity, is called the Tangent of the arch AC, or of the angle ABC. Con. The tangent of- half a right angle is equal to the radius. VII. The straight line BE, between the centre and the extremity of the tangent AE is called the Secant of the arch AC, or ofthe angle ABC. Con. to Def. 4, 6, 7, the sine, tangent and secant of any angle ABC, are likewise the sine, tangent, and secant of its supplements CBF. 224.PLANE TURIGONOMETRY. Ij is manifest, from Def. 4. that CD is the sine of the angle CB.F Let CB be produced till it meet the circle again in I; and it is also manifest, that AE is the tangent, and BE the secant, of the angle ABI, or CBF, from Def. 6, 7. Cor. to Def. 4, 5, 6, 7. The sine, versed sine, tangent, and secant of an arch, which is the measure of any given angle ABC, is to the sine, versed sine, tangent and secant, of any other arch which is the mea- sure of the same angle. as the radiusg of the first arch is to the -radius of the second. Let AC, MN be measures of the an- B M D gle ABC, according to Def. 1.; O;D) CB the sine, DA the versed sine. AE the tangent, and BE the:se" cant of the arch AC, according to Def. 4, 5, 6, T; NO the sine, OMT the versed sine, MIP the tangent, and BP the secant of the arch IMN,' according to the same definitions, Since CD, NO, AE, MP are parallel, CD: NO::rad CB: rad. I B, and AE: MP: rad. AB: rad. BM, also BE: BP: AB: BM; likewise because BC: -BD:: BNS BO, that is, BA: BD:: BM: BO, by conversion and alternation, AD: MO:: AB: MB. Hence the corollary is manifest. And therefore, if tables be constructed, exhibiting in numbers the sines, tangents, secants, and versed sines of certain angles to a given radius, they will exhibit the ratios of the sines, tagents, &c.,of the samne angles to any radius whatsoever. In such tables, which are called Trimornometrical Tables, the radius is either supposed I, or some number in the series 10, 100,.1000, S&c The use and construction of these tables are about to be explained. XlII. The difference between any angle and a righit angle, or between any arch and a quadrant, is called the Comnplemrent of that angle, or of that arch. Thus, if BI be h perpendicular to A1,- the angle CBRH is the complement of the angle ABC, and the arch HC the complement of AC; also ___ the complement of the obtuse A' angle FBC is the angle -IBC, its excess above a right angle; / and the complement of the arch iTx FC is HC,. /-= PLAN E TRIG4ONO1METR~Y r IXo The sine, tangent, or secant of the complement of any angle is called the Cosine, Cotangent, or Cosecant of that angle. Thus, let CI, or DB, which is equal to CL, be the sine of the angle CBH; IlK the tangent, and BR the secant of the same angle: C L or B D1 is the cosine, HK the cotangent, and BK the c osecant of the angle ABC" Con. 1. The radius is a mean proportional between the tangent and the cotangent of any angle ABC; that is, tan. ABC Xcot. ABCG R2. For, since - HlK, BA are parallel, the angles HKB, ABC are equal, and KHB, BAE are right angles; therefore the triangles BAEK.~KHB are similar, and therefore AE is to AB, as BH or BA to HK'. CoRi. 2. The radius is a mean proportional between the cosine and secant of any angle ABC; or cos. ABC Xsec. ABC-R'. Since CD, AE are parallel, BD is to BC or BA, as BA to BE. PROP. I, Irs a right angled plane triangle, as the hypote2nuse to either ogf the sides, so the radius to the sine of the angle opposite to that side; and as either of the sides is to the other side, so is the radius to the tangent of the angle opposite to that side. Let ABC be a right angled plane triangle, of which BC is the hypotenuse. From the centre C, with any radius CD, describe.the arch DE; draw DF at right angles to CE, and( from E draw EG touching the circle in Ei, and meetinrg CB in G; I)F is the sine, and EG the tangent of the arch DE, or of the angle C. The two triangles DFC, BAC are equiaDgular,- because the angles DFC, BAC are right angles, and the angle at C is common. There| fore, CB: BA:: CD:DF; but CD is the-radius, and DUF the sine of the angle C, (Def. 4.); therefore CB: BA:: R: sin Also, because EG touches the / circle in E, CEG is a right angle, and therefore equal to the angle' - BAC; and since the angle at C is common to the triangles CBA, CGE, these triangles are equiangular, wherefore CA AB:: CE: EG; but CE is the radius, and EG the tangent of the angle C; there, fore, CA: AB:: R:.tan C.o Con. 1. As the radius to the secant ()' the angle C, so th1e side adjacentt to that ang!e to the hypotenuse. For CG is the secanrt of y g 2126t -PLANE TRIGONOMETRY. the angle G (def. 7.), and the triangles CGE, CBA being equiangular, (CA: CB:: CE: CG, that is, CA: CB:: R:sec. C. Con. 2. If the analogies in this proposition, and in the above corollary be arithmetically expressed. making the radius = 1, they give in. c BCtan. AC se Also, since sin. C -- cos B, because B is the complement of C, cos. B =B and for the same rea. AC son, cos. C -AC BU FCon. 3. In every triangle, if a perperpendicular be drawn from any of the angles on the opposite side, the segments of that side are to one another as A. the tangents of the parts into which the opposite anglle is divided by the perpendicular. For, if in the triangle ABC, AD be drawn perpendicular to the base BC, each of the triangles CA.D, ABD being right angled, AD: DC:: Ri: tan. CAD, AB C, and A a: LDB:: R: tan. DAB; therefore, ex aequo, DC: DB:: tan. CAD: tan. BAD. SCHOLIIUM. The proposition, just demonstrated, is most easily remembered, by stating it thus: If- in a right angled triangle the hypotenuse be made the radius, the sides become the sines- of the opposite angles; and if one of'the sides be. made the radius, the other side becomes the tangent of the opposite ganle, and the hypotenuse the secant of it. PROP. ILo The sides of a plane triangle are to one another as the sinles of the. o6p.posite angles. From A any angle in the triangle ABC, A let AD be drawn perpendicular to BCo And because the triangle ABD is right angled at D, AB: AD:: R: sin. B; and for the same reason, AC: AD:: R: sin.,C, and inversely, AD: AC:: sin. C: XR therefore, ex aequo iaversely, AB AC:: sin. C:: sin. -B. In the same manner it may be demonstrated,t-hat AB: BC:: sin. C: sin. A. Therefore,. D E, Es.! JLN A, R N G( E OMETR- I.G N 0 M1r, y22 PROP. IMIo TVhe sum'of the sines of any two arches of a circle, is to the difference of their sine,%, os the tarvgt; of half the sum of the archis to the tongent of half their di(Jcrence. Let AB, AC be two arches of a circle ABCD; let E be the centre, and IEG the diameter which passes through A; sin. AC+sin. AB sin. AC- sin.:ABl:: tan, I (AC+AB): tan. - (AC- AB). Draw BF parallel to AG, meeting the circle again in F. Draw BH and CL perpendicular to AE, and thoy will be the sines of the arches AB and AC; produce CL till it meet the circle again in D; join LDF, FC, DE, EB, EC, DB. Now, since EL from the centre is perpendicular to CD, it bisects the line CD in L and the arch CAD - in A: DL is, therefore equal to LB, or to the sine of the arch AC; F. —-. and BH or LK being the sine of AB, DK is the sum of the sines ofthe arches AC and AB, and CK \ is the difference of their sines; i /-: I ]DAB also is the sum of the arch- L | es AC and AB, because AD is equal to AC, and BC is their difference. Now, in the triangle DFC, because FK is perpendicular to DC, (3. cor. 1.) DK: KGC: tan. DFK: tan. CFK; but tan. DFK —tan. 2 are. BD, because the angle DFK (20. 3. j is the half of DEB, and is therefore measured by half the arch DB. For the same reasbn, tan. CFK=tan.. arc..BCI and consequently, DK: KC:: tan. 1 are. BD: tan. 1 are. BC. But DK is the sum of the sines of the arlches AB and AC; and KC is the difference of the sines; also BD is the sum of the arches.;-B and AC, and BC the difference of those archcs. Therefore, &c. Q. E. D. CoR. 1. Because EL is the cosine of AC, and EH of AB, FK is the sum of these cosines, and KB-their difference; for FK:::~ F[B+ EL=EH+EL, and KB=Lll —EI -EL. Now, FK: KB:: tan. FDK: tan. BDK: and tan. DFK-cotan. FDK, because DFK is the complement of FDK; therefore, FK: KB:: cotan. DFK: tan. BDK, that is, FK: KB:: cotan. - arc. DB:tan. 2 arc. BC. The sum of the cosines of two arches is therefore to to the difference of the same cosines as the cotangent of half the sum of the arches to the tangent of half their difference. CoR. 2.In the right angled triangle FRD, FE: KD:: R: tan, DFK:; Now FKcs. AB+cos. AC,, KI!)min, AB+,-in. A., and IPLANE' TRIGONOMETRYl ian. DFK = tan. I (AB +- AC), therefore cos. AB+cos. AC. sin. AB + sin. AC:: R: tan. (AB + AC). In the same manner, by help of the triangle FKC, it may be shown that cos. AB co. AB C'- sios. AC sin. AB:: R: tan. i (AC-AB)U Coi. 3. If the two arches AB and AC be together equal to 90", the tangent of half thesir sum, that is, of 450, is equal to the radius. And the arch BC being the excess of DC above DB, or above 96,% the half of the areth BC will be equal to the excess of the half of DC above the half of DB, that. is, to the excess of AC above 450; there — fore, when the sum of two arches is 90~, the sum of the sines of those arches is to their difference as the radius to the tangent of the difference between either of-them and 45~. PROP. IVo The sum of any two sides of a triangle is to their difference, as the tar' gent of half the sumn of the angles opposite to those sides, to the tangent of half their diference. Ilet A.BC be any plane triangle; CA-AB: CA - B:: tan 1 (B+): tan;, (B -CAd Pro~ (!2.) -Ca: AB(~ For (2.) CA: AB:: sn. B sin, C; and therefore (E. 5.) CA+AB: CA-Al:: sin. B+sin. C: sin. B-sin. C,. But, by the last, sin. B+ sin. C: sin. B - sin. C:: tan. - (B+C): tan. (B- C); therefore also, ( 1. 5.t CA+A-AB: CA —AB tan - (B-+-C) tan. E (B - CE Q. E. De B C Otherwise, without- the 3d. Let ABC be a triangle; the sum of AB and AC any two sides, is to the difference of AB and AC as the tangent of half the sum of the an; gles AGB and ABC, to-the tangent of half their differen-ce; PLANE T'1'RGONOTMETRY. 29 About the centre A with the radius AB, the greater of the two sides, describe a circle meeting BC produced'in D, and AC produced in E and F. Join DA, EB, FB; and draw FG parallel to CB, meeting EB in G.: Because the exterior angle EAB is equal to the two interior ABCV: ACB, (32. 1.): and the angle EFB, at the circumference is equal to half the angle EAB at the centre (20. 3.): therefore EFB is half the sum of the angles opposite to the sides AB and AC. Again, the exterior angle ACB is equal to the two interior CADU ADC, and therefore CAD is the difference of the angles ACB, ADC, that is, of ACB, ABC, for ABC is equal to ADC. Whermefore also DBF, which is the half of CAD, or BFG, which is equal to DBF, is half the difference of the angles opposite to the sides AB, AC. Now because the angle FBE in a semicircle is a right angle, BE is the tangent of the angle EFB, and BG the tangent of the angle BFGC to the radius FB; and BE is therefore to BG as the tangent of half the sum of the angles ACB, ABC to the tangent of half their difference. Also CE is the sum of the sides of the triangle ABC, and CF their difference; and because BC is parallel to FG. CE: CF:: BE: BG, (2. 6.) that is, the sum of the two sides of the triangle ABC is to their difference as the tangent of half. the sum of the angles opposite to those sides to the tangent of half their difference. Q.E. D. PROP. V. THEORY If a perpendicular be drawn fron any any gle 4f a triangle to the bppositeeside, or base; the sum of the segments of the base is to the sum of the other twor sides of the triangle as the dgfference of those sides to the difference of the segments of the base. For (K. 6.), the rectangle under the sum and difference of the segs nmerits of the base is equal to the rectangle under the sum and differl 2PLANE TRIGo N0.tjHEI E. ence of the sides, and therefore (16. 6.) the sum of the segments ot the base is to the sum of the sides as the difference of the sides to the difference of the segmeints of the base, Q. E. D. PR.OP. VI. THiEOR. In an' triangle, twice the rectarngte contained by any two sides is to the difference between the sum (of the squares of those sides, and the square of the base, as the radius to the cosine of the angle included by the twa sides. Let ABC be any triangloe, 2AB.B(C is A. to the diffetence between ABW+BCa and AC2 as radius to cs. B. From A draw AD perpendicular to / \ BC, and (12. and 13. 2.) the difference / between the surm of the squares of AB / and BC, and the square on AC is equal / to 2BC.BD. / But BC.BA: BC.BC D:: BA: BD it: cos, B, therefore also 2BC.BA.: C aBC.BD:: R: cos. tB. Now 2BC.BD is the difference between AB3 +BC2 and AC2, therefore twice the rectangle AB.BC is to the dif- A ference between AtiU+- BC2, and AC2 as radius to the cosine of B. Wherefore, &c. Q. E. D. Con: If the radius=- I, BD=BA X cos. B,(K {,), and 2BC.BA Xcos. B=-tBC.BPD, and therefore when B is acute, 2BC.BA X cos.B - GaC2 +BA2-AC2, and adding A(C to 3 both; AC2+2 cos. BXBC.BA=D BC2+BA; and taking v2. cnos. BXBC.BA from both, AC2=BC2-S cos. BXBC.BA+BA2, Wherefore AC= (B2A —2 cos. BXBC. BA+BA2). If B is an obtuse angle, it is shown in the same way that AC — V (BC2.+2 cos. B XBC.B,+A A 2). PROP. VII. tbur times the rectangle contained by any two sides of a triangle, is to the rectangle containetd] by twlo straight lines, of wzhich one is the base o0r third side of the trian7gle increased by the difference of the two sides, and the other the bas'e diminished by the difference of the same sides, as the square iof the radi*us to the square of the sine of half the angle included between. the two sides of the triangle. Let ABC be a triangle of which BC is the base, and AB the greater of the two sides; 4AB.AC (BC+(ABB A)) X (BC - (AB A:C)) a R2: (sin. - BAC)2. PLANE TRIG ONOMlIETRY..3i Produce the side AC to D, so that AD=AB; join BD, and draw A AE, CF at right angles to it; fiom the centre C with the radius CD describe the semicircle GDH, cutting BD in K, BC in G, and meeting BC produced in H. It is plain that CD is the difference of the sides, and therefore that BIH is the base increased, and BG the base diminished by the differ. ence of the sides; it is also evident, because the triangle BAD is isosseles, that DE is the half of BD, and DF is the half of D1K, wherefore DE-DF-the half of BD-DK, (6. 5 ) that is, EF=- BK. And bce cause AE is drawn parallel to CF, a side of the triangle CFD, AC AD:: EF: ED, (2. 6.); and rectangles of the same altitude being as their bases AC.AD: AD2:: EF.E ) ED2, (6. 6.), and therefore 4AC.AD: AD2:: 4EF.ED: ED2, or alternately, 4AC.AD:,EF.ED: AD": ED2. But siuce4EF-=~2 BK,4EF.ED =2BK. ED 2ED.BK=DB.BK — HB.BG; therefore, 4AC.AD: DB.BK:: AD2: E)D2. Now AD e ED:: R: sin. EAC —sin. I BAC (I. Trig.,) and AD2: EDZ:: R3 (sin. - BAC)2: therefore, (1 I. 5.) 4AC..AD: IB.BG:: R2: (sin.. BAC)2, or since AB=-AD, 4AC. AB: HB.BG:: R2: (sin ~ BAC)2. Now 4AC. AB is four times the rectangle contained by the sides of.the triangle; HB.BG is that contained by BC+(AB-AC) and BC -~ (AB-AC).'Therefore, &c. Q. E. 1). COR. Hence 2 yV AC.AD: F B.BG:: P:. sin. 2 BAGC PROP. VIII, Four tines the rectangle containted by any twzo sides of a t-riangle, is to the rectangle contained by too straight lines, of w;hich one' is the suwn of those sides increased by the base of the triangle, and the other the sum of the same sid.es diminished by the base, as the square of the radius to the square of the cosine of half the angle included between the two sides of the triangle. Let ARBC be a triangle, of which BC is the base, and AB the great2 er of the other two sides, 4AB.AC: (AB+ AC-I-BC) (AB +AC - BC):: R: (cos. I BAC)3. From the centre C, with the radius CB, describe the circle BLIVI, meeting AC, produced, in L and AI. Produce AL to N, so that AN - AB let AD-AB;. draw AEB perpendiclTlar to BD; join B.N, and 232 PLANE TRIG ONOMETRY. let it meet the circle again in P; let CO be perpendicular to B:N; and let it meet AE in R. It is evident that 1MN=AB+AC+BC; and that LN=AB+AC -— BU. Now, because BD is bisected in E, (3. 3.) and DN in A, BN is parallel to AE, and is therefore perpendicular to BD, and the triangles DAE, DNB are equiangular; wherefore, since DN 2AD9 BN —AE, and BP=2BO:=2RE; also PN=2AR. But because the triangles ARC and AED are equiangular, AC AD:: AR: AE, and because rectangles of the same altitude are as A: B-3D D -their bases, (1. 6.), AC.AD: AD2:: AR.AE: AE2, and alternately AC.AD: AR.AE:: AD2: AE2, and 4AC.AD: 4AR.AE:: AD2' AEW. But 4AR.AE2AR X 2AE = NP.NB=MN.NL; therefore 4AC.AD: MN. NL:: AD)2:AEa. But AD: AE:: R:cos. DAE (1) -cos. (BAC): Wherefbre 4AC.AD: MN.NL::: cos. O BAC)2. Now 4AC.AD is four times the rectangle under the sides AC and AB, (for AD= AB), and MN.aNL is the rectangle under the sum of the sides increased by the base, and the sum of the sides diminished by the base. Therefore, &c. Q. E. D. Con. 1. Hence 2 / AC.-AB: x/ MN.NL R:: R: cos. 2 BACo GCo. 2. Since by Prop. 7. 4AC.AB::(BC- + (AB-AC)) (BC- (AB - BC)):: R2: (sin. - BAC)2; and as has been now proved 4AC.AB:(AB+-tAC +BC) (AB~AC-BC):: R2: (cos. V BAC)2: therefore ex aequo, (AB+AC+BC) (ABAC- BC): (BC -+ (AB -— AC)) (BC -(AB -AC)) (cos. i BAC)2: (sin. BAC). 2 But the cosirt: of any.rch is to tih sie:e as the ra~dius to the angeant of PLANE TRIGONOMETRY..S. the sane arch; therefore, (AB+AC+BC) (AB+AC BC. (B3C+ (ABl-AC)) (BC-(AB-AC)):: R2: (tan-: BAC)2; and V (AB+AC+BC) (AB+AC_-sBC):, (BC+AB —AC) (BU — (AB AL)O)': tan - BAC, LEMMA- IL I' there be two unequal mnaganitizdes, half their diierenee added to ha!' their sum is equal to the greater; and half their dference talkenfronp half their sumn is equal to the less. Let AB and BC be two unequal magnitudes, of which AB is the greater; suppose AC bisected in D, and AE equal to BC. It is manifest A E' DIt B C that AC is the sum, and EB the differ once of the magnitudes. And because AC is bisected in Dj AD is equal to DC: but AE is also equal to BC, therefore DE is equal to DB, and DE )or DB is half the difference of the magnitudes. But ABl is equal;to BD and DA, that is, to' half the difference added to half the.sum; and BC is equal to the excess of DC, half tie sum above DB, half the difference. Therefore, &c. Q. E. Do Cou. HepGe, if the sum and the difference of two magnitudes be given, the:magqnitudes themnselves may be round; for to half the sum add -half the difference, and it will give th.e greater: from half.the sum asubtract half the difference, and it will give-the less., SECTION II. OF THE RULES OF TRIGONOMETETIlCAL CALCULATIO N. THzE GENERAL PROBLEM which Trigonometry proposes to resolve is: In any plane triangle, of the three sides and the three angles, any.three being given, and one of these three being a side, to find any of the other- three. The things here said to be given are understood to be expressed by their numerical values: the angles, in degrees, minutes, &Po and the sides in feet, or any other known measured Gg PLANE TRIGON0OMETRYL The reason of the restriction in this problem to those cases inr which-at least one side is given, is evident from this, that by the angles alone being given, the magnitudes of the sides are nol determined. Innumerable triangles, equlangular to one another, may exist, without the sides of any one of them being equal to those of any other; though the ratios of- their sides to one another will be the same in them-all, (4. 6.). If. therefore, only the three angles are given, nothing can be determined of the triangle but the ratios of the sides, which may be found by trigonometry, as being the same with the ratios of the sines of the opposite angles. For the conveniency of calculaton, it is usual to divide the gelneral problem into two; according as the triangle has, or has not, one of its angles a right angle. PROB. I. In a.right angled triangle, of the three sides, and three angles, any tw& being given, besides the right angle, and one of those two being a side, it is required to find the other three. It is evident, that when one of the acute angles of a right angled tryi angle, is given, the other is given, being the complement of the former to a right angle; it is also evident that the sine of any of the acute angles is the cosine of the other. This problem admits of several cases, and the solutions, or rules for calculation, which all depend on the first Proposition, imay be conve. rniently exhibited in the form of a table; where the first column contains the things given; the second, the things required; and the thirds the rules or propositions by which they are found..IVEN. SOUGidT. SOLUTION. CB and B, the.. CBaend Bnthed AC. R sin B: CB: AC. ahypon ansd, A.: cos tB:C AR. 2 angle. C.. 1 AC and C, a C side and one ofC:R ACBC. 3 the acute angles. A CB and BA,::: sin C. the hypotenuse C. C BA R: sin C. 5 and a side. AC.R: cosC CB AC. 6 AC and AB, C. AC:AB R:tan C. 7 the two sides. CB. Cos C: R:: AC: CB. 8 P LANE TRIG? ON OILETRY.? JRemarks on the Solutions in the table. In the second case, when AC and C are given to find thi-hypotc~ muse BC, a solution may also be obtained by help of the secart, for CA: CB:: R: sec. C.;- if, therefore, this proportion be made R Sec. C:: AC: CB, CB will be found. In the third case, when the hypotenuse BC and the side AB are given to find AC, this may be done either as directed in the Table, br by the 47th of the first; for since AC' — BC-. BA2. AC, VBC2 - BA2. This value of AC will be easy to calculate by loga-. rithms, if the quantity BC-BA' be separated into two multipliers, which may be done; because (Cor. 5. 2.), BC2' — BA- (BC+BA) (BC —BA.). Therefore ACV/ (BC+BA) (BC(;-BA). When AC and AB are given, BC may be found from the 47th, as in the preceding instance, for B: - BA+'2. But BA+~AC2 cannot be separated into two multipliers; and therefore, when BA and AC are large numbers, this rule is inconvenient for computation by logarithms. It is best in such cases to seek first for the tangent of C, by-the analogy in the Table, AC: AB: R: tan. C; but if C itself is not required, it is sufficient, having found tan. C by this proportion, to take from the Trigonometric Tables the.cosine that corresponds to tan..C, and then to compute CB from the proportion cos. C: R:: AC: CB. PROBLEM II. In aln oblique angled triangle, of the three sides and three angles, any three being given, and one -o these three being a side, it is required to find the other three. This problem has four cases, in eachi of which the solution depends on some of the foregoing propositions. CASE L Two angles A and B, and one side' AB, of a triangle ABC, being given, to find the other sides. _ itI,. I'-,IG ONo' Al E T It V'o SOLUTION. Because the angles A and B are given, C is also given, being too supplement of A+B; and, (2.) Sin. C. sin. A:: AB: BC; also, Sin. C b sin B:- AB ~ ACCASE IT. two sides AB and AC, and-the angle B opposite to one of theni, being given, iofind the other angles A and. C, and also the other side SOLUTION6 The angle C is found from this proportion, AC: AB:: sih B: sin lb. Also, A=1 80~-B —C; and then, sin. B: sin A:: AC: CB, by Case 1. In this case, the ahgle C may.hve two values; for its sine being found by the proportion above, the angle belonging to that sine may either.b ihat which is found in the tables, or it may be the supplement bf it, (Cot. def. 4.). This ambiguity, however, does not arise from any defect it the solution, but from a circumstance essential to the problem, viz. that whenever AC is less than AB, there are two triangles which have the sides AB, AC, and the angle at B of the same magnitude in each, but which are nevertheless unequal, thil angle opposite to AB in the one, being the supplement of that which is opposite to it in the other. The truth 6f this appears by describing'rom the centre A with the raditus AC, an arch intersecting BC in C PLAXNE TRIGOINO0iMETR'Yo 23-7 fand C'; then, if AC and AC' be drawn, it is evident that the triangles ABC, ABC' have the side AB and the angle at B common, and the sides AC and AC' equal, but have not the remaining side of the one equal to the remaining side,of the other, that is, BC to BC', nor their other angles equal, viz. BC'A to BCA, nor BAGC' to BAC. But in these triangles the angles ACB, AC'B are the supplements of one another. For the triangle CAC' is isosceles, and the angle ACC'= the AC'C, and therefore, AC'B, which is the supplement of AC'C, is also the supplement of ACC' or ACB; and these two angles, ACB, AC'B are the angles found by the computation above. From these two-angles, the two angles BAC, BAC' will be found: the angle BAC is the supplement of the two angles ACB, ABC, (32. 1.), and therefore its sine is the same with the sine of the sum of ABC and ACB. But BAC' is the difference of the angles ACB, ABC: for it is the difference, of the angles AC'C and ABC, because AC'C, that is, ACC' is equal to the sum of the angles ABC, BAC', (32. 1.). Therefore, to find BC, having fbund C, make sin C: sin (C+B):: AB: BC; and again, sin C: sin (C-B):: AB: BC. Thus, when AB is greater than AC, and C consequently greater than B, there are two triangles which satisfy the conditions of the question. But when AC is greater than AB, the intersections C and C' fall on opposite sides of B, so that the two triangles have not the same angle at B common to them, and the solution ceases to be ambiguous, the angle required being necessarily less than B, and therefore an acute angle. CASE IIl. Two sides AB and AC, and the angle A, between them, being given to find the other angles B and C, and also the side BC. SOLUTION-' First, make ABB+AC: AB — AC:: tan I (C+B): tan ~ (C -B.), Then, since I (C+B) and l (C B) are both given, B and C may be found. ForB=- (C+B)+- (C- B,) and C — (C-+B) — (C-.B) (Lem. 2.). To find BCo Having found B, make sin B: sin A:: AC: BC. But BC may also be found mW-ithout seeking for the angle B and C; for BC=-/AB2a- 2 cos AXAB.AC+AC2, Prop. 6. This method of finding BC is extremely useful in many geometri-,cal investigations, but it is not very well adapted: for computation by logarithms, because the quantity under the radical sign cannot: be separated into simple multipliers., Therefore, when: AB and- AC are expressed by large numbers, the other solution, by finding the angles, and then computing BC, is preferable, S32 -PLAN1E TRIGONOMETRY. CASE IV. The three sides AB, BC, AC, being given, to find- the angles A,.B C. SOLUTION I. Take F such that BC BA+-AC:: BA -AC: F, then F is-either the sum or the difference of BD, DC, the segments of the base, (5.). If F be greater'than BC, F is the sum, and BC the difference of BD, DC; but, if F be less than BC, BC is the sum, and F the ditference of RD and DC. In either case, the sum of BD'and DC, and their difference being given, BD and DC are found. (Lem. 2.) Then, (I.) CA: CD:: R: cos. C; and BA: BD:: R: cos. B; wherefore C and B are given: and consequently A. Nvh a d are giv —en, and 3 1D Cn C] SOLUTION If. Let "D be the difference of the sides AB, AC. Then (Cor. 7.) ~ AB.gA.C:, / (BC+ D) (BC -D):: R: sin'BAC. SOLUTION 1It. Let S be the sum of the sides BA and AC. Then (I Cor. 8.':.n-'.AC: V/(-SS+BC) (S —BC):: R cos BAC, SOLUTION IV. S and D retaining the significations above, (2 Cor. 8.) V(S+BC) (S-BC): /(BC+D) (BC - D):: R: tan' BAC. It may be observed of these four solutions, that the first has the advantage of being easily remembered, but that the others are rather more expeditious in calculation. The second solution is preferable to the third, when the angle sought is less than a right angle; on the other hatd, the third -is preferable to the second, when the angle sought is greater than a right angle; and in extreme- cases, that is, when the angle sought is very acute or very obtuse, this distinction PILANE TRIGONOMETRY. 23i. -is very material to be considered. The reason is, that the sines of angles, which are nearly = 900, or the cosines of angles, which are nearly = 0, vary very little for a considerable variation in the corresponding angles, as may be seen from looking intor tihe tables of sines and cosines. The consequence of this.is, that when the sine or cosine of such an angle is given, (that is, a sine or cosine nearly equal to the radius,) the angle itself cannot be very accurately found. If, for instance, the natural sine.9998500 is given, it will be immediately perceived from the tables, that the arch corresponding is between 89a, and 89~, 1'; but it cannot be found true to seconds, because the sines of 89o and of 89Q, 1', differ only by 50 (in the two last places,) whereas the arches themselves differ by 60 seconds. Two arches, therefore, that differ by 1", or even by more than 1", have the same sine in the tables, if they fall in the last degree of the quadrant. The fourth solution, which finds the angle from its tangent, is not liable to this objection; nevertheless, when an arch approaches very near to 90, the variations of the tangents become excessive, and are too irregular to allow the proportional parts to be found with exactness, so.that when the angle sought is extremely obtuse, and its half of consequence very near to 90, the third solution is the best. It may always be known, whether the angle sought is greater or less than a right angle by the square of the side opposite to it being greater or less than the squares of the other two sides. SECTION IIL CONSTR UCTION OF TRIGONOMETRICAL TABLES. irt all the calculations performed by the precedingsrules, tables of s,ines and tangents are necessarily employed, the construction of which remains to be explained. The tables usually contain the sines, &c. to every minute of the quadrant from 1' to 90o, and the first thing required to be done, is to compute the sine of 1', or of the least arch in the tables. 1. If ADB be a circle, of which the centre is C, DB any arch of that circle, and the arch DBE double of DB; and if the chords DE, DB be drawn, and also the perpendiculars to them from C, viz. CF, 0G, it has been demonstrated, (8. 1. Sup.) that CG is a mean propor. X240 PLANE TRGONOMETRYL / / I'',\ / \ tional between AH, half the radius, and AF, the line made up of the ray dius -and the perpendicular CF. Now CF is tbh cosine of the arch B'D, and CG the cosine of the half of BD; whence the cosine of the -half of any arch BD, of a circle of which the radius = 1, is a mean proportional: between 1- and I +cos BD. Or, for the greater generality, supposing A — any arch, cos ~ A is a mean proportional.between ~ and 1+cos A, and therefore (cos - A)2-= (1 +cos A) or cos ~ A2- (1 + cos A).'2. From this tlieorem, (which is the same that is demonstrated (9. 1. Sup.), only that it is here expressed trigonometrically), it is.evident, that if the cosine of any arch be given, the cosine of half that arch may be found. Let BD, therefore, be equal to 600, so that the chord BD -- radius, then the cosine or perpendicular CF was shovKn (9. 1. Sup.) to be -~, and therefore cos. - BD, or cos 3Q0~ /.lq-~) —v/; —-—. In the same manner, cos 1-5Q~f (1+'cos 30o), and cos 70, 30'-V/ I (I+-cos 15~), &c. In thisway the cosine of 3~, 45', of l-0, 52', 30", and so on, will be computed, till after twelve bisections of the arch of 60~, the cosine of 52". 44"'. 93"". 45v. is found. But from the cosine of an arch its sine may be found, for if from the square of the radius, that is, from 1, the square of the cosine be taken away, the remainder is the square of the sine, and its square root is the sine itself. Thus the sine of 52". 44"'. 03"". 45,.- is found. -3. But it is manifest, that the sines of very small arches are to Q0e another nearly as the arches themselves. For it has been shown that the number of the sides of an equilateral polygon inscribed in a 9ircle may be so great, that the perimeter of the polygon and the cirt PLANE TRIGONO METR'Y'4 i cum}eronce of the circle may differ by a line less tian any given line, or, which is the same, may be nearly to one another in the ratio of equality. Therefore their like parts will also be neally in the ratio of equality, so that the side of the polygon will be to the arch which it subtends nearly in the ratio of equality; and therefore, half the side of the polygon to half the arch subtended by it, that is to say, the sine of any very small arch will be to the arch itself, nearly in the ratio of equality. Therefore, if two arches are both very small, the first will be to the second as the sine of the first to the sine of the second. Hence, from the sine of 52". 44"'. 03"". 45,. being found, the sine of 1' becomes known; fort ats 52". 44"'. 03"'. 45v. to t, so iS the sine of the former arch to the sine of the lattcr.'Thus the sine of 1' is found=O.0002908882. 4. The sine 1' being thus found, the sines of P', of 3', or of any number of minutes, may be found by the following propostion. T HEOR E. Let AB, AC, AD be three such ar ches, thIat BC the dif-brcn-ce ot 4he first and second is equal to CD the difference of the soecond aiu. third; the radius is to the cosine of the common edififrence BC as the. sine of AC, the rmiddle arch, to half the sum of the sines of 3 alnl' AD, the extteme. arches. Draw CE to the centre: let BF, Cd9 acn D' pCi'Pe-no; i D' tO AE, bie the sines of the arches AB, AC, AD. Join 3D.:no iet it meet CE in I; draw 1K perpendicular to AE, also BL ad I'at pr pendicular to DIH. Then, because the arcil BD is bisected in C, EC is at Fig It angles to B D, and bisects it in 1; als o BI is the sine, and El tin e cosine of' BC of' or CD. And, since B1D is biscetetd itln, -and IM is parallel to BL, (. 6.), LD is / \ aliso bisected in MI.'Zow BF is equal to HlL, the refore, P t:-D H=D- j -HL:= {.DL+2LF =2, - -F1 ti 2MtI =r' /9 o 2KI; and ltherefore ITK is halK the sunm. of BF and 0) H.'ut bsecause th. t1,i-' /- 28 macs tG~, [-I~KE are qtiangla;, ea.' angles CG.FJ B n ic Ie' uar E i:: (G: IKE, anrd it Ias been shown i hal: Ei —: C~s 1"", ari I (B3F+~DU)2; theirefore.'cos "'' eia AC s A'3'- yi n A, Q. E.- D. Con. Hence, if ithe poinet B coincide vith L, i.: cos. BC: sin. BC: sin. B ), that is, the radius is'o the cosine of any (arch, as the sine of the arch is to half the sine of twice the arch; or if any arch=A, sin. 2A= —sin, Acos. A or sin:. 2i =-.-,sin. AXcos. A. Therefore also, sin')2' sil i (ccos 1'; so th at from 1the ie si n: a,-scsine of one minute the, sine of 2' is found. 242: TRG-N N),.OMEi2-F Again,.1,:, 3, einog three such arches that the difference between the'first and second is the same as between the second and third, R:: os 1-' sin 2 (sin l'+sin 3'), or sin l'+sin 3'-2 cos'-+ sin 2', and taking sin 1' from both, sin 3'-2 cos 1' Xsin. 2'-sin 1. In like manner, sin 4'=2' cos 1' Xsin 3'-sin 2', sin 5'=2' cos 1'iXsin 4'-sin 3', sin 6'=2' cos'X sin- 5'- sin 4', &c. Thus a table containing the sines for every minute of the quadrant may be computed; and as the multiplier, cos 1' remrains always the same, the calculation is easy. For computing the sines of arches that differ by m.ore than 1', the method is the same. Let A, A+B, A+2B be three such arches, then, by this theorem, R: cos B:- sin (A+B): 2 (sin A+sin (A+ 2B)); and therefore making the radius 1, sin A+sin (A+2B)=2 cos B >xsin (A~B), or sin (A+2B)=2 cos BXsin (A+B)-sin A. By means of these theorems, a table of the sines, and consequently also of the cosines, of arches of any number of degrees and minutes, sin A from 0 to 90, may be constructed. Then, because tan A= - cos A' the table of tangents is computed by dividing the sine of any arch by the cosine of the same arch. When the tangents have been found in thlis manner as far as 45~, the tangents for the other half of the quadrant may be found more easily by another rule. For the tangent of an arch above 45~ being the co-tangent of an arch as much under 45'; and the radius being a mean proportional between the tangent and co-tangent of any arch, (1 Cor. def. 9 —), it follows, if the differq. ence between any arch and 45~ be called D, that tan (45~- D): I I: tan (45"+D), so that tan (45-'+D)= —.. tan (45 -D)) Lastly, the secants are calculated fiom (Cor. 2. det] 9.) where it is shown that the radius is a mean proportional between the cosine and the secant of any arch, so that if A be any arch, see A. — cos A' The versed sines are found by- subtracting the cosines from the radius. 5. The preceding Theorem is one of fou', which, when arithmreti cally expressed, are frequently used in the application of trigonometry to the solution of problems. ltno, If in the last Theorem, the arch AC:=A, the arch BC=:B, and the radius EC= —, then AD=-A-b, and AB = A- B; and by what has just been demonstrated, 1: cos B sin A: sin (A~-B)+ — sin (A —B), and therefore sin AXcos B = sin (A+B) + - (A —B). 2do, Because BF, 1K, DHI are parallel, the straight lines BD and P.'1 are cut proportiona!!y, and. tlerofcore FHI. the dfl,fernce of tih P-LANI'li ({-OtN 0 3.:[TRY, 43 straight lines FE and HIE, i;. bisected in K; and therefore, as was shown in the last Theorem, KE is half the sum of FE and HE, that is, of the cosines of the arches AB and AD. But because of the similar triangles EGC, EKI, EC: E: GME: EK; now, GE is-the cosine of AC, therefore, RPt: cos BC::cos AC:,cos AD -, cos AB, or I:cos B:: cos A: cos (A+B) + cos (AB); and therefore, cos A X cos B=- cos (A+B) + 2- cos (A — B) 3tio, Again, the triangles IDM, CEG are equiangular, for the alb gles KIM, EID are equal, being each of them right angles, and therefore, taking away the angle EIM, the angle DIM is equal to the angle EIK, that is, to the angle ECG; and the angles DMI, CGE are also equal, being both right angles, and therefore the triangles IDMCGtE have the sides. about their equal angles proportionals, and consequently, EC: CG:: DI: IM; now, IT[ is half the difference of the cosines FE and EtI, therefore, R:sin AC:: sin B:: cos AB cos AD, or 1: sin A:: sin B: -:cos (A- B) -- cos (A+B); and also; sin AXsin B= —- cos (A —B)- 2 cos (A-t-B) 4to, Lastly, in the same trianogles E(G, DIM, EC; EG:: ID DM'; now, lDM is half the difference of the sines DH and BE, there-. fbre, R: cos AC:: sin BC: a sin AD-1 sin AB, or 1: cos A:; sin B:- sin (A+B) -~ sin (A+ B); and therefore, cos A X sin B —= sin (A - B) - sin (A —B). 6. If therefore A and 3 be a'.y two arcShes- whatsoever, the radits being supposed 1;. sin A Xcos BB —- sin (A+B)+' sin (A-B). I. cos AXcos B-= cos (A- B) -+ cos (A+B). ITI. sif A Xsin B=- cos (A- B) cos (A-+B>) IV. cos AXsin B=1 sin (A+B)- sin (A-B). From these four Theorems are also deduced other four-. For adding the first and fourth together, sin AXcos B4+cos AXsin B=sin ('+i-B) Also, by taking the tburth from. the first sin AXcos B —cos AXsin B =sin (A-B). Again, adding the second and third, cos AX cos B — sin A Xsin B — cos (A- B); And, lastly, subtracting the third firom the second,:cosA X co ir X sin B=-os (A+B)? 244 P!;t L NE -}'IF IGO.WJ g ffi L To MR i 7. Again, since by the first of' the above theoremls, sin AxcosBstsin(A+B)+- - In(A - B),iL A-+B=S, and A -B= D S + D D S + D then (Lemr. 2.) A —,and B -,; whereforesin- X cos S D sin S-I-i sin D. But as S and D may be any arches what* 2 2 over, to preserve the forlmer notation, they mnay be called A and B, which al1o express any arches -whatever' thus, A+-B A- B sin- XG c,,- 2 sin A+1 sin B, or 2 2? A+B A-B 2 smin * o3 -=sin A-sin-B. In the same nmanner, from Theor. 2 is derived, A-+B A - B Cos X -— C-=CXs B T-I-cos A. Froom the o3d 2 sim —- -- Xin — — =cos B —cos A; and from the 4thg A+B.- B 2 cos -- Xsin — - sin A.- sin B. In all these Theeoreins, the arch B is supposed less than A. 8.'Theorems of the same kind with respect to the tangents of arches may be dedluced fiom the preceding. Because the tangent of any arch is equal to the sine of the arch divided by its cosine, sin (A+B) tans (A tB) Bu it has just been shown, that cos (A-lB) sin (A.B) -- sin AXeos B+cos A Xsin B, and that cos (A-+-B) cos A Xcos B-sin A Xsil B; therefore tan (A+-B) sin A X co;s B3 -cos A X sin B cos AXc(os B -sn AXsin B ~ and dividing both thie numerator and delnominator of' thL;s fi'tation by cos A X cos B, tan (A -B) - tan A tan B tan A tan B - n e Hn ie mEendnelr, tan (A - B,) I tan A ta n lB +tan AXtar B 9. If the theorem demonstrated in Prop, 3. be expressed in the s.ame nanner with those above, it gives sin A+sin B tan J (A+B) sin A-sin B tan, (A-B) Also by Cor. 1, to the 3d, cos A+cos B _cot I (A +B) eCos A -- cos B tan { (A - B) 'PLANEl TiR [ L4 0 N4' 0 Ai ME Tt1,t. 2. 5B And by Cor'. 2, to the same proposition, sin A+ -sin B tan'- (A-B) cos A- ~ B Or —-5 or siince R is here supposed sin A+sin B -—: tan ~- (Aq- B). cosA+cos B a 10. In all the preceding theorems, 1, the radius is supposed-19 because in this way the propositio1ns are most concisely expressed, and are also most readily applied to trigonometrical circulation. But if it be required to enunciate -any of tUhem geometrically, the mtrultiplier R, which has disappeared, by being made == 1, must be restored, and it will always be evident from inspection in what terms this multiplier is wanting. Thus, Theor. 1, 2 sin AXcos B sin (A+-B)+ sin (A-B), is a true proposition, taken arithmetically; but taken geometrically, is absurd, unless we supply the radius as a multiplier of.the ternms on the right hand of the sine of equality. It then becomes 2 sin A x cos B t=R (sin (A+l B) +- sin (A —B)); or twice the rectangle under the sine of A, and the cosine of B equal to the rectangle under the radius, and the sumin of the sines of A+B and'A — B la general,'the number of lineoar mnultipliers, that is, of lines whose numerical values are multiplied togelther, must be the same in every term, otherwise we will compare unlike magnitudes with one another. T'he propositions in this section are useful in many of the higher branches of'the Mathematics, anld are the foundation of what is call, e -the.4ritftinetic of sines. EfLEMENTS SPHERICAL TRIGONOMETRY. PROP. I. )f a sphere be cult by a plane through the centre, the section isa circles havintg the same centre with the sphere, and equal to the circle by the revolutioin of which the sphere waus described. For all the straight lines drawn from the centre to the superficies of the sphere are equal to the radius of the generating semicircle, (Def. 7. 3. Sup.I. Therefore the common section of the spherti cal superficies, and of a plane passing through its centre, is a line, lying in one plane, and having all its points equally distdnt from the centre of the sphere; therefore it is the circumference of a circle, (Def. 11. 1.), having for its centre the centre of the sphere, and for its radius the radius of the sphere, that is, of the semicircle by which the sphere has been described. It is equal, therefore, to the circle, of which that semicircle was a part. Q. E. Do DEFINITIONS. I. ANY circle, which is a section of a sphere by a plane through its centre, is called a great circle of the sphere. Coa. All great circles of a sphere are equal; and any two of themn bisect one another. They are all equal, having all the same radii, as has just been shown and any two of them bisect one another, for as they have the same centre, their common section is a diameter of both, and therefore bisects botht SPHERICAL TRIGONOMETRY. 2 4, II. The pole of a great circle of a sphere is a point in the superficies of the sphere, from which all straight lines drawn to the circumference of the circle are equal. III-. A spherical angle is an angle on the superficies of a sphere, confained by the arches of two great circles which intersect one another; and is- the same w:ith the inclination of the plane~s of these great cire, cles. iV. A spherical triangle is a figure, upon the superficies of a sphere, comna prehended by three arches of three great circles, each of which is less than a semicircle. PROP. II. Thwe arch of a great circle, between the pole and the circurmflerence ot another great circle, is a quadrant. Let ABC be a great circle, and D its pole; if DC, an arch of a great circle, pass through D, and meet ABC in C, the arch DC is a quadrant. Let the circle, of which CD is an arcl, meet ABC again in A, and let AC be the common section of the planes of these great circles, which will pass through E, the centre of the -~f -, sphere: Join DA, DC. Because AD=DC,(Def. 2.), and equal straaiht lines, in the same circle, cut off equal arches, (28. 3.) the arch AD -the = A t arch DC; but ADC is a semicircle, therefore the arches AD, DC are B each of them quadrants. Q. E. D. Cor. 1. If DE be drawn, the angle AED is a right angle; and DE being therefore at right angles to every line it meets with in the plane of the circle ABC, is at right angles to that plane, (4. 2. Sup.). There.. fore the straight line drawn from the pole of any great circle to the centre of the sphere is at right angles to the plane of that circle; and, conversely, a straight line drawn frorn the centre of the sphere perpendicular to the plane of any greater circle, rmeets the sucperficies of the sphere in the pole of that circle. Con. 2. The circle ABC has two poles, one on each side of its plane, which are the extrelmities of a diameter of the sphere perpendicular to the plane A BC and no other points but these tw' c an hr poles of the ci ircle..... 248 SPHERICAL TRIGONOMETRP-Y PROP. Ill. If the pole of a great circle be the same with the interseclion of other Izwo great circles: the arch oj the/first. mentioned circle intercepted between the other two, is the measure of the spherical angle which the same two circles make with. one another. Let-the great circles BA, CA on the superficies of a sphere, of which the centre is D, intersect one another in A, and let BC be an arch of another great circle, of which the pole is A; BC is the mea. sure of the spherical angle BAC. Join AD, DB, DC; since A is the pole of BC, AB, AC are quadrants, (2.), and the angles ADB, ADC are right angles: therefore (4. def. 2 Sup.), the angleCGDB is the inclination of the planes of the circles AB, AC, and is (def. 3.) equal to: the spherical angle BAC but the arch BC( measures the angle B/DC, therefore it also measures the spherical angle ]B BAC.* Q.E.D. Cos. If two arches of great circles, hAH and AC, which intersect one another in A, be each of them quadl;ants, A will be the pole of the great circle which passes through E and C the extremities of''those arches. For since thle arches AB and AC are quadrants, the angles ADB, A:DC are right annles, and AD is therefore perpendicular to the plane BDC, that is, to the plane of the great circle which passes through B and C. The point A is therefore (Cor. 1. 2.) the pole of the great circle whtch rasses through B1 and C, if the planes of tw'o /geit crcles of a sphere oe at right anjies to one another, the circeunfcre~nce oJ each of tie circles palsses through the poles of the otherT; anld if the cir'zcfer'nce of one great circle pass.throuzgh the poles (f c"nthc'r, i zie 9 pla.e's cf tttesc crclucs are at s'ieF't anlgles. Let ACBD, AEBF be two gre'at circles, fhe planes of which are iight angles'to one anot1her. lhe poIles ef' the cirl!e BA EBF are in tlle circumference ACBD, and the poles of the circle ACBD in the cirecumference AEBF. From G4 the centre of the sphlere, drawv GC in the plane ACBD perpendicular to AB. Then because GC in the plane, ACBD, at. * When in any reference no rnenticn is made of a Book, or of the Plane Trigonometry. he c Spherical Trigonom.(etry is mean!l SPHERICAL TRIGONOMI ETRY. 24,}ight angles to the plane AEBE, C is at right angles to the common section of the two planes, it is (Def. 2. 2. Sup.) also at right an-.les to the plane AEBF, and there. F fore (Cor. 1. 2.) C is the pole of the circle AEBF; and if CG be A produced to D, D is the other pole of.the circle AEBF. In the same manner, by drawing GE Win the plane AEBF, perpen. dicular.to AB, and producing it to Fi, it has shown that E and F are the poles of the circle ACBD. Therefore, the poles of each of these circles are in the circumference of the other. Again, If C be one of the poles of the circle AEBF, the great circle ACBD which passes through C, is at right angles to the circle AEBF. For, CG being drawn from the pole to the centre of the circle A.EBF,.is at right angles (Coi. 1. 2.) to the plane of that circle.; and therefore, every plane passing through CG (17. 2. Sup.) is at right angles to the plane AEBF; now, the plane ACBD passes through CG. Therefore, &c. Q. E. D. Con. 1. If of two great circles, the first passes through the poles of the second, the second also passes through the poles of the first. ~For, if the first passes through the poles of the second, the plane of the first must be at right angles to the plane of-the second, by the second part of this proposition; and therefore, by the first part of it, the circumference of each passes through the poles of the other. COR. 2. All greater circles that have a common diameter have their poles in the circumference of a circle, the plane of which is perpendicular to that diameter. PROP. V. In isosceles spherical triangles fhle angles at the base are equal. Let ABC be a spherical triangle, hlaving the side AB equal to the side AC; the spherical angles ABU and ACBare equal. Let C be the centre of the sphere-; kA join DB, DC, DA, and from A on the straight lines D1B, DC, draw the perpendiculars AE, AF; and from the points E and F draw in the plane DBC the straight lines EG, FG perpendicular.4o DB and DC, meeting one another in G: Join AG. Because D-E is at right angles to each of the straight lines AE, EG, it is at ~,ighlt ag!les to Lthe plane AEG, which 3 =.-,.''"''il';~~ 2O6B SSPHElRICAL TRtCGONOMETRY. passes through AE, EG (4. 2. Sup.); and therefore, every piano that passes through DE is at right angles to the plane AEG (17. 2 Sup.); wherefore, the plane DBC is at right angles to the plane AEG. For the same reason, the plane DBC' is at right angles to the plane AFG, and therefore AG, the common section of the planes AFG, AEG is at right angles 18 2. Sup.) to the plane DBC. and the angles AGE, AGF are consequently right angles. But since the arch AB is equal to the arch AC, the angle ADB is equal to the angle ADC. Therefore the triangles AIlE, AD3, havt the angles EDA, FDA, equal, as also the angles AE D, A FD, which are: right angles; and they have the side AD common, therefore the other sides are equal, viz. AE to AF, (26. 1.), and DE to DF. Again, be. cause -the angles AGE, AGF are right angles, the squares on AG and G.E are equal to the square of AE; and the squares of AG and GF to the square of AF. But the squares of AE and AF are equal, there: fore the squares of AG and GE are equal to the squares of AG and GF, and taking away the common square of AG, the remaining squlares of GE and GF are equal, and GE is therefore equal to GF. WhereLore, in the triangles AFG, AEG, the side GF is equal to the side GE; and AF has been proved to be equal to AE, and the base AG is comrn xnon; therefore, the angle AFG is equal to the angle AEG (8. 1.)I iBut.the angle AFG is the angle which the plane A DC makes with the plane DBC (4. def. 2. Sup.) because FA and FG, which are drawn in these planes, are at right angles to DF, the common section of the planes. The angle AFG (3. def.) is therefore equal to the spherical angle ACB; and, for the same reason, the angle AEG is equal to thl spherical angle ABC. But the angles AFG, AEG are equal. Therec fore the spherical angles ACB, ABC are also equals Q. E. D. PROP. VI. If the angles at the base of a spherical triangle be equal, the triangle is isosceles. Let ABC be a spherical triangle having the angles ABC, ACB equal to one another; the sides AC and AB are also equal. Let D be the centre of the sphere; join DB, DC, DA, and from A on the straight lines DB, DC, draw the perpendiculars AE, AF; and from the points E and F, draw in the plane DBC the straight lines LG, FG A perpendicular to DB and DC, meeting one another in G; join AG. -Then, it may be proved, as was done f in the last proposition, that AG is at right angles to the plane BCD, and that, therefore the angles AGF, AGE are right angles, and also that the angles AFG, AEG are equal'to the angles Jid which the planes DAC, DAB make - SPHERICAL TRIGONO~METR-Y 21. with the plane DBC. But because the spherical angles ACB, ABC are equal. the angles which the planes DAC, DAB make with the plane DBC are equal, (3. def.) and therefore the angles AFG, AEG are also equal. The triangles AGE, AGF have therefore two angles of the one equal to tlo angles of the other, and they have also the Mide AG common, wherefore they are equal, and the side AF is equal to the side AE. Again, because the triangles ADF, ADE are right angled at F and E, the squares of DF and FA are equal to the square of DA, that is, to the squares of DE and DA; now, the square of AF is equal to the square of AE, therefore the square of DF is equal to the square of DE, and the side DF to the side DE. Therefore, in the triangles DAF, DAFE, because DF is equal to DE and DA common, and also AF equal to AE, the angle ADF is equal to the angle ADE; therefore also the arches AC and AB, which are the measures of the angles ADF and ADE, are equal to one another; and the triangle ABC is isosceles. Q. E. D. PROP. VII. Jiny two sides of a spherical triangle are greater than the third. Let ABC be a spherical triangle, any two sides AB, 13BC are great er than-the third side AC. Let D be the centre of the sphere; join DA, DB, DC. Tile solid angle at D is contained by'three plane angles ADB, ADC, \ ]BDC; any two of which, ADB, BPC are greater (20. 2. Sup.) D -| than the third ADC; and therefore -A any two of the arches AB, AC, BC, Which measure these angles, as AB and BC, must also be greater than the third AC. Q. E. D. PROP. VIIL The three sides of a spherical triangle are less t'han the circulmference of a great circle. Let ABC be a spherical triangle as before, the three sides AB, BC, AC are less than the circumference of a great circle. Let D be the centre of the sphere: The solid angle at D is contaidied by three plane angles BDA, 1BDC, ADC, which together are less than four right angles (21. 2. Sup.) therefore the sides AB, BC, ACt which are the measures of these angles, are together less than fout quadrants described with the radius AD, that is, than the circuitb fegrnce of a great circle. Q4. E. D. 2-2 SP:HERICAL TRIGONOMETRY. PROP. IX. In a spherical triangle the greater angle is opposite to the greater side; and conversely. Let ABC be a spherical triangle, the greater angle A is opposed to the greater side BC. Let the angle BAD be made equal to the angle B. and ther BED, DA will be equal, (6.), and therefore AD, DC are equal to BC;but AD, DC. are greater than AC (7.), therefore, BC is greater than AC, that is,the greater angle A is opposite to the greater side BC. The converse is demonstrated as Prop. 19. 1,:lem, Q. Ed UD PROP. X..4Icco ding as the sum oJf two of the sides of a spherical triangle, is grenaz er than a semicircle, equal to it, or less, each of the.interior angles at the base is greater than the exterior and opposite angle at the basej equal to it, or less; and also the sum of the two interior angles at the base greater than two right acrles, equal to two right angles, or less than two right anglesosl Let ABC be a spherical triangle, of w-lich the sides are AB and BC; produce any of the two sides as AB, and the base AC, till they meet again in D; then, the arch ABD is a semicircle, and the spheri. cal angles at A and D are equal, because each of them is the inclination of the circle ABD to the circle ACUD. 1. If AB, BC be equal to a semicircle, that is, to AD:, iC will be equal to BD, and therefore (5.) the angle D, or the angle A will be equal to the angle BCD, that is, the interior angle at the base equal to the exterior and opn C posite. 2. If AB, BC together be greater than a semicircle, that is, greitei than ABD, BC will be greater than BD; and therefore (9.), the angle D, that is, the angle A, is greater than the angle BCD 3. In the same manner it is shown, if AB, BC together be less thak' a semicircle, that the angle A is less than the angle BCD, SPHERICAL TRIGONOMETRYo r25i Now, since the angles BCD, BCA are equal to two right angles, if the angle A be greater than BCD, A and ACB together will be greater than two right angles. If A be equal to BCD, A and ACB toge. ther, will be equal to two right angles; and if A be less than BCID, A and ACB will be less than two right angles. Q. E. D. PROP. XI. If the.angular ponlts of a spherical triangle be made the poles of three great circles, these three circles by their intersections willform a triangle, which is staid to be supplemental to the former; and the two triangles are such, that the sides of the one are the supplements oJ the arches which measure the angles of' the other. Let ABC be a Spherical triangle; and from the points A, B, and C das poles, let the great circles FE, ED, DF be described, intersecting one another inr F, D and E;i the sides of the triangle FED are the supplements of the measures of the angles A, B, C, viz.;FE of the angle BAG, DE of the angle ABC, and DF of the angle ACB: And again, AC is the supplement of the angle DFE, AB of the angle ED, and BC of the angle EDF. Let AB produced meet DE, EF in G, M; let AC meet FD, FE in K, L; and let BC meet FD, DE in N, H. Since A is the pole of FE, and the circle AC passes through A, EF will pass through the pole of AC (1. Cor. 4.) and since AC passes through C, the pole of FD, FD will pass through the pole of AC; therefore the pole of AC T is in the point F, in which the arches DF, EF intersect each other. In the D s E same rmanner, D is the pole of BC, and E E the' pole of AB. And since F, E are the poles of AL, AM, the arches FL and ElII (2.) are quadrants, and FL, EM together, that is, FE and ML together, are equal to a semicircle. But since A is the pole of MLI, ML is the measure of the angle BAC (3.), consequently FE is the supplement of the measure of the angle BAC. In- the same manner, ED, DF are the supplements of the measures of the angles, ABC, BCA. Since likewise CN, BH are quadrants, CN and BR: together, that is, NH and BC together, are equal to a semicircle; and since D is the poie of NH, NH is the measure of the angle FDE, therefore the measure of the angle FDE is the supplement of the side BC. In the same manner, it is -shown that the measures of the angles DEF, EFD are the supplements of the sides AB, AC in the triangle ABC. Q. E. D. Air4 SPHERICAL TRIGONOMETRYX PROP. XII. The three angles of a spherical triangle are greater than two, and less than six,, right angles. The measures of the angles A, B, C, in the triangle ABC, together with the three sides of the supplemental triangle DEF, are (II11.) equal to three semicircles; but the three sides of the triangle FDE, are (8.) less than two semicircles; therefore the measures of the angles A, B; C are greater than a semicircle; and hence the angles A, B, C are greater than two right angles. And because the interior angles of any triangle, together with the exterior, are equal to six right angles, the interior alone are less than, six right angles. Q. E. D. PROP. XIII. If to the circumference'of a great circle, from a point, in the surface of the sphere, which is not the pole of that circle, arches of great circles be drawn; the greatest of these arches is that which passes through the pole of the first-mentioned circle, and the supplement of it is the least; and of the other archtes, that which is nearer to the greatest is greater than that which is more remote. Let ADB be the circumference of a great circle, of which the pole is H, and let C be any other point; through C and H let tih semicircle ACB be drawn meeting the circle ADB in A and B; and let the arches CD, CE, CF also be described. From C draw CG perpendicular to AB, and then, because the circle AHCB which passes through H, the pole of the circle ADB, is at right angles to ADB, CG is perpendicular to the plane ADB. Join GD, GE, G]F? CA, CD, CE, CF, CB. Because AB is the diameter of the circle ADB, and G h point in it, which is not the centre, (for the centre is in the point where the perpendicular from H meets 1 / AB), therefore AG, the part of the diameter in which the centre is, is the greatest, (7. 3.), and GB N / the least of all the straight lines that can be drawn from G to the circumference; and GD, which is E nearer to AB, is greater than GE, which is more remote.; But the triangles CGA, CGD are right angled at G, and therefore AC2AG-+GC2, and DCa - DG' GC2; but AG-'+GC+7 DG2+G C2; because AG7DG; therefore AC7DC2, and AC 7DC. And because the chord AC is greater than the chord C:, the arch AC X -SPHERICAL TRIGONOMEThRY. b5-greater than the arch DC. In the same manner, since GD is greater.than GE, and GE than GF, it is shown that CD is greater than CE, and CE than CF. Wherefore also the arch CD is greater than the arch CE, and the arch GE greater than the arch CF, and CF than CB, that is, of all the arches of greater circles drawn from C to the circumference of the circle ADB, AC which passes through the pole H, is the greatest, and UB its supplement is the least; and of the others, that which is nearer to AC the greatest, is greater than that which is more remote. Q. E. D. PROP. XIV. il' a right angled spherical triangle, the sides containing the right angle are of the same affection with the angles opposite to them, that is, if the sides be greater or less than quadrants, the opposite angles zzeil be greater or less than right angles, and conversely. -Let ABC be a spherical triangle, right angled at A, any side AB will be of the same affection with the opposite angle ACB. Produce the arches AC, AB, till they meet again in D, and bisect AD in E. Then ACD, ABD are semicircles, and AE an arch of 90". Also, because CAB is by hypothesis a right angle, the plane of the circle ABD is perpendicular, -to the plane of the circle, r A.CD, iso that the pole of C ACD is in ABD, (cor.. 4.), andis therefore the point E.'Let EC be an arch of A7N a great circle passing throiagh E and- C. Then because E is the pole of the circle ACD, EC E is a (2.) quadrant, and the plane of the circle EC (4.) is at right angles to the plane of the circle ACD, that is, the spherical angle ACE is a right angle; and therefore, A. when AB is less than AE, the angle ACB, being less than ACE, is less than a right angle. But when AB E isgreaterthanAE, the angle ACB is greater than ACE, or than a right angle. In the same way may the converse be demonstrated. Therefore, &c. Q.E. D. at~Di SPHERICAL TR[GONOMETRYL PROP; XY. If the t, o sides of a right angled sph7erical triangle about the right alit gle be of the same affection, the hypotenuse will be less than a quadrant; and if they be of different affection, the hypotenuse will be greater than a quadrant. Let ABC be a right angled spherical triangle; according as the two sdes AB, AC are of the same-or of different affection, the hypoteo -nyse BC will be less, or greater than a quadrant. The construction of the last proposition remaining, bisect the semicircle ACD in G, then AG will be an arch of 90Q, and G will be the pole of the circle ABD. I. Let AB, AC:be each less than 90o. Then, because C is a point on the surface of the sphere, which is not the pole of the circle ABD, the arch CGD, which passes through G the pole of ABD is greater than CE,: (13.), and CE.greater than CB. But CE is a quadrant,' as was before shown, therefore CB is less than a quadrant. Thus also it is proved of the right angled triangle CDB, (right angled at D), in which each of the sides CD, DB is greater than a quadrant, that the hypotenuseBC is less than a quadrant. 2. Let AC be less, and AB greater than 900. Then because CB falls between CGD and CE, it is greater (13.) than CE, that is, than a quadrant. Q. E. D. Con. 1. Hence conversely, if the hypotenuse of a right angled trio angle be greater or less than a quadrant, the sides will be of different or the samie affection. Con. 2. Since (14.) the oblique angles of a right angled spherical triangle have the same affection with the opposite sides, therefore, according as the hypotenuse is greater or less than a quadrant, the oblique angles will be of different, or of the same affection. CoR. 3. Because the sides are of the same affection with the oppo., site angles, therefore when an angle and the side adjacent are of' the same affection, the:hypotenuse is less than a quadrant -and converselyv PROP. XVL, in any spherical triangle, if the perpendic-ular upon the basefrointh6 op, posite angle fall within the triangle, Ithe angles at the base are- of the same affection,; atd if the perpendicular fall without the triangle, the angles at the base are of differenat ctS/ction, Leet ABC be a spherical triangle, and let the arch CD be drawX, from C perPendicu!ar to the base AB, SPHERICAL TRIGONOMETRY. 257 i. Let CD fall within the triangle; then, since ADC, BDC are iight angled spherical triangles, the angles A, B must each be of thp rsameaffection with CD, (14.). A. 2. Let CD fall without the triangle; then (14.) the angle B is of'the same affection with CD; and the angle CAD is of the same affec(ion with CD; therefore the angle CAD and B are of the same af~ection, and the angle CAB and B are therefore of different affictions. 4. E. D. Cor., Hence, if the angles A and B:be of the same affection, the perpendicular will fall within the base; for if it did not, A and B would be of different affection. And if the angles A and B be of different affection, the perpendicular will fall without the triangle; for, if it did not, the angles A and LB would be of the same affection, contrary to the suppositiono PROP. XVIL if to the base of a stherical trian'gle a perpendicular be drawn from the opposite angle, which either falls within the triangle, or is the nearest of th-e two thatfall without; the least of the segfrments of the base is adjacent to the least of the sides of the triangle, or to the greatest, ac. -cording as the sum of the sides is less or greater than a semicircleo Let ABEF be a great circle of a sphere, H its pole, and GHD any circle passing through H, which therefore is perpendicular to the circle ABEF. Let A and B be two points in the circle ABEF, on oppcsite sides of the point D, and let. D be nearer.to A than to B, and let C be any point in the circle GHD between H and D. Through the points A and C, B aind C, let E the arches AC and BC be drawn, and let them be produced till they meet the circle A BEF in the points D E and'F, then the arches AUE, BCF are semicircles. Also ACB, ACF, CFE,:ECB, are four spherical triangles contained by arches Aa:of the same circles, and having the. same perpendiculars CD and CG, t5S SPiHERICAL TRIG' ONOM JE2Rio 1. Now because CE is nearer to the arch CtIG than CB is, Ci is greater than CB, and therefore CE and CA are greater than CB and CA, wherefore CB and CA are less than a semicircle; but because AD, is by supposition less than DB, AC is also less than CB, (13.), and therefore in this case, viz. when' the perpendicular falls within the triangle, and when the sum of the sides is less than a semicircle, the last segment is adjacent to the least side. 2. Again, in the triangle FCA the two sides FC and CA are less than a semicircle; for since AC is less than CB, AC and CF are less than BC and CF. Also, AC is less than CF, because it is more remote from CHG than CF is; therefore in this case also, viz. when the perpendicular falls without the triangle, and when the sum of the sides is less than a semicircle, the least segment of the base AD is adjacent to the least side. 3.- But in the triangle FCE the two sides FC and CE are greater than a semicircle; for, since FC is greater than CA, FC and CE are greater than AC and CE. And because AC is less than CB, EC is greater than CF, and EC is therefore nearer to the perpendicular MCHG than CF is, wherefore EG is the least segment of the base, and is adjacent to the greater side. 4. In the triangle ECB the two sides EC, CB are greater than a semicircle; for, since by supposition CB is greater than CA, EC and CB are greater than EC and CA. Also, EC is greater than CB, wherefore in this case, also, the least segment of the base EG is adjacent to the greatest side of the triangle. Therefore, when the sum of the sides is greater than a semicircle, the least segment of the base is adjacent to the greatest side, whether the perpendicular fall within or without the triangle:: and it has been shown, that when the sum of the sides is less than a semicircle, the least segment of the base is adjacent to the least of the sides, whether the perpendicula. fall within or without the triangle. Wherefore, &c. Q. E. D. PROP. XVIIIo Eni right angled spherical triangles, the sine of either of the sides aboug the right angle, is to the radius of the sphere, as the tangent,f the ~retnaining side is to the tangent of the angle opposite to that side. Let ABC be-a triangle, having the right angle at A; and let AB be either of the sides, the sine of the side AB will be to the radius, as the tangent of the other side AC to the tangent of the angle ABC, opposite to AC.., Let D be the centre of the sphere; join AD, BD, CD, and let AF be drawn perpendicular to BD, which therefore will be the sine of the arch AB, and from the point F, let there be drawn in the plane BDC the straight line FE at right angles to BD, meet-'g DC in E, and let AE be joined, Since therefore the straight SPHERICAL TRIGONOMETRY- 259!ine DE is at right angles to both FA E and FE, it will also be at right angles to the plane AEF (4. 2. Sup.); where/ fore the plane ABD, which passes - through DF, is perpendicular to the plane AEF (17. 2. Sup.), and the plane AEF perpendicular to ABD: But the plane ACD or AED, is also perpendicular to the same ABD), be- cause the spherical angle BAC is a right angle: Therefore A E, the common section ofthe planes A ED:, A EF, is at right angles to the plane ABD, (18. 2. Sup.), and EAF, EAD are right angles. Therefore AS is the tangent of the arch AC; and ia the rectilineal triangle AEF, having a right angle at A, AF is to the radius as AP to the tangent of the angle AFE, (I. P1. Tr.); but AF is the sine of the arch AB, and A E the tangent of the arch AC;- and the angle AFE is the inclination of the planes CBD, ABD, (4. def. 2. Sup.), or is equal to the spherical angle ABC: Therefore the sine of the arch AB is to the radius as the tangent of the arch AC to the tans gent of the opposite angle ABC. Q. E. D. Con. Since by this proposition, sin AB: R:: tan AC: tan ABC and because R: cot ABC:: tan ABC: R (I. Cor. def. 9, PI. Tr.) by equality, sin AB: cot ABC:: tan AC: R. PROP. XIX.:t right angled sphiericul triangles the sine of the hypotenuse is to the radius as the sine of either side is to the sine of the angle opposite to that side. Let the triangle ABC be right angled at A, and let AC be either of the sides; the sine of the hypotenuse BC w1ll be to the radius as the sine of the arch AC is to the sine of the angle ABC. Let D be the centre of the sphere, and let CE be drawn perpend"i cular to )B, which will therefore be the sine of the hypotenuse BC, and from the point E let there be drawn in the plane ABDthe straight line EF perpendicular to DB, and let CF be joined: then CF will be at -right angles to the plane ABD, because as was shown of EA in the preceding proposition, it is the comrnmon section of two planes -DCF, ECF, each perpendiculat to the plane ADB. Wherefore CFD,CFE - are right angles, and CF is the sine of the arch AC; and in th.e triangle CFE having the right angl.e CFE s~to260;SPHERICAL trIGONOMETRi. CE is to the radius, as CF to the sine of the angle CEF (1. PI. Tr. But, since CE, FE are at right angles to DEB, which is the commoln section of the planes CBD, AB D, the angle CEF is equal to the inclination of these planes, (4. deft 2. Sup.), that is, to the spherical angle ABC. Therefore the sine of the hypotenuse CB, is to the radius, as the sine of the side Ax. to the sine of the opposite angle ABC. Q. E. I) PROP. XX. in right angled s.hercial triangles, the cosine of the hypotenuse is to tehi radius as the cftanmgent of either of the angles is to the tangent of the remaining angie& Let ABC be a spherical triangle, having a right angle at A, the co. sine of the hypotenuse BC is to the radius as the cotangent of the an. gle ABC to the tangent of the angle ACB. Describe the circle DE, of which B is the pole, and let it meet AC in F, and the circle BC in E; and since the circle BD passes through ihe pole B, of the circle DF, DF must pass through the pole of' BP. (4.). And since AC is perpendicular to BD, the plane of the circle AC is perpendicular to the plane of the circle BAD, and therefor6 AC must also (4.) pass through the pole of BAD; wherefore, the pole of the circle BAD is in the point F, where the circles AC, DE, intersect. The arches FA, FD are therefore quadrants, and likewise the arches BD, BE. Therefore, in the triangle CEF, right angled at the point E, CE is the complement of BC, the hypotenuse of the triangle ABC! EF is the complement of the arch ED. the measure of the angle ABC, and FC, the hypotenuse of the triangle CEF, is the complement of ACj and the arch AD, which is the measure of the angle CFE, is the complement of AB. But (18.) in the triangle CEF, sin CE: R:: tan EF: tan ECF, that is, in the triangle ACB, cos BC:R:: cot ABC: tan AGBC Q. Elk Do SPH ER ICL TRiGONOM i TR,' 2Ct6 Cox. Because cos BC: R:: cot ABC: tan ACB, and (Cor. 1. deft 9. PI. Tr.) cot ACB::: R: tan ACB, ex coquo, cot ACB cos BC R: cot ABC. PROP. XXI. In right angled spherical triangles, the cosine of an anagle is to the radius as the tangent of the s:d'e adjacent to that angle is to tde tangent of the hypotenuse. The same construction remaing; In the triangle CEF, sin FE: U(:: tan CE: tan CFE ( 8.); but sin EF-=cos ABC; tan CE-cot 1C, and tan CFE = cot AB, therefore cos ABC R:: cot BC: cot AB. Now, because (Cor. 1. def. 9, PI. Tr.) cot BC: R:: R: tan BC, and cot ABS::: R: tan AB, by equality inversely, cot BC: cot AB:: tan ABE: tan BC; therefore (11. 5.) cos ABC: R:: taa AB; tan BC. Therefore, &c. Q. D. \ Con. 1. From the demonstration it is manifest, that the tangents 6o any two arches AB, BC are reciprocally proportional to their cotatn~ gents. COR. 2. Because cos ABC: R:: tan ABI: tan BC, and R: cos BC:. tan BC: R, by equality, cos ABC: cot BC:: tan AB: R, That is, the cosine of any of the oblique angles is to the cotangent of the hypotenuse, as the.angent of the side adjacent to the angle is to the radius. PROP. XXII. In right angled spherical triangles, the cosine of either of the sides is to the.radius, as the cosine of the hypotenuse is to the cosine of the other side. The same construction remaining: In the triangle CEF, sin CF R::sin CE: sin CFE, (19.) but sin CF = cos CA, sin CE=cos WBC, and sin CF —co, AB ~ therefore cos CA ~ R:: Cos BC: cos _ ED 26(2 SPHISERICAL TRIGONOMETRY. PROP. XXIII. In right angled spherical triangles, the cosine of either of the sides is ti the radius, as the cosine of the angle opposite to that side is to the sine of the other angle. The same construction remaining: In the triangle CEF, sin CF R:: sin EF: sin ECF, (19.); but sin CF = cos CA, sin EF-cos ABC, and sin ECF-sin BCA: therefore, cos CA: R:: cos ABC sin BCA. Q. E. D. PROP. XXIY. 7n spherical triangles, whether right angled or oblique angled. the sines of the sides are proportional to the sines of the angles opposite to them. First, Let ABC be a right angled triangle, having a right angle at A; therefore, (19 ) the sine of the hypotenuse BC is to the radius4 (or the sine of the right angle at A), as the sine of the side AC to the C sine of the angle B. And, in like manner, the sine of BC is to the sine of the angle A, as the sine of AB to the sine of the angle C; wherefore (11. 5.) the sine of the side AC is to the sine of the angle B, as the sine of AB to the sine of the angle C. Secondly, Let ABC be an oblique angled triangle, the sine of any of the sides BC will be to the sine of any of the other two AC, as the side of the angle A opposite to BC, is to the sine of the angle B opt posite to AC. Through the point C, let there be drawn an arch of a great circle CD perpendicular to AB; and in the right angled trian. C' 3C B A k gle BCD, sin BC:R: sin CD: sin B, (19.); and in the triangle ADC, sin AC: R:: sin CD: sin A; wherefore, by equality inversely, sin BC: sin AC:: sin A: sin B. In the same manner, it may be proved that sina BC;,sin AB:; sin A v -in C, &e. Thrfoere, & Qb EJ SPHERICAL TRItGONOMETRY.o 263 PROP. XXV. ka oblique angled spherical triangles, a perpendicular arch being drawzs from any of the angles upon the opposite side, the cosines of the angles at the base are proportional to the sines of the segments of the vertical angle. Let ABC be a triangle, and the arch CD perpendicular to the base BA; the cosine of the angle B will be to the cosine of the angle A,.as the sine of the angle BCD to the sine of the angle ACD). For having drawn CD perpendicular to AB, in the right angled triangle BCD, (t23.) cos CD:: cos B: sin DCB; and in the right angled triangle ACD, cos CDI R:: cos A: sin ACD; therefore (11. 5.) cos B: sin DCB:: cos A: sin ACD, and alternately, cos B cos A:: sin BCD sin ACD. Q.E. D. PROP. XXTI. The same things remnaiing, the cosines of the sides BC, CJl, are proport tional to the cosines of BD, DA, the segments of the base. For in the triangle BCD, (22.), cos BC: cos BD': cos DC: R and in the triangle ACD, cos AC: cos AD:: cos DC: R; therefore (11. 5.) cos BC: cos BD:: cos AC: cos AD, and alternately, cos BC: cos AC:: cos BD: cos AD. Qo E. D. PROP. XXVII.'he same construction eremaining, the sines of BD, DA the segments otf the base are reciprocally proportional to the tangents of B and., the angles at the base. In the triangle BCD, (18.), sin BD: PtF tan DC e tan B; and iza the triangle ACD, sin AD: R:: tan DC: tan A; therefore, by equal: ity inversely, sin BD ~ sin AD:.: tan A: tan B. Q. E. D. CC / /1\\ _ Z 264 SPHERICAL TRIGONOMETRY PROP. XXVIII. Dse same construction remcaining, the cosines of the segmnentls of the.'ver tical angle are reciprocally proportional to the tangents of the sides. Because (21.), cos BCD: R:: tan CD tan BC, and also cos ACD: R: tan CD: tan AC, by equality inversely, cos BCD: cos ACD tan AC: tan BC. Q. E. D. PROP. XXIXo if from an angle of a spherical triangle there be drawn a perpendicu' lar to the opposite side, or base, the rectangle contained by the tangents of half the sum, and of half the diference' of the segments of the base is equal to the rectangle contained by the tangents of half the sum, and of half the difference of the two sides of the triangle. Let ABC be a spherical triangle, and let the arch CD be drawn from the angle C at right angles to the base AB, tan - (m-+n) xtan t (rn-n) —{ tan (a+b) X- tan (a-b). Let BC=a, AC= —b; DD=m, AD-=n. Because (26.) cos a: cos b-. cos m: cos n, (E. 5.) cos a+b cos a-cos b:: cos rm+cos n cosm -cos n. But ( I. Cor. 3. P!. Trig.), cos a+cosb: cos a-cos b cot (a+b): tan - (a-b), and also, cos rn+cos n: cos m-cos rb cot I- (m+n): tan 1- (rn-n). Therefore, (l1. 5.) cot 2 (a4-b) tan - (a-b): cot I (m-in): tan I (m-n). And because rectangles of the same altitude are as their bases, tan 2 (a+b) X cot 4 {(a+b): tan - (a+b)X tan 2 (a-b):: tan a (m'+n)X cot 4 (rn+n) tan - (.mXn)+tan - (m —n). Now the first and third terms of this proportion are equal, being each equal to tihe square of the radius, (I. Cor. P1. Trig ), therefore the remaining two are equal, (9. 5.) or tan -s (m a-n)X tan 4- (m-n) tan n- (a-b)X tan t (a —b); that is, tan { (BD+AD) X tan - (BD -AD) - tan 4 (BC-+AC) X tan (BC —AC). Q.E.D. Con., 1. Because the sides of equal rectangles are reciprocally proportional, tan I (BD+AD): tan - (BC+AC) - tan - (BC-AC) tan I (BD- AD). CoR. 2. Since, when the perpendicular CD falls within the triangle, BD+AD = AB, the base; and when CD falls without the triangle BD- AD-AB, therefore in the first case, the proportion in the last corollary becomes tan - (AB): tan - (BC + AC):: tan - (BC-AC): tan 4- (BD -AD); and in the second case, it becomes by inversion and alternation, tan k (ARB) tan4-(0C+ AC) ta tan,BC.-A A: tan 4 (BD~:AD),." I t J- J 3 4'a SPHERICAL TRIGONOMET RY. C; A.s A SCHOLIUM. THE preceding proposition, which is very useful in spherical trigonometry, may be easily remembered from its analogy to the propossi. tion in plane trigonometry, that the rectangle under half the sum, and half the difference of the sides of a plane triangle, is equal to the rectangle under half the sum, and half the difference of the se mnents of the base. See (K. 6 ), also 4th Case PI. Tr. We are indebted to NAPIER for this and the two following theorems, which are so well adapted to calculation by Logarithms, that they must be considered -g, three of the most valuable propositions in Trigonomntly? PROP. XXX. f't a perpendicuiar be draws froma an a gle of a sphericol triaglee the,opposite side or base, the sine of the sum of the angles at the base is to the sine of their difference as the tangent of haIf the base to the, tangent of half the differen ce of its segments, when the perpendicularfalls with. in; but as the co- tangent l o half the base to the co-tangent of half the sum of the segments, wz~hen the perpendicular falls?litho ut the triangte: Jlnd the sine of itk sum of the twzo sides is to the sine of their dfference as the co-tangent of half the angle contained by the sides' to the tangent of half the difference of the angles which the perpendi. cular makes with the same sides, w'hen it falls within, or to the tangent of half the sum of these ang.leS, wh'en itfalls without the triang-le If ABC be a spherical triuagle, and AD a perpendicular to the base BC, sin (C+B): sin (C.-B): tan - BC tan ~ (BD-DIC), when AD falls within the triangle; but sin (C-l-B): sin (C —B): cot 7 BC'.cot (BD+-DC), hiion AC fhlls wihout. u And algain, L! 266t SPIfERCAL TRIGONOMElTRY D sin (AB+AC): sin (AB —AC):: cot 2 BAC: tan - (BAD-CAID) when AD falls within: but when AD falls without the triangle, sin (AB+AC): sin (AB-AC):: cot BAC: tan i (BAD+CAD). Fori in the triangle BAC (27.), tan B: tan C:: sin CD: sin BDI and therefore (E,; 5.), tan C-+tan B: tan C-tan B:: sin BD,+-sin CD: sin BD -sin CD. Now, (by the annexed Lemma) tan C+tan B: tan C —tan B:: sin (C+-B): sin (C-B), and sin BD+-sin CD: sin BD -sin CD:: tan - (BD+CD): tan I- (BD —CD), (3. P1. Trig.)) therefore, because ratios which are equal to the same ratio are equal to one another (11. 5.), sill (C-+B): sin (C -B):: tan - (BD+CD) i tan ~ (BD.CD)). e s t tC iNow when AD is within the triangle, BD-CDBC, and therefore in (C-B): sin (C -B):: tan BC: tan - (BD-CD). And again when AD is without the triangle, BD CD=BC, and therefore sin (C'+B): sin (C -LB):: tan (BD+CD): tan 2 BC, or because the tangents of any two-arches are reciprocally as their co-tangents, in (C+B): sin (C-B):: cot ~ BC: cot - 12UD+CD)-The second part of the proposition is next to be demonstrated. Be= cause (Sf3.) tan AB: tan AC:: cos CAD: cos BAD, tan AB-tan AC tan AB tan AC:: cos CAD+cos BAD: cos CAD-cos BAD~ But (Lemma) tan AB+tan AC: tan AB-tan AC:: sin (AB+AC): sin (AB- AC), and (1. cor. 3. P1. Trig.) cos CAD+-cos BAD: cos CAD-ecos BAD:: cot I (BAD+CAD): tan- (BAD -CAD). Therefore (11.5.) sin (ABl-AC): sin (AB.-AC):: cot (BAD+CAD): tan (BAD- CAD). Now, when AD is within thetriangle, BAD+I CAD=BAC, and therefore sin (AB+AC): sin (AB —AC):: c- ot/: -1AC: tan I (BA D-. CAD9\' ,SPHERICAL TRIGONOlIETRi L 2Y.6 But if AD be without the triangle, BAD - CAD =BAC, and therefore sin (AB+AC): sin (AB-AC):: cot 2 (BAD+CAD): tan - BAC; or because cot _ (BAD+CAD): tan BAC:: cot I BAC tan f (BAD+CAD), sin (AB+AC): sin (AB —AC) cot 2 BAC: tan ~ (BAD+CAD). Wherefore, &c. Q. E. D. LEMMA. ~~'!he sum of the langents of any two arches, is to Ihe dierence of thei;r tangents, as the sine of the sulm of the arekes, to the sine of their dif. fereice. Liet A and B be two -r,4es, tan A - tan B: tan A - tan B:: sin (A+B)::(A-B). For, by 6~ p. )' 243, sin AXcos B+cos AXsin B=sin (A+B), sin A sin B and theref —- dividing all by cos A cos B, s + B n..(A+B) sie a.co-SXcosB' that is,-because tan A, tan A+tan B cos A X ~os B' cos A $in (A+ B):-ie A.+... In the same manner it is proved that tan A-tan B cos A Xcos-B' sin C(A-'B)' Therefore tan A + tan B: tan A-tan B: dn cos A-X cos B (A+JB): sin (A-B) Q. E. D. PROP. XXXI. _The sine of half the sum of any ttwo angles of a spherical triangRe is to the sine of'half their difference, as the tangent of half the side adjacent to these angles is to the tanrgent of half the difference of the sides op. posite to them.; and the cosine of half the sufm of the same angles is to the Cosine of half their djlerence, as the tanlgent of half the side adacet to them, to the talgent of half the sum of the sides opposite. Let C+B=2S, C -B=2D, the base BC=2B, and the difference of the segments of the base, or BD —CD=2X. Then, because (30.) si (C+B): sin (C- B):: tan - BC: tan } (BD —CD), sin 2S sin 2D:: tan B:. tan X. Now, sin 2S-sin (S+S)=2 sin S X cos S, (Sect. III. cor. P1. Tr.). In the same manner, sin 2D -2: sin D X eos D:. Therefore sin S X cos S: sin D X cos D;:n tan Xt C, Again, in the spherical tliangle ABC it has been proved, that si4 C(+sin B:sin C-sin B: si AB-sin AC: sin AB-sin AC, and since sin C +-sin B=2 sin (C~+B)+cos, (C- B), (Sect. II. 7. Pl. Tr.)=G2 aia S Xcos D; arid sin C-sin B=2 cos, (C+B) x sin 4 (C-B) = 2 cos S X sin D. TI lrefore 2 sin S Xcos D: 2 cos SX sin D:: sin AB + sin AC; sin AB -,;u AC. But (3. P1. Tr.) sin AB+sin AC:sin AB -sin AC:: tan (AB- XC) tan (AB - AC) -2,) -a, (AB -'AC): tan:. tan A, I beirngequal to I (AB+AC) )a ato ~ (AB-AC). Therefore sin S Xcos D: cos S Xsin D:: tan I: Lt a. Since then tan X sin D Xcos D tan a cos $Xsin D tan Bsin S > cos n tan sin os D' by rmUlltvinsg X an X tan A (sin D)j2'Xcos S'Xcos D (sin DBy quals by equals, tan tan B tan (sin S)2' Xcos S Xcos D (sin S)2' tar - (BD-DC)_ tan (AB+AC) tan X tan: u tan - (AB- AC) tan 1- BC tisiA taBtan9 tan X tan X tanA tan X tan A tan Aa and therefore, tas also tan;B (tan B)2 atan'B tan X (tal B)2 tan X tan A) (sin ) (tan _(sin tan But a(-X (ahence sin and tan B ta-nY ('sin 8)2 c (tan B) (sin S)2 tan B sin D.,irsinS: sin D tan B:tani A, that is, sin (C+B): sin sin S (C -B) tan BC ta.n - (AB-AC); which is the first part of tile proposition. tanA cos SXsin D tan s sin SXcos D A'.gain, since -r inversel tan I sin SXcos D d tanA cos S sinD - tanX sin DX cosD. tan X.and since t X therefore by multiplication, X tan B sin 8 X cos S Y tan B tan I: (cos D)S lan a (css,)d, But it Was already shown that tan X tan fXtan a tan B (tan B)2, wherefoe alsi.an X tan I (tan y)2 tan B tan A (tan B)2 tanX tan X_(cos_'ow n, i Xtaa (eos S)' as has just been shown, _erfore(cos D)' (tan )a2 Cos D tan I eos.-) - (tan Ba anc tonsequently o-' or 0 S'HER1ICAL TITGI ONorMd:fETRYL.ai S': cos D:: tan B: tan y, that is, cos (C-I-B): cos (C-B)::tan 2 BC: tan (C + B); which is the second, part of the proposition. Therefore, &c Q. E. D. COR. 1. By applying this proposition to the triangle supplemental. t6 AB J (I. 3.), and. by considering, tlrhat the sine of half the sum or half the difference of the supplements of two arches, is the same with the sine of half the sum or half the difference of the arches themselves: add'that the same is true of the cosines, and of the tailgents of half the sum or half the differernce of the supplements of two arches: but that the tangent of half the supplement of an arch is the same with the cotangent of half the arch itself; it will follow, that the sine of half the sum of any two sides of a spherical triangle, is to the sine of half their difference as the cotangent of half the angle contained between them, to the tangent of half the difference of the angles opposite to them: and also that the cosine of half the sum of these sides, is to the cosine of half their difference, as the cotangent of half the angle contained between them, to the tangent of half the sum of the angles opposite to them. Con. 2. If therefore A, B, C be the three angles of a spherical triangle, a, b,'c the sides opposite to them, 1T. sin (A+B): sin I (A —B):' tn I c: tan ~ (a —-b) II. cos 2 (A+B): cos I (A-.B):: tan' c: tan I (a+b). IIt. sin - (a+b): sin I (a-b)::tan C: tan Ct (A-B). IV. cos' (a+b): cos 4 (a- b): tan I C: tan (A+-B) 70 $i1tPHERICAL TRL(ONtOi LETRJI, PROBLEM Io in a right angled spherical triangle, of the three sides and thn:ee angle s any two being given, besides the right angle, to find the other.three. This problem has sixteen cases, the solutions of which are contained in the following table, where ABC is any spherical triangle right angled at A. GIVEN. SOUGHT. i SOLUTION. AC. R: sin BC sin B' sin_ AC, (19). BC and B AB. R: cos B': tan BC: tan AB, (21). 2 C. R: cos BC:: tanB: cot C, (20). 3 AB. R: sin AC::tanC: tan AB, (18). 4 AC and C. BC. cos C: R': tan AC:tan BC, (21). 5 B. R: cos AC'sin C cos B, (.23) 6 AB. tan B- tan AC:: R: sin AB, (lS). 7 AC and B. BC. sin B: sin AC:: R: sin BC, (19). 8 C. cos AC cos B R: sin C, (23). 9 AB. cos AC: cos BC: R: cos AB, (22). 10 AC andBC. B. sin BC: sin AC::.R: sin B, (19). 11. C. tan BC-: tAC:: R: cos C, (21). 12 BC. R: cos AB:: cos AC: cos BC, (22). 13 IB andW. B. iin AB R: tan AC: tan B, (18). 14 C. sinrAC: R: tan AB: tan C, (18). 14 AB. sin B: cos C:: Pt: cos AB, (23). 15 B and C. AC. sin,C: cos B:: R: cos AC, (23). 15 BC. tan B cot C R cos BC, (20). 6 1),, _. _ -----; --- - I /1 SPHERICAL TRIGONOMETRY. 271 TABLE for determining the affections of the Sides and Angles found by the preceding rules. AC and B of the same affection, (14). 1 If BC L. 90', AB and B of the same affection, otherwise different, (Cort 15). 2 If B C 90~, C and B of the same affection, otherwise different, (15.) 3 AB and C are of the same affection, (14.) 4 If AC and C are of the same affection, BC Z.90~; otherwise BC 90, (Cor. 15.) B and; AC are of the same affection, (14.) 6 Ambiguous. 7 Ambiguous. 9 Ambiguous..9 When BCL90~, AB and AC of the same; otherwise of different affection, (15.) 10 AC and B of the same affection, (t4.) I When BC L90", AC and C of the same; otherwise of different affection, (Cor. 15.) 12 BC / 90~, when AB and AC are of the same affection, (I. Cor. 15.) 13 B and AC —of the same affection, (14.) 14 C and AB of the same affection, (14.) 14 AB and C of the same affection, (14.) 15 AC and B of the same affection, (14.) 15! When B and C are of the same affection, BC. 900, otherwise, BC 7 90~0 ( 15.) 16 The cases marked ambiguous are those in which the thing sought has two values, and may either be equal to a certain angle, or to the supplement of that angle. Of these there are three, in all of which the thigs given are a side, and the angle opposite to it; and accordg ingly, it is easy to show, that twVo right angled spherical triangles may always be found, that have a side and the angle opposite! to it the same in both, but of which the renmaining sides, and the remaining angle of the one, are the supplements of the remaining sides and the remaining angle of the other, each of each. Through the gflection of the arch or angle found may in all the other cases be determined by the rules in the second of the preceding tables', it is of use to remark, that. all these rules exc?.pt two, may.!e r:educ, 272 SPHERICAL TRIGONOMETRYI ed to one, viz. that zhen the thing found by the rules in thefirst table ig either a tangent or- a cosine; and when, of the tangents or cosines: emn ployed in the computationt of it, one only belongs to an obtuse angle, the angle required is also obtuse. Thus, in the i15th case, when cos AB is found, if C be an obtuse anw gleo, because of cos C, AB must be obtuse; and in case 16, if either B or C be obtuse, BC is greater than 90~, but if B and C are either both acute, or both obtuse, BC is less than 90'. It is evident, that this rule does not apply when that which is found is the sine of an arch; and. this, besides the three ambiguous cases, happens also in other two, viz. the 1st and 11th. The ambiguity is obviated, in-these two cases, by this rule, that the sides of a spherical Tight angled triangle are of the same affection with the opposite angles. Two rules are therefore sufficient to,remove the ambiguity in all the cases of the right angled triangle, in which it can possibly be ret moyed. SPHERICAL TRIGONOMETRMY 273 It may be useful to express the same solutions as in the annexed table. Let A be at the right angle as in the figure, and let the side opposite to it be a; let b be the side opposite to B, and c the side opposite to C. GIVEN. SOUGHT, SOLUTION. b. sin b -- sin a X sin B. I a-and B. c. c tan c - tan a X cos B. 2 C. cot = cos a X tan B. s c. tan c = sin b X tan C. 4 B. cOS B = cos b X sin C. 6 __,. _ _ _ tan b!c. - sin c -,. 7 tan B i~ sl~ sin b b and B. a.. si n Zn - 8 sin b = _ cos b c; b an C. B sin C 9i C. sin c 1=sin b a.and b. ca. csin B - 1 sin a tans b @.; cos C -- t-:a C2 C CO5C rtan B a. cos a -- cos b X cos c. 13 Bn tan b b and c. B tan B s=nn c' 1 tan c C. tan C s' 16 Li I @o s! Co C cos B B and C. cnos b a. c os a1.)le.... Cer;-. —~- tan.M nr. M 274 SPHERICAL rjIGONOLIEThRY PROBLEM ILo iJn any oblique ac ged spherical triangle, of the three sides and thret angles, any three being given, it is required to find the other three. In this Table, the references (c. 4.), (c. 5.), &c. are, to the cases -inthe preceding Table, (16.), (27.) &c. to the propositions in Spherical Trigonornetry GIVEN. sUI SOLUT'ION I Let fall the perpendicular CD fiont the unknown angle, not requir. ed on AB. ne of theR:R oi A tan AC: tan AD, (c. 2.); therefore B)b is known, fotherangles, and sin BD): sin AD:: tan A: tan B, (27.); B and AI are of B. the samne or difeirent affection, Two sides according as AB is greater'or less than BD, ( i 6.). ABA, -AC,..- Let fall the perpendicular CD fiom and the in- one of the unknrown angles on the side AB. 2 eluded angle The third R: cos A: tan AC:tan AD, (c. 2.); therefore BD is known. A. side and cos AD: cos BD -. cos AC:-cos B(-, (26.); according BC. as the rsertlents A D and DB are of the same or different affection, AC -and ( oB - ill he of the same or different affect/on. T.ABLE continued.. GIVEN. O'UG'..OLUTION. From C the extremity of A C neart the side souiaht,-let fall the per-i pendicular CD on AB. 3 - gThe side 1.{: cos LNC tan A: cot ACDi (c. 3.); therefore BCD is know-n, BC. and cos BCD: cos ACD tan A: -: tan BC, ( 18.). BC is lessi Two angles -. or greater than 90, accordigi as the angles A;and BCD iae' A Anad AC of. the, same, or f i.gcgirAt vIlr^-IIrr *I:and...aC n a.' L.t f'all the paerpend-;ClU"P:I; 5-' oret' -: A_, - one of' the given aorles wi: o t,-c.: -. opposite side A}-3o-:'the side be-: Cos AC:an A ct ACD-; aH t R' 0: @' "r'h al: ci B D 1T1ho third C(c. 3) tilcrcfro tla a::gle.BCD'! tween them.'.'. s given, a sin A D sMii BCD angle os k: c(s 3 (5.);7 B'...7and LA fire of'the same o1' dif4 Bf. ierel~t affezction, according as CO falls within or without the triangle, that is, according as:ACB is greater; or less than,.......... -- (:.). - 0 /,''-\N \'.\ /,> D < [) 2a76 Y-SPRE RIC AL TRLGO3NON IETW:. TABLE continued. GIVEN BOUGHT. SOLUTION. I __ ~-~_. —.~_ _......:. The angle Sin BC:,sin AC:: sin A:sin B, 5 -B (24.) The affection of B is amopposite to biguous, unless it can be deterthe other gi- mine:: by this rule, that accordven side ing as AC + BC is greater or Two sides AC. less than 180~, A+ B is greater or less than 180~, (i0.) AC and BC, From ACB the angle sought draw andanangle, The angle CD perpendicular to AB; thenO ACB R cos AC:tan A: cot ACD, I A contained by (c.;o.); and tan B:': tan AC:-, the given cos ACD: cos BCD, (28.) A3CD 6 AC and BC. ambiguous, because of the ambi" one of them, guous aign +- or -. BCS Let fall the perpendicular CD from The third the angle C, contained by the given sides, upon the side AB. side.tR: cos A:: tan AC: tan- AD, B (c. 2); cos AC: cos BC:: cos A.B. AD:Coa BD, (26.) ABAD -A BD, wherefore AB _.__I- is:ambiguous. C A SP11ERICAL TALG UNOMETRY1 TABLE continued.d 6J olr~N S7UGiSIJ. SO.LUTION. The side Sin B:sin A: sin AC-: sin BC, BC (24.); the aflection of BC is un~ opposite certain, except when it can be deB I I to the termined by this rule, that accordother ina as A+lB is (reater or less than given an- 180~, AC + -BC is also greater or gle A. less than 180~, (10.). /Two angles —- -- ---— 7 — From the unknown angle C, draw AB, The side CD) perpendicular to AB; then, L: OS A:: tan AC. tani AD, ands-side (c. 2.); tan B:tan A:: sin AD: adjacent sin BD. BD is ambiguous; and to tho 9 A toC the therefore AB —A D D ~ BD mayv -gniven l have four values, some of which opposite to angle -Will be excluded by this conditlion, A, *B. that AB must be less than 180". one o f them, From the. angle required, C, draw - B. CD perpendicular to- AB. The third H R:cos AC::tan A -cot ACD, (c. 3.),cos A: cos B:: sin ACe): angle sin BCD, (25.). The affection 10 of BCD is uncertain, and thereACB. fore ACB= ACD B~ BD, has four values, some of which may be excluded by the condition, that ACB is less than 180~. The three From C one of the angles not required, draw CD perpendicular to AB. Find an arch 1 such'thatl sides, si}des, tan I AB; tan. (AC+BC): tan! -t1~AB, AC,. One of the t - (AC BC): tan 4 E; then, if AB be greater than E, AB is the and't angles sum, and E the difference of AD anld DB; but if AB be less than I r1 /.1..... A, E, E is the sum and AB the diffe-!v. $ rence of AD, DBJ, (29.). In ei-J ther case, AD and DB are known,, ____ _ i i__ and taAC:tanAC:tanD::R:cosA. F~st~P""-~'-~cc.i"~ac,~ _ _ 5 i SPHERICAL TRIhGONOMETRY.. TAITTLE eontinuand GIVEN. SOUGHT- E BOLUTION, Suppose the supplemsents of the three given angles, A, B, C, to be, a, b, c, and to be the sides The three One of the of a spherical triangle. Find, by the last case, the angle of 21 angles sides this triangle, opposite to the side a, and it will be the supA, B, C. BC. plemnent of the side of the given triangle opposite to the angle A, that is, of BC, (11I.); and therefore BC is found. In the foregoing table, the rules are given for ascertaining the affection of the arch or angle found, whenever it can be done: Most of these rules are contained -in this one rule, which is of general application, viz. that uvhen the thing found is either a tangent or a cosine;, and of the tangents or cosines employed its the computation of it, either one or three belong to obtuse angles, the angle found is also' oebtuse. This rule is particularly to be attended to in cases 5, and 7, where it. removes part if eW ^)wXbtiuity. It,nay itt necessary to remark with respect to the i th ecase, that the segments of the base complted there are those cut off' by the nearest perpendicular; and also, that when the sum of the sides is less than,80o, the least segment is'adjacent to the least side of the triane gtI: otherwise to the greatest, (17.). SPHERICAL TRIGONOMIETRY. 270 The last table may also be conveniently expressed in the following manner, denoting the side opposite to the angle A, by a, to B by b, and to C by c; and also'the segments of the base, or of opposite an? gle, by x and y. GIVEff. f5UGIT, SOLUTION. Find x, so that tan x=-tan bXcos A; thenj 1 b and c, and B. sin x XXtan A. tan B —= the angle sin (c - x) between - Find x, as above, ahem B... cos b X cos (c - xj 2 hthem A. Aen eos a=c —between —-"I...... o. ___Cos X. Find x, so that - 3 Angles cot x=cos b Xtan A: then tan bXcos x A and C tan a b - -- cos -(c - ) and..... 4 Find x, as above, side b. Bc os cos AXsin (c-x) sin x sin b X sin A:6 tB sin B= Sid.es cot Find.x sin a Find x, so that ~Sl~des ~ cot x=cos b X tan A: then 6 and coC cos xXtan b a and b Cos -GOS =atan a and Find x, so that |7 fangle A. tan x=tanb X cos A; and find y, so that, C!cos a X o x 4 i i cos Y'' y-' cos b' i! L TRIGONOMETRYo TABLE continued. sin bXsin A sin; aD-sin Find x, so that h tan xtan bXos A; andyso The angles,,g t t' 1 a i nd B sin xXtan A iand B sin y= —--— o - } and tlh; 5 " ta B Find x, so that side b| cot x —cos b X tan A; and also y, i j 6so that sin aXcos B sin ~': cos A C=X:j +y. Let a+b+c=s. si n (-s b ) X 9in (4 @) I" A /sin nb sin C,sin bt sin c in a~vi 4 12 a /.sin sx sin C.,!2~.}~ ~~ T A ~/s'in B Xsill. C i'V.'2 - _- r — i -~4"~~~-~-'~-iT:T-T —C~-~ i~l ir.~-~~~ —— ~ —;-^=-I-I~~I_~-~ ~j= APPENDIX SPHERICAL TRIGNOOMETRY CONTAINING NAPIER'S RULES OF THIE ClBRCULAR PART, THE rule of the Ctr2cular Parts, invented by NAPIPR,, is of greal use in Spherical Trigonometry, by reducing all the theorems em. ployed in the solution of right angled triangles to two. These twc are not new propositions, but are merely enunciations, which, by helf of a particular arrangement and classification of the parts -of a triangle, include all the six propositions, with their corollaries, which have beer demonstrated above from the 18th to the 23d inclusive. They arl perhaps the happiest example of artificial memory that is known. DEFINITIONS. if in a spherical triangle, we set aside the right angie, and consider only the five remaining parts of the triangle, viz. the three sides and the two oblique angles, then the two sides which contain the right angle, and the complenments of the other three, namely, of the two angles and the hypotenuse, are called the Circular Parts. Thus, in the triangle ABC right angled at A, the circular parts are AC, iB with the complements of B, BC, and C. These parts are called circular; because, when they are named in the natural order of their succession, they go round the triangle. II. When of the five. circular parts any one is taken, for the niiddle para. then of the remaining four, the two which are immediately adjacenit to it, on the right and left, are called the adjacent parts; and the other two, each of which is separated from the middle by an ad, iaces part, are called opposite parts. N nr APP:ENDIX TO Thus in the right angled triangle ABC, A, being the right angle, AC AB, 90~-B, 90~- BC, 90" —C, are the circular parts, by Def. 1.; and if any one as AC be reckoned the middle part, then AB and 900 -C, which are contiguous to it on different sides, are called adjacent parts; and 90~-B, 90 — BC are the opposite parts. In like manner f' AB is taken for the middle part, AC and 90~ — r` are the adjacent parts: 90~-Bga and 90~0C are the opposite. Or if 90~=-BC be the middle part, 90O-:B, 90~-='C are adjacent; AC and AB opposite, &co This arrangement being made, the rule of the circular part is contained in the following PROPOSITION' In a right angled spherical triangle, the rectangle under the radius and the sine of the middle part, is e,- ual to the -ectangle under the tangents of the adjucent parts; or to the rectangle under the cosines of the opposite parts. The truth of the two theorems included in ibs enunciation may be easily proved. by taking each of the five circular parts in succession for the middle part, when thei- eneral propositiin will be found to coincide with some one of- the analogies itn the table alr;ead given for the resolution of the cases of right angled spherical triangles. Thus, in the triangle ABC, if the complement of the hypotenuse BC be tak. en as the middle part, 90~0-.B, and 900-oC, are the adjacent parts, AB and NC the opposite. Then the general rule gives these two theo orems, RXcos BC=cot BXcot C; and R Xcos BC=cos AB Xcos AC. The former of these coincides with the coro to the 20th; and the latter witeh the 22d. To apply the foregoing general proposition, to resolve any clise of a right angled spherical triangle, consider which of the three quantities named (the two things given and the one required) must be made the middle term, in order that the other two may be equidistant from it, that is, may be both adjacent, or both opposite; then one or other of the two theorems contained in the above enunciation will give the value of the thing required. Suppose, for example, that AB and BC are given, to find C; it is evident that if AB be made the middle part, BC and G are the oppQe SPHERICAL TRiGONOMETRY. P 283 Oite parts, and, therefore R Xsin AB=sin C Xsin BC, for sin C=cos (90~-C), and cos (90"-BC)=~in BC, and consequently sin AB sin BC' Again, suppose that BC and C are given to find AC; it is obvious that C is in the middle between the adjacent parts AC and (a oQ —BC)d therefore R Xcos C=tan AC Xcot BC, or tan AC - cos C cot BC tan BC; because, as has been shown above, t tan Bi., In the same way may ail the other cases be resolved. One or two trials will always lead to tihe knowledge of the part which in any given case is to be assumned as the middle part; and a little practice will make it easy, even without such trials, to judge at once which of them is to be so assumed. It may be useful for the learner to range the.names of the five circular parts of the triangle rotund the circnumfe rence of a circle, at equal distances fromn one another, by which means the middle part will be imimediately detetnained Besides the rule of the circular parts, Na pier derived from the last of the three theorems ascribed to himllab,,ve, (schol.:oj the solutions of all the cases of otlique angled triangles These solutions ma'e as follows: A, B, C, denoting the three triangles of a spherical triangle, and a, b, c, the sides opposite to theme Given two sides b, c, and the angle A between them. To find the angles B and C. sin ~ (b — c) tn I (B-C)=cot 2 A1 si2 n (b (1.) cort. 1 cos ~ (b —c) tan ~ (B+C)=Cot - AX 2,o ( c)- (31.) cor. l. To find the third side a, sin B: sin A: sin b: sin a. IL. Given the two sides b, c. and the angle B opposite to one of them, To find C, and the angle opposite to the other side. sin b: sin c:" sin B: sin C. To find the contained angle A. cot',A=tan (B C)X i s- - (b.c 284 APPENDIX TO To find the third side ac Si/r B: sin A sin b: sin cc. III. Giviit two angles A and B, and the side c between themt, Td find the other two sides a, b. sin. (A B) in (b-a) tan cX i - 2). (31.),f2 cos ~ (A- B) ran 2 (b+-a) tan 2 cXe- (A B) (31.) Cos -(AB) To find the third angle C. sin a: sin c;e sin A: sin (C Given two angies A and B, anid the side a, opposite to one of' theim4 To find b, the side opposite to the other. sin A sin B:: sin: sin b. To find c, the side between the giveL angles 6 i. -(A-t-Bj(31. -tan } z c — ( b sin (A. -B)' (3 To find the third angle C. sin a sin c sin A: sin GC The other two cases, when the three sides are given to find the ajn:gles, or when the three angles are given to find tie -,ides, are rc:solved by the 29th, (the first of NAP:RE's Propositi)n.s,) in the sa:e way as in the table already given for the case of the oblique angled triangle, There is a solution of the case of the three sides being given, i hich it is often very convenient to' use, and which is set down here, thought the proposition on which it depends has not been dvmonstrated, SPHERICAL TRIGONOMETRY. 285 Let a, b, c, be the three given sides, to find the angle A, contained between 6, and c. If Rad. = 1, and a + b + c. s sin X sin I (s -c) sin A ___ or, /sin b X sin c Vsnos A1 - X sin (s-a) / sin b X sin c In like manner, if the three angles, A, B, C are given to find c, the side between A and B. LetA+B C =S, sin - C cogi SX cos (I S-A) sin -l c = -; or, vsin B X sin C V /os (1 S -B) X Cos (9 S-U) V/sin B x sin C These theorems, on account of the facility with which Logarithms are applied to them, are the most convenient of' any for resolving the two cases to which they refer. V hen A is a very obtuse angle, the second theorem, which gives the value of the cosine of its half, is to be used; otherwise the first theorem, giving the value of the sine of its half-is preferable. The same is to be observed with respect to the side c, the reason of which was explained, Plane Trig. Schol. RaND OF 2PHE;RICAL TRIGONOMIETRY I I i 1 1 L —I F. j tm M I a 1!4 0 -i mi .4 111.11 M M "I I lc I I li I A I I. it li I X I a I I NOTES,N THEI FIRST BOOK OF THE ELEMENTS.. DEFINITIONS. I. SN thie definitions a few changes have been made, of which it is nie cessary to give some account. One of these changes respects the first definition, that of a point, which Euclid has said to be, 6 That ~which has no parts, or which has no magnitude.' Now, it has been objected to this definition, that it contains only a negative, and that it is not convertible, as every good definition ought certainly to be. That it is not convertible is evident, for though every point is unextended, or without magnitude. yet every thing unextended or without magnitude, is not a point. To this it is impossible to reply, and therefore it becomes necessary to change the definition altogether, which is accord. ingly done here, a point being defined to be, that which has position but not magnitude. Here the affirmative part includes all that is essential to a point, and the negative part excludes every thing that is not essential to it. I am indebted for this definition to a friend, by whose judicious and learned remarks I have often profited. After the second definition Euclid has introduced the following, "C the extremities of a line are points."' Now, this is certainly not a definition, but an inference from the definitions of a pointand of a line. That which terminates a line can have, no breadth, as the line in which it is has none; and it can have no length, as it would not then be a termination, but a part of that which it is supposed to terminate. The termination of a line can therefore have no magnitude, and having necessarily position, it is a point. But as it is plain, that in all this we are drawing a consequence from two definitions already laid down, and not giving a new definition, I have taken the liberty of putting it down as a corollary to the second definition, and have added, that the int.ersectieons oF one line with anothe~ 0 o ~NOTES. arepoints, as this affords a good illustration of the nature oft a poiwl and is an inference exactly of the same kind with the preceding. The same thing nearly has been done with the fourth definition where that which Euclid gave as a separate definition, is made a corollary to the fourth, because it is in fact an inference deduced from comparing the definitions of a superficies and a line. As it is impossible to explain the relation of a superficies, a line and a point to one another, and to the solid in which they all originate, better than Dr. Simson has-done, I shall here add, withvarery littlechange, the illustration given by that excellent Geometer. " It is necessary to consider a solid, that is, a magnitude which has length, breadth., and thickness, -in order to understand aright the definitions of a point, line and superficies; for these all arise from a solids and exist in it; Trhe boundary, or boundaries which contain a solid, are called superficies, or the boundary which is common to two solids which are contiguous, or which divides one solid into two contiguous parts, is called a superficies; Thus, if BCGF be one of the boundaries whicth contain the solid ABCDEFGH, or which is the commnnon boun. dtaiy of this solid, and the solid BKLCFNMG, and is therefore in the one as well as the other solid, it is called a superficies, and has no thickness; For if it have any, this thickness must either be a part of the thickness of the solid AG, or the solid BM, or a part of the thick. hess of each of them. It cannot be a part of the thickness of the solid BMIl; because, if' this solid be removed from the solid AG, the superfi. Cies BCGFG; the bhundary,f the solid AG, remains still the same as it was. Nor can it be a part of the thickness of the solid AG: because if this be removed from the solid-BM, the superficies BCGF, the boun. dary of the solid BM, does nevertheless remain; therefore the superfcien BCGF has no thickness, but only length and breadth. "' The boundary of a superficies is called a line; or a line is the common boundary of two superficies that are contigutous, or it is that which divides one superficies into two contiguous parts: Thus, if BC be one of the boundaries which contain the superficies ABCD, or which is the, common boundary of this superficies, and of the superficies, KBCL, which is contiguous to it, this boundary BC is called a line, and has no breadth, For, if it have any, this must he part either of the breadth of the superficies ABC D or of the superficies KBCL, or part of E _ 3 each of them. It is not part of the breadth of the superficies KBCGL-; for if this superficies El /j/ / be removed from the superficies ABCD, the line BC which is-the bo'unrdary of the superficies A BC D termains the same as it was. Nor D can the breadth that BC is supposed to- have, be a part of the breadth of: the superficies ABCD becausea if thi be- removed from 3 r INOTES29 A hile superficies KBCL, the line BC, which is the boundary of the superficies KBCL, does nevertheless remain: qTherefore the line BC has no breadth. And because the line BC is in a superficies, and that a superficies has no thickness, as was shown; therefore a line has nei~ ther breadth nor thickness, but only length. " The-boundary of a line is called a point, or a point is a common boundary or extremity of two lines that are contiguous: Thus, if B be the extremity of the line AB, or the common extremity of the two lines AB, KB, this extremity is called a point, and has no length: For if it have any, this length must ei- & ther be part of the length of the line AB, or of the line KB. It is not part of the length of KB; for if the line KB be removed from E AB, the point B, which is the extremity of the line AB, remains the same as it was; Nor is it part of -the length of the line AB; for X if AB hbe removed fronm the line / KB, the point B, which is the ex-V/ tremity of the line IKB, does ne- A G vertheless remain: Therefore the point B has no lengtht And because a point is in a line, and a linO has neither breadth nor thickness, therefore a point has no lengths breadth, nor thickness. And in this manner the definitions of a point, line, and superficies are to be understood." 11I. Euclid has defined a straight line to be a line which (as we translate it) ".lies evenly between its extreme points." This definition is obviously faulty, the word evenly standing as much in need of an expla.. nation as the word straight, which it is intended to define. In the original, however, it must be confessed, that this inaccuracy is at least less striking than in our translation; for the word which we render evenly is sigss, equally, and is accordingly translated ex caquo, and equaliter by Commandine and Gregory. The definition, therefore, is, that a straight line is one which lies equally between its extreme points: and if by this we understand a line that lies between its extreme points so as to be related exactly alike to the space on the one side of it. arnd to the space on the other, we have a definition that is perhaps a little too metaphysical, but which certainly contains in it the essential character of a straight line. That Euclid took the definition in this sense however, is not certain, because he has not attempted to deduce firom it any property whatsoever of a straight line; and indeed, it should seem not easy to do so, without employing some reasonings of a more metaphysical kind than he has any where admitted into his ElementsO To supply the defects of his definition, he has therefore introduced the Axiom, that two straight lines cannot inclose a space; on-which Axiom it'is, ahd not on his'definition of a straight line, that his de. monstrations are founded. As this manner of proceeding is certainly not so regular and scientific as that of laying down a definition, from which the properties of the thing defined mnay be logically de, duced, I have substituted another definition of a straight line in the room of Euclid's. This definition of a straight line was suggested by a remark of Boscovich, who, in his Notes on the philosophical Poem of Professor Stay, says, "Rectam lineam rectm congruere totam toti i' in infinitum productumn si bina puncta unius binis alterius congruante "patet ex ipsa admodum clara rectitudinis idea quam habemus."9 (Sulpplementurnm in lib. 3. ~ 550.) Now, that which Mr. Boscovich would consider as an inference from our idea of straightness, seems itself to be the essence of that idea, and to afford the best criterion for judging whether any given line be straight or not. On this principle we have given the definition above, If there be two lines which cannot coincide in two points,'without coinciding altogether, each of them is called a straight line. This definition was otherwise expressed in the two former editions it was said, that lines are straight lines which cannot coincide in part, without coinciding altogether. This was liable to an objection, viz. that it defined straight lines, but not a straight line; and though this its truth' is but a mere cavil, it is better to leave no room for it. The definition in the form now given is also mnore simple. From the same definition, the proposition which Euclid gives as an Axiom, that two straight lines cannot inclose a space, follows as a new cessary consequence. For, if' two lines inclose a space, they must intersect one another in two points, and yet, in the intermediate part, must not coincide; and therefore by the definition they are not straight lines. It follows in the same way, that two straight lines cannot have a common segment, or cannot coincide in part, without coinciding al. together. After laying down the definition of a straight line, as in the first Edition, I was favoured by Dr. Reid of Glasgow with the perusal of a MS. containg many excellent observations on the first Book of Eurclid, such as might be expected from a philosopher distinguished for the accuracy as well as the extent of his knowledge. HIe there-defined a straight line nearly as has been done here, viz. 6 A straight "line is that which cannot meet another straight line in more points "than one, otherwise they perfectly coincide, and are one and the "same." Dr. Reid also contends, that this must have been Euclid's own definition; because, in the first proposition of the eleventh Bool, that author argues, " that two straight lines cannot have a common seg"' ment, fbr this reason, that a straight line does not meet a straight "line in more points than one, otherwise they coincide." Whether this amounts to a proof of the definition above having been actually Euclid', I will not take upon me to decide: but it is certainly a proof NOTES. S23% that the writings of that geometer ought long since to have suggested this definition to his commentators; and it reminds me, that I might have learned from these writings what I have acknowledged above to be derived; from a remoter source. There is another characteristic, and obvious property of straight lines,.by which I have often thought that they might be very conveniently defined, viz. that the position of the whole of a straight line is determined by the position of two of its points, in so much that, when two points of a straight line continue fixed, the line itself cannot change its position. It might therefore be said, that a straight line is one in 2which, if the position of two points be determined, the position of the wzhole line is determined. But this definition, though it amount in fact to;the same thing with that already given, is rather more abstract, and not so easily made the foundation of reasoning. I therefore thought it best to lay it aside, and to adopt the definition given in the text. V. The definition of a plane is given from Dr. Simson, Euclid's being liable to the same objections with his definition of a straight line; for, he says, that a plane superficies is one which "lies evenly between " its extreme lines." The defects of this definition are completely removed in that which Dr. Simson has given. Another definition differeant from both might have been adopted, viz. That those superficies are called plane, which are such, that if three points ofi the one coins cide with three points of the other, the whole of the one must coincide with the whole of the other. This definition, as it resemlbles that of a straight line, already given, might, perhaps, have been introduced with some advantage; but as the purposes of demonstration cannot be better answered than by that in the text, it has been th oughbt best to make no farther alteration. VI. In Euclid, the general definition of a plane angle is placed before that of a rectilineal angle, and is meant to comprehend those angles which are formed by the meeting of the other lines than straight lines. A plane angle is said to be " the inclination of two lines to " one another which meet together, but are not in the same direct " tion." This definition is omitted here, because that the angles formned by the meeting of curve lines, though they may become the subject of geometrical investigation, certainly do not belong to the Elements; for the angles that must first be considered are those made by the intersection of straight lines with one another. The angles formed by the contact or intersection of a straight line and a circle, or of two circles, or two curves of any kind with one another, could produce nothing but perplexity to beginners, and cannot possibly be understood till the -properties of rectilineal, angles- have been fuly 294 N4OTES. explained. On this ground, I am of opinion, that in an elementary treatise, it may fairly be omitted. Whatever is not useful. should, in explaining the elemetnts of' a science, be kept out of sight altogether; for, if it does not assist the progress of the understanding, it willgcertainly retard it. AXIOMS. AmONG the Axioms there have been made only two alterations. The 10th Axiom in Euclid is, that two straight lines Cannot inclose "' a space;" which, having become a corollary to our definition of a straight line, ceases of course to be rankled with self-evident propositions. It is therefore removed from among the Axioms, and that which was before the 1 i th is accounted the 10th. The 12th Axiom of Euclid is, that " if a straight line meets two "straight lines, so as to make the two interior angles on the same "side of it taken together less than two right angles, these straight "clines being continually produced, shall at length meet upon that "side on which are the angles which are less than two right angles." Instead of this proposition, which, though true, is by no means selfevident; another that appeared more obvious, and better entitled to be accounted an Axiom, has been introduced, viz. " that two straight -" lines, which intersect one anothet, cannot be both parallel to the "' same straight line." On this subject, however, a fuller explanation is necessary, for which see the note on the 29th Prop. PROP. IV. and VIII. B. I. The fourth and eighth propositions of the first book are the foundation of all that follows with respect -to the comparison of triangles. They are demonstrated by what is called the method of supraposition, that is, by laying the one triangle upon the other, and proving that they must coincide. To thissorme obtjections have been made, as if it were ungeometrical to suppose one figure to be removed from its place and applied to another figure.' The laying," says AIr. Thomas Simson iai his Elements, " of one figure uplon another, whatever evi. "dence it may afibrd, is a mechanical consideratiion, and depends on "no postulate." It is not clear what Mr. Sirason meant here by the word mechaanical: but he prob.ably intended only to say. that the method of supraposition involves the idea oi' motion, which belongs rather to mechanics than geometry; for I think it is impossible that such a.Giometer as he was could mean to assert, that the evidence -derived from this method is like that which arises firom the use of instruments, and of the same kind with what is furnished by experience and observation. Thle demonstrations of the fourth and eighth, as they are given by Euclid, are as certainly a process of pure reasoning, depending solely on the idea of equality, as established in the -ith Axiom, as any thing in geometry. But, if still the removal of the triangle from its place be considered as creating a difficulty, and as inelegant, because-it involves an idea, that of motion, not essential to geometry, this defbct may be entirely remedied, provided that, to Egiclid's three postulates, we be'allowed to add the following, viz. That if therie be two equal straight lines, and if any figure whatsoever be constituted on the one, a figure every way equalto it may be constituted on the other. Thus if AB and DE be two equal straight lines, and ABC a triangle on the base AB, a triangle DEF every way equal to ABC may be supposed to be constituted on DE as a base. By this it is not meant to assert that the method of describing the triangle DEF is actually known, but merely that the triangle DEF may be conceived to exist in all respects equal to the triangle ABC.:Now, there is no truth whatsoever that is better entitled than this to be ranked among thie Postulates or Axioms of geometry; for the straight lines AB and:DE being every way equal, there can be nothing belonging to the one that may not also belong to the other.' On the strength of this postulate the fourth Proposition is thus demonstrated. If ABC, DEF be two tri-anglcs, sucE that the two sides AB and AC of the one are equal to the two ED, DF of the other, and the angle BA.,; contained by the sides AB, AC of the one, equal to the angile tED/F,! containeid by the sides ED, DF of the other; the triangles ABC aiid EDF are every way equal. / \ \ On AB. let a bec dever/ w a' I C r 011On AB let a triangle be constituted every way equal to tile triai'ge DEF; then if this triangle coincide with the triangle ABC, it is evident that the proposition is true, for it is equal to DEF by hypothesis and to ABC, because it coincides with it; wherefore ABC, DEF are equal to one another. But if it does not coincide with ABC, let it have the position ABG; and first suppose G not to fall on AC; then the angle BAG is not equal to the angle BAC. But the angle BAG is equal to the angle EDF, therefore EDF and ABC are not equal, and they are also equal by hypothesis, which is impossible. Therefore the point G must fall upon AC; now, if it fall upon AC but not at C, then AG is not equal to AC; but AG is equal to DF, therefore DFI and AC are not equal, and they are also equal by supposition, which is impossible. Therefore G must coincide with C, and the triangle AGB with the triangle ACB. But AGB is every way equal to DEF.:.herefore, ACB a-nd DE are also every way e qual Q. E,.. iN OTES. By help of the same postulate, the 5th may also be very easily demostrated. Let ABC be an isosceles triangle, in which AB, AC are the equal sides; the angle ABC, A%1B opposite to these sides are also equal. Draw the straight line EF equal to BC, and suppose that on EF the triangle DEF is constituted every way equal to the triangle ABC, that is, having DE -equal to AB, DF to AC, the angle EDF to the angle BAC, the angle ACB to the angle DFE, &c. A -I) / \\ // / E:B G x -I, Then because DE is equal to AB, and AB is equal to AC, DE is equal to AC; and for the same reason, DF is equal to AB. And because DF is equal to AB, DE to AC, and the angle FDE to the angle BAC, the angle ABC is equal to the angle DFE, (4. 1.). But the angle ACB is also, by hypothesis, equal to the angle DFE; therefore the angles ABC, ACB are equal to one another. Q. E. Do. Thus also, the Sth proposition may be demonstrated independently of the 7th. Let ABC, DEF be two triangles, of which the sides AB, AC are equal to the sides DE, Dli each to each, and also the base BC to the base EF; the angle BAC is to equal the angle EDF. /'k BK\ D