George Davidson 
 1825-1911 
 
 Professor of Geography 
 University of California 
 
TEX'T-BOO 
 
 OF 
 
 ELEMENTARY PLANE GEOMETRY 
 
 BY 
 
 JULIUS PETERSEIM, 
 
 PROFESSOR AT THE ROYAL POLYTECHNIC SCHOOL AT COPENHAGEN 
 AND MEMBER OF THE ROYAL DANISH SOCIETY OF SCIENCE. 
 
 TRANSLATED BY 
 
 R. STEENBERG JR. 
 
 (AUTHORIZED ENGLISH EDITION) 
 
 LONDON 
 
 SAMPSON LOW, MARSTON, SEARLE & RIVINGTON 
 COPENHAGEN: ANDR. FRED. HOST & SON 
 
 1880 
 (All rights reserved) 
 
Printed by Bianco Ltmo. 
 
INTRODUCTION. 
 
 1. livery object which occurs in nature has numerous 
 properties; to facilitate the investigation of which, one prelim- 
 inarily endeavours to consider one at a time, independently 
 of the others; the different investigations of the same kind 
 are then collected, and thus the different sciences, nat^tral 
 sciences, are formed. Thus, if we examine a piece of chalk, 
 we might ask about its origin and occurrence, its specific 
 gravity and colour, the combination of its elements, its form, 
 &c. , and these questions will be answered respectively by 
 Geology, Physics, Chemistry, and Geometry. 
 
 2. Geometry treats of the form without regard to the 
 substance; when we speak of a sphere, we do not consider 
 of what it is made, but only of the space which it occupies ; 
 every object occupies a space, which has extension in all di- 
 rections; this is a geometrical body. 
 
 The boundary between a body and the surrounding space 
 is called a superficies or the surface of the body ; a superficies 
 has no thickness. 
 
 If a part of a superficies be cut off from the surrounding 
 part, the boundary is called a line\ a line has no breadth. 
 If a part of a line be cut off, the limit is termed a point \ it 
 has no magnitude, but only denotes position. 
 
 3. When a point moves, it produces a line, similarly a 
 superficies may be imagined as being produced through the 
 
 M513332 
 
movement of a line, a body through the movement of a 
 superficies. 
 
 4. Let us imagine a body revolving round two fixed 
 points, that is, moving so that two points in it do not change 
 their places in space, we could then suppose a line to be 
 drawn through these two points and through all the other 
 points in the body, which during the movement do not change 
 their places. Such a line is called a straight line. The straight 
 line, therefore, is determined by two points, that is, that through 
 two given points there can only be drawn one straight line. 
 From this it follows again, that two different straight lines 
 can only have one point in common (a point of intersection), 
 for if they had two points in common, they must coincide. 
 A straight line can be imagined to be infinitely produced by 
 the addition of other straight lines, each having two points 
 in common with the preceding ones. Where it cannot be 
 misunderstood line is often used for straight line. 
 
 A line of which no part is straight is called curved, a 
 line composed of several straight lines is called a ^broken 
 line". 
 
 5. A plane superficies or plane is one in which every 
 straight line lies wholly, when two of its points lie in it; the 
 position of a plane is determined, when it contains three given 
 points which are not in a straight line. A straight line and 
 a plane are supposed infinitely produced, when the opposite 
 is not stated; a straight line may, without being changed, be 
 imagined to be made to slide along itself; a plane may, with- 
 out being changed, be made to slide along itself or turned 
 in itself round one of its points. 
 
 6. A figztre is a limited part of a plane, 
 its boundary is called Perimeter, when it is 
 broken, Periphery, when it is curved. A tri- 
 angle is contained by three straight lines 
 (sides), a quadrilateral by four, a multilateral 
 figure or polygon by an indefinite number; 
 a line joining the extremities of two sides of a polygon, with- 
 out itself being a side, is called a diagonal (A G). A polygon 
 
is called convex, when the prolongations of the sides fall 
 outside the polygon; in the opposite case it is said to be 
 not convex. 
 
 7. When a straight line of fixed length (OA) revolves in 
 a plane round one of its extremities, its other extremity de- 
 scribes a closed curved line called a circle\ the fixed point 
 is called the centre and OA the 
 
 radius', all radii of the same circle 
 are equal', a line joining two points 
 of the circumference is called a 
 chord (BG)\ a diameter (DA) is a E 
 chord through the centre; all dia- g / # 
 
 meters are equal', a chord produced 
 
 is called a secant (EF)\ it lies partly inside and partly out- 
 side the circle and cuts it in two points; if the secant be 
 moved, so that the two points coincide in one point (7), this 
 is called a point of contact and the line a tangent (GH)\ the 
 tangent has only one point in common with the circle and 
 lies wholly outside of it. A part of the circumference is 
 called an arc (^ AC). A part of the circle contained by a 
 chord and an arc (B2C) is called a segment', a part contained 
 by two radii and an arc (BOC) is called a sector. 
 
 A figure is said to be inscribed in a circle, when its 
 sides are chords, circumscribed, when they are tangents. 
 
 8. Two figures are said to be congruent, when they are 
 the same in everything, but only lying in different places; 
 two congruent figures may be supposed to be placed on each 
 other, so that they cover one another; the parts which then 
 coincide are called corresponding. The sign for congruence 
 is 23. Circles with equal radii are congruent. 
 
 9. That which is given in a proposition is called Hypo- 
 thesis, that which is to be proved, Thesis. 
 
 10. Plane geometry only treats of figures in the same 
 plane, chiefly the straight line and the circle; Stereometry 
 treats of lines, superficies, and bodies in general. 
 
1. 
 
 THE SITUATION OF STRAIGHT LINES. 
 
 I. THE MUTUAL DEPENDENCE OF ANGLES. 
 
 II. When a straight line revolves round one of its points, 
 until it again arrives at the position from whence it started, 
 it is said to have completed a whole revolution; if it has not 
 revolved so much, its position is determined by stating through 
 how great a part of the revolution it has turned. We say 
 that the line forms a certain angle with its former position, 
 and the angle between two lines is thus that part of a revo- 
 lution, which one line must perform in order to cover the 
 
 other. The two lines are called the legs or 
 sides of the angle, their point of intersection 
 its vertex. The symbol for angle is /L ; 
 where it cannot be misunderstood, an angle 
 is denoted by a single letter at the vertex 
 (/_ x or Z. A)', otherwise by three letters, 
 
 one at the vertex and one at each of the legs, the first in 
 
 the middle (Z. BAG). 
 
 12. A whole revolution is divided into 360 (degrees), 
 each degree into 60' (minutes)^ and each minute into 60" 
 (seconds)] an angle of 180, corresponding to half a revolution, 
 is called an angle of continuation ; its legs lie in a straight 
 line\ an angle of 90, corresponding to -J- revolution, is called 
 a right angle (symbol R.); angles less than 90 are called 
 acute, between 90 and 180 obtztse. Angles which are not 
 right angles are called oblique. Two lines which form a 
 right angle are said to be at right angles or perpendicular to 
 one another. The symbol for perpendicular" is -L- . 
 
 13. One angle is said to be the complement of another 
 angle, when their sum is equal to iR. , and the supplement, 
 
when their sum equals 2R. Thus, when an angle is o, its 
 complement is 90 - a, its supplement 180- . 
 
 Two angles mtist therefore be equal, when their comple- 
 ments or supplements are eqiial. 
 
 14. Adjacent angles are those which 
 have one leg in common and whose other 
 legs are continuations of each other 
 (b and a); two such angles are together 
 equal to 180, and are therefore supple- 
 mentary angles. 
 
 15. Vertical angles are those whose legs are continu- 
 ations of each other; they are equal, as the revolution which 
 makes the legs of the one angle coincide also makes the legs 
 of its vertical angle coincide; the angles formed by the inter- 
 section of two lines can therefore be found, when the one is 
 known; for example, if a = 45, then also a' = 45, b = b' = 135. 
 
 16. The sum of the exterior angles 
 
 of a polygon is ^R. For if a line be / "v 
 
 imagined to be placed on one of the 
 sides of the polygon, from thence turn- 
 ed on to the next side and so on, till 
 it again falls on the first side, it will 
 by degrees have turned through all the 
 exterior angles of the polygon; but it 
 has thereby performed a whole revolution, that is 4R. *). 
 
 17. The sum of the angles of a polygon is forind by 
 taking as many times 2R., as the polygon has sides, and 
 
 from that subtracting ^R. (2//R. 4R., when it has n sides). 
 For the sum of two adjacent angles is 2R. ; therefore the angles 
 
 *) If some angular points of the polygon turn inwards, the proofs in 16 
 and 17 still hold good, when the revolution in backward direction is 
 reckoned as negative. 
 
 We say in 16 that the line has performed a whole revolution, although 
 it has arrived back on itself in a different way than in 1 1 ; we really hereby 
 supplement the definition of the plane; the same reasoning could, for ex- 
 ample, not be applied to arcs lying on the surface of a sphere. 
 
of the polygon together with their adjacent angles will make 
 as many times 2R., as the figure has sides (2/iR.); from this 
 must be subtracted the sum of the adjacent angles, which 
 is 4R. 
 
 The s^t1n of the angles of a triangle will be 2R. , of a 
 quadrilateral 4R., of a seventeen sided figure 3oR., &c. 
 
 18. From 17 it follows that only one of the angles in a 
 triangle can be a right angle or obtuse; in the right-angled 
 triangle the side opposite to the right angle is called the 
 hypothenuse. 
 
 19. The angle adjacent to one angle 
 
 of a triangle equals the sum of the other 
 
 two. 
 
 For M = = 180 c 
 
 and -j- b = 180 c 
 
 therefore u == a ~\- b. 
 
 20. A triangle which has two sides equal is termed 
 isosceles*, the intersection of the equal sides is termed the 
 vertex, the third side the base, and its opposite angle the 
 vertical angle. 
 
 The angles at the base of an isosceles 
 triangle are eqzial. For if the triangle be 
 lifted up and placed on itself in an inverted 
 position, so that Z B covers itself, BC falls 
 along BA, and BA along BC, then A must 
 fall in C, and C in A, and consequently A C 
 coincides with CA ; as the angles A and C 
 thereby coincide they are equal. 
 
 21. When two angles of a triangle are eqztal, the tri- 
 angle is isosceles. This is proved in a similar way; when 
 AC covers CA, AB must fall along CB, because the angles 
 are equal, and CB along AB\ now as the sides fall on each 
 other, B must coincide with B, so that the sides coincide. 
 
 22. When all the sides of a triangle are equal, it is termed 
 eqtiilateral', the angles of this must all be equal (20), therefore 
 each 60, and conversely. 
 
23. When one of the angles of an isosceles triangle is 
 known, the others can be found; if one of the angles at the 
 base is g, the other will be the same, and the vertical angle 
 1 80 2g] if the vertical angle is <, the others will together 
 be 180 <, therefore each 90 it . 
 
 II. THE DEPENDENCE OF ANGLES AND ARCS. 
 
 24. An angle having its vertex at the centre of a circle, 
 and its legs being radii, is called an angle at the centre; by 
 superposition it may be shewn that eqtial angles at the centre 
 of the same circle or of equal circles stand on equal arcs 
 and conversely, and that equal arcs are S2ibtended by equal 
 chords. 
 
 The circle is like the whole revolution divided into 360, 
 so that an angle at the centre contains the same number of 
 degrees as the arc on which it stands; we therefore say that 
 the angle at the centre is measured by its arc. 
 
 An angle at the circumference of a circle has its vertex 
 in the circumference, and its sides are chords ; it contains half 
 as many degrees as the arc on which it stands. 
 
 a) If the one leg of the angle 'passes through the centre, 
 a radius is drawn to the extremity of the other leg; we 
 then have 
 
 y = x + z (19) 
 
 or, as x = z (20) 
 
 y = 2x, therefore x \y . 
 
As y is measured by the arc AB, x will be measured by 
 half the arc. 
 
 b) If one leg lies at each side of the centre, this case is 
 made to refer to the preceding 1 one by drawing a diameter 
 from the vertex; we then have 
 
 x == \^AB\ y == |^BG\ ..... (24, a) 
 therefore 
 
 c) In the same manner we get, if both legs lie at the same 
 side of the centre, 
 
 x + y = = \^AC\ y = = ^BC, .... (24, a) 
 therefore by subtraction 
 
 x = ^AB. 
 Thus the proposition holds good in all cases. 
 
 25. As the proposition in 24 holds good however small 
 the one chord becomes, it must also hold when the chord 
 grows infinitely small, that is, when it (produced) becomes a 
 tangent. Therefore an angle contained by a chord and a 
 tangent is measiired by half the arc which the chord czits of. 
 
 26. From these propositions it follows that an angle at 
 the circumference, which stretches over a diameter, is a 
 right angle, for the arc on which it stands is 180, and that 
 the opposite angles of an inscribed quadrilateral are supple- 
 mentary angles as they together stand on the whole circum- 
 ference, which is 360. 
 
 27. An angle, having its vertex inside the circle, is mea- 
 sured by half the sum of the arcs, which it and its vertical 
 angle intercept. 
 
 For x = y + z ........ (19) 
 
 and y = \^B\ z = \^b, .... (24) 
 
 therefore 
 
 x = -J-fl + \t> = -J (B + b). 
 28. An angle, having its vertex oiitside 
 the circle, is measured by half the difference 
 of the arcs, which it intercepts. 
 
 The legs of the angle can be two secants, a secant and 
 a tangent, or two tangents. 
 
In all cases we have 
 
 y = x + z, or x = y z\ (19) 
 
 but y = = -J-5; z = = J, (24 and 25) 
 
 therefore x = -J (B b). 
 
 In the last case, instead of #, we can put 360 &, from 
 which we get 
 
 x = i -(360 2b) 180 6. 
 
 Therefore an angle contained by two tangents is mea- 
 sured by subtracting the mimber of degrees in the smaller 
 arc from 180. 
 
 From 27 and 28 it follows that of any angles standing 
 on an arc AB, those which have their vertices outside the 
 circle are less, and those having their vertices inside, greater, 
 than an angle at the circumference standing on AB. 
 
 Note. If the legs of an angle C pass, the one through A, 
 the other through 5, a circle can always be described through 
 A and B, so that C shall lie within this circle. Describe a 
 semicircle on AB] if /_ G ~> R., G'will lie within this semicircle; 
 if not then with the middle point of the circumference of the 
 semicircle as centre, describe a circle through A and B\ the 
 angles at the circumference of this circle, standing on AB, are 
 -JR. ; if Z C> iR., then C will lie within this second circle; but 
 if not we proceed in this manner, and shall in succession get 
 circles, the angles at the circumferences of which are |R., JR., 
 &c. , we must then at last get a circle in which the angles 
 at the circumference standing on AB are less than Z. C, and 
 this circle will pass outside C. 
 
IO 
 
 29. Two tangents, drawn from any one point to a circle, 
 are eq^ial, reckoned from this point to the points of contact. 
 For the triangle formed by joining the points of contact is 
 isosceles, as the angles at the base are measured by the same 
 arc (25). (See the last figure to 28). 
 
 30. A tangent is perpendicular to the radius to the point 
 of contact, for the angle they contain is a right angle, as the 
 arc is 180. (25). 
 
 III. PARALLEL LINES. 
 
 31. If a straight line cuts two other straight lines and 
 makes the exterior angle equal to the interior, opposite angle, 
 
 on the same side of the line (as x 
 and y), the two lines are said to be 
 
 \x parallel. Parallel to " is denoted 
 
 \ thus ^. When the angles formed 
 
 \ u by the intersection of one straight 
 
 \line with the parallel lines are equal, 
 they will be eqzial for all straight 
 lines. 
 For as z = x 4- ? 
 
 then z = u, when x = y. 
 
 32. At each of the points of intersection there are four 
 angles; when the lines are parallel, each of the acute angles 
 in the one group is equal to each of the acute angles in the 
 other group, but is the supplement of each of the obtuse 
 angles. Conversely, when one of these conditions is fulfilled, 
 the lines are parallel. 
 
 33. Parallel lines can never cut each other. 
 
 For if they did, we would, by drawing a line cutting 
 them both, get a triangle, in which the angle adjacent to one 
 angle of the triangle would be equal to one of the other 
 angles of the triangle, which is in opposition to 19. 
 
1 1 
 
 Note. Conversely, we see that the lines must meet if x 
 and y are not equal; for let A be a point in the one line, 
 B a point in the other, and AB the line joining them ; we can 
 then always (28, Note) in the line through A, within a certain 
 finite circle find a point 6', so that AGB = y x (or x y); 
 CB must therefore just be the line through B, as its angle 
 with AB becomes y\ that is, the two lines meet in a point 
 C at a finite distance. 
 
 34. By observing the angles formed by an intersecting 
 line, it is evident that through one given point, there can 
 only be drawn one line parallel to a given line, and that 
 
 When two lines are parallel to the same third line, they 
 are parallel to one another. 
 
 35. Angles with parallel legs 
 are eqital, when the legs both extend 
 in the same direction or both in oppo- 
 site directions. 
 
 If one of the legs of the one 
 angle be produced till it cuts a leg 
 of the other angle, we get 
 
 therefore z = x. 
 
 EXAMPLES. 
 
 1. In a triangle one angle is 75, another 42; how great 
 is the third? 
 
 2. In a triangle the two angles are each 42 12' 42"; how 
 great is the third? 
 
 3. In a triangle one angle is A, the other 5; how great 
 is the third? 
 
 4. In an isosceles triangle the vertical angle is 60 ; how 
 great is the angle at the base? 
 
 5. How great is the angle between perpendiculars on two 
 of the sides of an equilateral triangle? 
 
 6. is a point within a triangle ABC] prove that /L AOB 
 > Z ACB. 
 
12 
 
 7. The vertical angle of an isosceles triangle is 40 ; prove 
 that the line bisecting its adjacent angle is parallel to 
 the base. 
 
 8. How great is the angle between the lines bisecting two 
 adjacent angles) 
 
 9. In a triangle the two angles are 40 and 80 ; bisect the 
 third angle, and from its vertex draw a perpendicular 
 to the opposite side; how great is the angle between 
 this and the bisecting line? 
 
 10. In a right-angled triangle one af the acute angles is r, 
 how great is the other and how great are the angles 
 in which the right angle is divided by a perpendicular 
 from it to the hypothenuse? 
 
 11. Prove that the perpendicular from one end of the base 
 of an isosceles triangle to the opposite side, forms an 
 angle with the base, half as great as the vertical angle. 
 
 12. In a right-angled triangle the one angle is 60; this is 
 bisected, and from the vertex of the right angle a per- 
 pendicular is drawn to the hypothenuse. Prove that one 
 of the triangles, thus formed, is equilateral. 
 
 13. The legs of one angle are perpendicular to the legs of 
 another angle; prove that the two angles are equal or 
 supplementary. 
 
 14. One of the angles at the base of an isosceles triangle, 
 whose vertical angle is 36, is bisected; prove that the 
 two small triangles will be isosceles, and find the lengths 
 of all the lines in the figure, when the sides of the given 
 triangle are r and the base t. 
 
 15. In a triangle one angle is v\ how great is the angle 
 between the lines which bisect the two other angles? 
 
 1 6. Prove that an angle of a triangle is a right angle, when 
 a line from its vertex to the middle point of the oppo- 
 site side is half as great as this. 
 
 17. Prove that, when an angle of a right-angled triangle is 
 30 , then the lesser of the sides containing it is half 
 as great as the hypothenuse. 
 
1 8. The side AB of a triangle ^50 is produced to D, so 
 that #/) = .Z?G\ Prove that the line bisecting the angle B 
 is parallel to the one joining DC. 
 
 19. Prove that in a right -angled isosceles triangle, the per- 
 pendicular from the vertex of the right angle to the 
 hypothenuse is half as great as this. 
 
 20. The exterior angles of a triangle are bisected, thereby 
 three triangles are formed, each having a side in common 
 with the given triangle. Prove that the three triangles 
 contain the same angles. 
 
 21. On one leg of any angle ABC, mark off any part AB, 
 thereupon draw a line AD^= BC, and make AD = AB. 
 
 Prove that the line BD bisects the given angle. 
 
 22. In a triangle ABO the lines bisecting the angles A and 
 B intersect in 0. Through is drawn DE ^ AB. Prove 
 that DE = = AD + RE. 
 
 23. The demonstration of the proposition in 31 cannot be 
 applied, if the two lines xy and za cannot by prolongation 
 be made to cut each other. How may it be demon- 
 strated in this case? 
 
 24. Two angles have parallel legs; shew that the lines bi- 
 secting them are parallel or perpendicular to each other. 
 
 25. Ho^v many diagonals can be drawn in a polygon with 
 
 n sides? (Ans. - n - j 
 
 26. In a quadrilateral the opposite sides of which are par- 
 allel the one angle is v\ how great are the others? 
 How great must the angle v be, in order that a circle 
 can be described about the quadrilateral, and where will 
 the centre of this circle lie? 
 
 27. Prove that the chord ol an arc of 60 equals the radius. 
 
 28. In a circle two chords issue from the same point, and 
 cut off arcs of 120 and 80. How great is the angle 
 between the two chords? 
 
 29. In a circle a diameter AB is drawn, and a chord CD 
 equal to the radius. How great is the angle between 
 AC and BD, and between AD and BC1 
 
30. In a triangle ABC, BC < BA. With B as centre and 
 BO as radius a circle is described, cutting CA in 2, BA 
 in ZX Shew that Z Z>^.4 - = i#. 
 
 31. In a triangle ACB, /. C = jfr. With centre describe 
 a circle through the middle point of AB, cutting the 
 hypothenuse or its prolongation the second time in D. 
 Shew that the acute angle which CD makes with the 
 hypothenuse is double of one of the angles of the tri- 
 angle. 
 
 32. A circle has its centre on the circumference of another 
 circle; the two circles cut one another in A and B\ prove 
 that the one arc AB contains half as many degrees as 
 one of the other arcs AB. 
 
 33. In a circle two equal angles at the circumference BAG 
 and DEF are drawn; shew that EF^h CD. 
 
 34. A triangle ABC is inscribed in a circle with centre 0\ 
 D is the middle point of the arc BC. Shew that the 
 angle ADO is half as great as the difference between 
 B and 0. 
 
 35. Two chords EA and EB in a circle are produced to 
 and />, so that CD is parallel to the tangent at E. Prove 
 that the opposite angles of A BCD are supplementary. 
 
 36. A is one of the points of intersection of two circles, of 
 which the one passes through a point B, the other 
 through a point C. Through A a line is drawn, cutting 
 the first circle in Z>, the other in E. Prove that the 
 angle between the lines BD and CE is constant (the 
 same, in whatever way the line through A is drawn). 
 
 37. A circle touches one side of a triangle and the two 
 other sides produced. Prove that the distances from 
 the points of contact with the latter to their point of 
 intersection equals half the perimeter of the triangle. 
 
 38. A triangle ABC, the angles of whieh are known, is in- 
 scribed in a circle. How great is the angle which the 
 tangent touching the circle at A makes with BC! 
 
 39. Three small triangles are cut off from a triangle by tan- 
 gents to its inscribed circle. Shew that the perimeters 
 
of the three triangles are together equal to the perimeter 
 of the given triangle. 
 
 40. Prove that the four lines bisecting the angles of any 
 quadrilateral bound a quadrilateral, the opposite angles 
 of which are supplementary. 
 
 41. Two circles cut each other in A and B. From A the 
 diameters AC and AD are drawn; shew that C, B, and 
 D lie in a straight line. 
 
 42. An angle has its vertex outside the circumference; one 
 of its legs passes through the centre, and the part out- 
 side the circle of the other one equals the radius. Prove 
 that the greater of the intercepted arcs is three times 
 as great as the lesser. 
 
 43. Through each of the points of intersection of two circles 
 a straight line is drawn. Prove that the chords, joining 
 the other points of intersection of these with the circles, 
 are parallel. 
 
 44. Two circles with equal radii cut each other; shew that 
 they divide each other into arcs which are respectively 
 equal. 
 
 45. With one of the points of intersection of tw r o equal 
 circles as centre describe a circle cutting the given circles. 
 Shew that two and two of the four points of intersection 
 are in a straight line with the other point of intersection 
 of the given circles. 
 
 46. Two equal circles cut each other in A and B. Through 
 A a line is drawn cutting the circles in D and E. Shew 
 that DE is bisected by a circle on AB as diameter. 
 
 47. Between the circumferences of two circles, two lines are 
 drawn through one of the points of intersection of the 
 circles. Prove that the chords joining the extremities 
 of the lines make the same angle in whatever way 
 the two lines are drawn. 
 
i6 
 
 IV. RELATIONS BETWEEN THE LENGTH OF 
 STRAIGHT LINES. 
 
 36, The length of a straight line is measured, by stating 
 how many times it contains another straight line, the unit, the 
 length of which is supposed known; the foot it used as unit; 
 it is divided into 12 inches, the inch into 12 lines (Duodecimal 
 measure) or the inch into 10 parts (Decimal measure). 5 feet 
 7 inches 3 lines is written 5' 7" 3'". 
 
 37, The greater side of a triangle has the greater angle 
 
 opposite to it. 
 
 From the greater side BC cut off 
 BD = BA. 
 
 Then Z DAB = = Z EDA, 
 
 but Z A > Z BAD, 
 
 Z BDA > Z C (19), 
 
 therefore Z A > Z 0. 
 
 38, Z? greater angle of a triangle is subtended by a 
 greater side. 
 
 When Z ^4 > Z C, then 5C> BA\ for if not, then either 
 BC == BA or BC<BA, but from the first case it would 
 follow that Z A = Z C, and from the second that Z A < Z C, 
 and both are in opposition to what was given. 
 
 39. One side of a triangle is less 
 than the sum of the other two. 
 
 Make BD = BC, then x == y, 
 therefore Z ACD > Z x, but from 
 that it follows (38) that AD > AC, 
 
 ^ or AB + BC > AC. 
 
 A. C This result may also be written 
 
 AB ->AGBG, so that in any 
 
 triangle one side is greater than the difference of the 
 other two. 
 
40. A straight line is shorter 
 than any broken line joining 
 its extremities. 
 
 ACDEB > ADEB > AEB 
 
 > AB . . . . . (39) 
 
 41. From a point there can B 
 only be dropped one perpendic- 
 
 ular to a line, and this is shorter than any other line from 
 the point to the line. 
 
 When BC -L AC, then we cannot have 
 BA -L- AC, as there cannot be two right angles 
 in a triangle (18); further by 38, BA > BC 
 as /. C > Z A. 
 
 The point C, where the perpendicular from 
 B meets the line A C, is called the projection 
 of B on AC. If D is the projection of another 
 point E, CD is called the projection of the line BE. 
 
 42. Two lines which diverge eqitally 
 from the perpendicular are equal. 
 
 That AB and BC diverge equally 
 either signifies, that Z x == /Ly or that 
 AD F= DC. In both cases the one side 
 of the figure must coincide with the 
 other, when it is turned round the per- 
 pendicular. 
 
 43. When two lines diverge ^ineq^ially from the perpen- 
 dicular, that is greatest which diverges 
 
 most. 
 
 If both lines lie at the same side of 
 the perpendicular, the proposition follows 
 from 38, as Z x > Z. y\ if they lie one on 
 each side, the one is turned round to the 
 same side as the other. 
 
 44. Every point in the perpendicular on the middle point 
 of a line is equidistant from the extremities of the line. 
 
i8 
 
 \C 
 
 B 
 
 The proposition follows from 42, as 
 AB = EG. 
 
 45. A point, lying outside the per- 
 pendicular on the middle point of a line, 
 is nearest to that extremity of the line, 
 which is on the same side. 
 
 For if the perpendicular BD be drawn, 
 then AD > DC, therefore AB > BC (43). 
 
 The perpendicular on the middle 
 point of a line therefore just contains 
 all the points eqitidistant from the ex- 
 tremities of the line, and no others. 
 
 46, If two triangles have two sides 
 equal, bitt the angles contained by them 
 
 uneqiial, the third side is greatest in the triangle which 
 has the greatest angle. 
 
 Let ACB and ADB be the tri- 
 angles; they are placed, so that the 
 equal sides AB coincide, further AD 
 = AC] a perpendicular on the middle 
 of CD will pass through A (45, last 
 piece); therefore CB > DB. (45). 
 
 47. The altitude of a triangle 
 is a line, drawn from an angular 
 
 point, perpendicular to the opposite side, which is then called 
 the base. The altitude must coincide with one of the sides, 
 if one of the angles at the base is a right angle, and fall out- 
 side the triangle, if one of the angles is obtuse; besides this 
 line, two others of particular interest issue from each angular 
 point, namely, the median line, going to the middle of the 
 opposite side, and the line bisecting the angle. 
 
 In the isosceles triangle the altitude, median line and 
 the line bisecting the vertical angle, coincide in one line ; 
 for if the altitude did not bisect the vertical angle or the 
 base, the sides of the triangle would be unequal (43). 
 
 E. 
 
48. The middle point of a chord, the 
 middle points of the corresponding arcs, 
 and the centre all lie in a line, perpen- 
 diczilar on the middle point of the chord. 
 
 For all the points lie equidistant from 
 the extremities of the chord. 
 
 EXAMPLES. 
 
 48. Prove that the diameter of a circle is the greatest chord. 
 
 49. Prove that the arcs between two parallel chords in a 
 circle are equal. 
 
 50. Which is the greatest and which the least line, that can 
 be drawn from a point to the circumference of a circle? 
 
 51. From a point within a triangle ABC lines are drawn 
 to B and C. Prove that OB + OC < AB -f AC. 
 
 52. Prove that in a circle, when ^ AB < ^ AC, then also 
 AB < AC. (The arcs are both supposed less than 180). 
 
 53. Prove that each side of a triangle is less than half the 
 perimeter. 
 
 54. Prove that in a circumscribed qziadrilateral , the sum 
 of the one pair of opposite sides is eqzial to the sum 
 of the other pair. (29). 
 
 55. With a point A outside a circle as centre, a circle is 
 described through the centre of the given circle, 
 and two chords are drawn from 0, equal to the di- 
 ameter of the given circle. Shew that these chords cut 
 the given circle in the points of contact of the tangents 
 from A. 
 
 56. Shew that any triangle, which lies wholly inside another 
 triangle, has less perimeter than this. 
 
 57. From a point within a triangle lines are drawn to the 
 three angular points; shew that the sum of these three 
 lines is greater than half the perimeter of the triangle, 
 but less than the whole perimeter. 
 
20 
 
 58. From the angular points of a triangle three lines are 
 drawn in such a manner, that any two of them are legs 
 of an isosceles triangle, of which one of the sides is the 
 base. Prove that the three lines intersect in the same 
 point. (Suppose that the three lines form a triangle 
 and express the sides of this in terms of the legs of 
 the isosceles triangles). 
 
 II. 
 
 I. CONSTRUCTION, CONGRUENCE AND SYMMETRY. 
 
 49. For the construction of figures with certain given 
 properties are used only the ruler, by which a straight line 
 can be drawn through two given points, and the compasses, by 
 which a circle, with given centre and given radius, can be 
 described. All constructions must therefore be a combination 
 of these two. In order that a point may be determined, there 
 must be given two conditions, which it must fulfil; if there 
 is only one condition given, there will be an infinite number of 
 points, which satisfy the problem, but these will all lie in a 
 certain straight or curved line, called the loczis of the point; 
 thus we have from the preceding: 
 
 The locus of the points vvhich are at a given distance 
 from a given point is a circle, with the given point as centre 
 and the given distance as radius. 
 
 The locus of the points which are equidistant from two 
 given points is a straight line, perpendictilar on the middle 
 point of the line joining the two given points. (45). 
 
 More loci will be mentioned hereafter. 
 
 Now if the problem is to find a point, we examine, which 
 two loci correspond to the two conditions, which the point, 
 according to its position, must fulfil; if the two loci can be 
 constructed by compasses and ruler, the point can be found, 
 
21 
 
 for, as it is to lie in them both, it must lie where they inter- 
 sect; the problem has therefore as many solutions, as the 
 loci have points of intersection. 
 
 50. When certain parts only in one way can be put to- 
 gether to make a figure. , then two fibres , both containing 
 these parts, imist be congment, for if they were not, the parts 
 would be put together in two ways; but if the parts can be 
 put together in different ways, then two figures containing 
 these parts do not require to be congruent. 
 
 51. Two figures are said to be symmetrical with regard 
 to a straight line, when to each point in the one figure, there 
 is a corresponding point in the other, so placed, that it is 
 found by dropping a perpendicular from the first point on to 
 the line, and producing it equally far on the other side; thus 
 A corresponds to a, B to //, &c. Symmet- 
 rical figures are congruent, as the one, by 
 
 being turned over, round the straight line 
 
 (axis of symmetry), will coincide with the 
 
 other; conversely, two figures, which, by 
 
 being turned over, round a straight line, 
 
 would coincide, must lie symmetrically with 
 
 regard to it. As examples of symmetrical 
 
 figures, we may mention that the altitude 
 
 from the vertex divides an isosceles triangle symmetrically, 
 
 that every diameter divides a circle symmetrically, &c. 
 
 When two pair of figures lie symmetrically with regard 
 to the same line, their points of intersection must also lie 
 symmetrically, for when the figures, by being turned, coincide, 
 their points of intersection must also coincide. 
 
 In constructing a figure, we often get two symmetrical 
 solutions, which therefore are congruent. 
 
 52. To draw a straight line, perpendicular on the middle 
 point of a given straight line. 
 
 With the extremities A and B of the line given as centres, 
 describe arcs with equal radii, then their points of intersection 
 must lie in the line required (45) ; in this manner two points in 
 the required line are found, which is then drawn through these. 
 
22 
 
 A- 
 
 The radii must be taken greater than 
 half AB, so that the arcs can cut each other. 
 By the same construction a line may 
 be bisected. 
 
 53. To describe a circle, passing through 
 three given points A, B and C. 
 
 Find the centre; it must be equidistant 
 from A and B, therefore in a perpendicu- 
 lar on the middle of AB, likewise in a 
 perpendicular on the middle of BC, and 
 must therefore lie where these intersect; 
 as there is only one point of intersection, 
 there can only be described one circle 
 passing through three given points. If the 
 three points lie in a straight line, the two 
 perpendiculars will be parallel and have 
 no point of intersection, so that the problem is impossible; 
 we usually say, that the parallel lines cut one another at an 
 infinite distance, and that therefore, through three given points 
 lying in a straight line, there can be described a circle, of 
 which the centre is infinitely distant and the radius of which 
 therefore is infinitely great; by this is only meant, that by 
 taking the centre sufficiently far away, a circle can be got 
 passing through the two points, and as near as we please, 
 to the third point. 
 
 If the centre of a given arc is to be found, we employ 
 the same construction, by taking any three points in the arc. 
 From this it follows, that two circles only can cut each other 
 in two points, for if they had three points in common, they 
 must coincide. The line joining the centres is called the line 
 of centres', if it is produced, it divides both circles symmetri- 
 cally, and therefore is perpendicttlar on the middle of the 
 line joining the points of intersection of the circles. (51). 
 
 If radii be drawn to one of the points of intersection, 
 they will be sides of a triangle, of which the line of centres 
 is the third side, and this must therefore be less than the 
 sum of the radii and greater than their difference, when the 
 
2 3 
 
 circles cut one another (39), whilst 
 it is greater than the S2im of the 
 radii, when the circles lie outside 
 each other, and less than their 
 difference, when one circle lies 
 within the other, 
 
 If the chord, which the circles 
 
 have in common, becomes infinitely small, so that the points 
 of intersection coincide in one point, this is called a point of 
 contact', it must lie in the line of centres, for, so long as 
 there is one common point on the one side of the line of 
 centres, there must also be one symmetrically w r ith this on 
 the other side. 
 
 From this we see, that the line of centres is equal to 
 the sum of the radii, when the circles touch one another 
 externally, and eqztal to their difference, when they toitch 
 internally. 
 
 Conversely, if the line of centres equals the sum or 
 difference of the radii, the circles must touch one another. 
 
 54. To raise a perpendic2ilar on a given line at a given 
 point. 
 
 a) From the given point cut off the equal 
 parts OA and OB. Next determine a 
 point equidistant from A and B, this must 
 lie in the required line, which now can 
 be drawn. 
 
 b) If the given point is the extremity of the 
 line, and this cannot well be produced, 
 a circle is described with any centre 
 
 G passing through the given point 0; 
 from the other point of intersection of 
 the line with the circle draw the di- 
 ameter AD; DO will then be the per- 
 pendicular. For Z. is a right angle, 
 as it stands on a diameter (26). 
 
 o 
 
55. From a given point A to draw 
 a perpendicitlar to a given line BC. 
 Take any two points in the line as 
 centres, and describe circles through 
 the given point; the line joining the 
 points of intersection of the circles will 
 be the required line (53). 
 
 56. To bisect a given angle or arc. 
 With the vertex of the angle C as 
 
 centre, describe the arc AB; next find a 
 point D equidistant from A and B. Then 
 CD bisects both the angle and the arc, 
 as it by construction is perpendicular on 
 the middle point of the chord AB. Any 
 angle cannot be divided by compasses and 
 ruler into equal parts in any other way than by continued 
 bisection, therefore only into 2, 4, 8, 16, &c. equal parts. 
 
 57. Throiigh a given point 
 in a line to draw a line, making 
 an angle with the given line 
 equal to a given angle. 
 
 Let be the given point 
 and A the given angle, with 
 
 and A as centres describe the arcs BC and DE with equal 
 radii; now when ^ D E is made equal to ^ BC, then Z. 
 - /- A. (24). 
 
 58. Throtigh a given point to draw a 
 line parallel to a given line. From the 
 given point A draw any line AB, cutting 
 the given line 50; then if we make Z A = 
 ~C /_ B, then DA = BC. (31). 
 
 59. To construct a triangle, having given 
 the three sides. 
 
 Mark off the one side A B ; the vertex of the triangle will 
 then be the point of intersection of two circles, with centres 
 A and B, and with the other given sides as radii. The prob- 
 lem is possible, when the circles cut each other, therefore, 
 
 D 
 
 B 
 
when the one side is less than the sum 
 of the other two and greater than their 
 difference (53). The circles cut each 
 other in two points, but the two solu- 
 tions are symmetrical, so that three lines 
 only in one way can be' put together 
 to form a triangle. From this it follows, 
 that two triangles are congruent, when 
 they have the three sides respectively equal. 
 
 60. To construct a triangle, having given two sides and 
 the angle contained by them. Mark off the given angle, and 
 mark off parts on its legs, equal to the given sides. As the 
 problem only has one solution, two triangles are congruent, 
 ivhen they have tzuo sides and the angle contained by them 
 respectively equal. 
 
 61. To construct a tri- 
 angle, having given an angle, 
 an adjacent side, and the 
 opposite side. 
 
 Mark off the given angle 
 A, and on one of its legs, 
 the adjacent side AB. The 
 point C will be determined 
 by a circle with centre B 
 and radius BC] different cases can now occur: 
 
 a) If BC is less than the perpendicular BD, the circle will 
 not cut AD, and the problem is impossible. 
 
 b) If BC == BD, the circle will touch AD at D, and there 
 will be one solution, namely, the right-angled triangle ADB. 
 
 c) If BC > BD but < BA, the circle will cut. AD in two 
 points, so that the problem has two solutions. 
 
 d) If BC == BA, the one point of intersection falls in A, so 
 that the one solution gives no triangle. 
 
 e) If BC > BA, the one point of intersection falls on the 
 opposite side of A, and the corresponding triangle does 
 not contain the given angle A, but its adjacent angle; 
 there is therefore in this case only one solution. 
 
26 
 
 From this it follows, that two triangles, having one angle, 
 an adjacent side, and the opposite side respectively equal, 
 are congment , if the opposite side be greater than or equal 
 to the adjacent side, but that, if this is not known, it is not 
 certain that the triangles are congruent. 
 
 62, To construct a triangle, having given a side and the 
 angles at the side. 
 
 Mark off the given side and on it the given angles, and 
 produce their legs till they intersect. 
 
 The problem is always possible, when the sum of the 
 given angles is less than 2R. As there only is one solution, 
 two triangles are congruent, when they have one side and 
 the angles at that side respectively equal. 
 
 63. To construct a triangle, having 
 given a side, an angle at the. side, and 
 the opposite angle. 
 
 Mark off the given side AB and 
 on it the given /L A ; thereupon make 
 Z x = Z. C; then Z y equals Z B, as 
 the sum of the angles is 2R. Then 
 draw EG parallel to AD. 
 
 The conditions of possibility for 
 the construction of this is the same as 
 
 for the preceding one. As there only is one solution, two 
 triangles are congruent, when they have one side, an angle 
 at the side, and the opposite angle equal respectively. 
 
 An isosceles triangle can be constrticted, having given the 
 base and the vertical angle. Draw any isosceles triangle with 
 the given vertical angle D, and on the base AB of this triangle 
 mark off AC equal to the given base; thereupon through C 
 draw a parallel to BD. 
 
 64. To describe a circle, touching three given straight 
 lines. 
 
 If is the centre, Oc and Ob radii to the points of contact, 
 then A AbO m A AcO (61, e; for we have Ob = = Oc\ AO = 
 A0\ Z&=/lc=R. and AO > Ob], so that the centre must 
 lie in the line bisecting Z. A. Similarly the centre must lie 
 
in the line bisecting Z. C, therefore 
 in the point of intersection of these 
 lines ; the line bisecting the third angle 
 also passes through this point. This 
 circle is called the inscribed circle of 
 the triangle; besides this one there 
 can, in a similar way, be constructed 
 three other circles, touching the three 
 lines, which will lie in the spaces P, 
 Q and R. These are called the escri- 
 bed circles of the triangle. 
 
 The length of the distances, between the points of con- 
 tact and the angular points of one of the lines, may be 
 expressed in a simple manner by the sides of the triangle. 
 
 If AB = = c, AC = = b, BC = a, and if we for the sake 
 of brevity call the perimeter 2s, and Ac = x, Be = y, Cb 
 = z, we have (29), 
 
 x 
 
 y = 
 
 y + z = 
 
 
 wherefore by addition and division by 2, 
 
 x + y + z = , 
 
 therefore 
 
 x = s a; y = s b\ z = s c. 
 
 Of the other distances may be mentioned the one from C 
 to the points of contact of the escribed circle with CA or CB. 
 These two distances, it will be seen, are together equal to 
 the perimeter of the triangle; therefore each equal to s. (29). 
 
 The determination of above also shews, that the locus 
 of the points which are equidistant from two given lines 
 is two lines, which bisect the angles between the given lines, 
 and therefore are perpendicular to one another. 
 
 65. On a given line as chord to describe an arc, so 
 that all the angles, standing on the chord and having their 
 vertices in the arc, shall be equal to a given angle. 
 
 Let AB be the given chord and v the given angle; as the 
 angles at the circumference, standing on AB, are to be equal 
 to v, the same must be the case with the angle, formed by 
 
28 
 
 the chord and the tangent at A, as 
 it is measured by the same arc; 
 therefore make Z. BAD = Z v, and 
 the centre may then easily be found. 
 For as AD must be a tangent, the 
 centre must lie in the perpendicular 
 AF on AD, and therefore lie where 
 this cuts a perpendicular on the 
 middle point of the chord. If the 
 given angle is a right angle, the arc 
 will be a semicircle. 
 
 We say that the arc or segment contains the given angle, 
 and that the arc is the locus of the points, at which the line 
 AB subtends the given angle, or from which A B is seen under 
 the given angle. 
 
 66. To draw tangents from a 
 given point P to a circle. 
 
 As the tangent must be per- 
 pendicular to the radius and there- 
 fore Z. CEP be a right angle, the 
 points of contact must lie in a 
 circle on CP as diameter. If the 
 given point P lies in the circum- 
 ference, a perpendicular is drawn on 
 
 the radius to the point ; if it lies within the circle, the problem 
 is impossible. 
 
 II. POLYGONS. 
 
 67. The propositions regarding the congruence of triangles 
 are often applied to prove that two parts of a figure are 
 equal. We then endeavour to find , or by the help of lines 
 to form, two triangles, in which these parts occur as corre- 
 sponding parts; if we can prove that the two triangles are 
 congruent, the problem is solved; in the following, when we 
 state that two triangles are congruent, we will place the 
 letters in such an order, that the corresponding parts come 
 
2 9 
 
 in the same place; thus, if we write A ABC & A ale, then 
 /.A == a, B == b, C == c, J5 == a, ^4C == ac, BC = be. 
 
 68. A trapezium is a quadrilateral, which has a pair of 
 sides parallel. 
 
 A parallelogram is a quadrilateral, which has both pair 
 of sides parallel. 
 
 69. The opposite sides and angles 
 of a parallelogram are equal. 
 
 CD, therefore Z x == y; 
 
 OB, therefore Z v == 2; 
 therefore 
 
 A ^156 T 23 CZM .... (62) 
 from this it follows v 
 
 AB=-CD] BC == DA-, Z 5 = = Z Z>. 
 
 From this proposition it follows that parallels all over 
 
 are at the same distance from each other, and that the locus 
 
 of the points which are at a given distance from a given line 
 
 is trvo lines parallel to the given line, at the given distance. 
 
 70. A quadrilateral, having the opposite sides equal, is 
 a parallelogram. 
 
 We have AB = = CD, 
 
 BC == DA, 
 
 AC == AC, 
 
 therefore A ABC & A CD A, 
 
 from which it follows 
 
 Z. x = y ; z = v 
 or AB^CD', BC^AD. 
 
 71. A quadrilateral, having a pair of opposite sides 
 equal and parallel, is a parallelogram. 
 
 Here we also get A ABC m A CD A. (60). 
 
 From this proposition it follows that tivo lines are par- 
 allel, when two points in the one line are at the same dis- 
 tance from the other line (and lie at the same side). 
 
 72. A quadrilateral, having the opposite angles equal, 
 is a parallelogram. 
 
 We have /_ A + B+ C+ D == 4R., 
 
 but A = C; B = Z); (Hyp.), 
 
therefore 
 
 from which it follows 
 
 Similarly 
 
 therefore 
 
 2 A 
 A 
 
 28 == 4R., 
 B = 2 R. , 
 
 73. 
 
 A + D = 2R., 
 AD j= CD. 
 
 diagonals of a parallelogram bisect each other. 
 For we have 
 
 C A BOG & A DO A 
 
 therefore 
 
 BO = D0\ OC == OA. 
 
 A D 74. Of special parallelograms may be 
 
 mentioned the rectangle, containing right 
 angles, and the rhombus, the sides of which are all equal. 
 A sqiiare is both rhombus and rectangle. 
 
 75. A reg^{,lar nrayed figure is one, which coincides with 
 
 itself, when it revolves through an angle equal to - - of 4 R. 
 
 round a certain point, the centre. 
 
 Corresponding points, lines, &c. are such, which, by the 
 said revolution, take each others places. 
 
 If we join the corresponding points in order , a regiilar 
 polygon is formed, which thus is an n rayed n sided figure; 
 the regular polygon is convex; the n rayed polygon with 
 re-entrant angles is called a star shaped polygon. 
 
 76. All the sides and angles of a regular polygon are 
 eqztal; for they all by degrees coincide by being turned round 
 
 the centre. For the same reason the 
 lines from the centre to the angular 
 points, the greatest radii (0 A), are equal, 
 and similarly the perpendiculars from 
 the centre on the sides, the least radii 
 (OD). The greatest radii divide the 
 polygon into congruent, isosceles tri- 
 angles, central triangles (AOB)] the 
 angles of these at the centre are called central angles ( AOB). 
 If n congruent, isosceles triangles, the vertical angles of 
 
 which are R., be laid side by side, with their vertices at 
 
one point, a regular n sided figure is formed; having obtai- 
 ned this, we can draw another regular n sided figure, with 
 sides equal to a given line (63, last piece). 
 
 77. The angles of a regular n sided figure are each 
 
 2 R. 4-R. 
 n 
 
 This is found by dividing the sum of the angles 2n R. 
 4 R. by their number. 
 
 78. A polygon is regular, when all its sides and all its 
 angles are equal. 
 
 For if it be compared to a regular polygon (76, last piece), 
 the sides of which are of the same length, and which has the 
 same number of sides, the two polygons will be seen to be 
 congruent, as they also have the angles equal (77), and but 
 one polygon can be constructed of these sides and angles. 
 
 Of regular figures we have already mentioned the equi- 
 lateral triangle and the square. 
 
 79. If a circumference be divided into a certain number, 
 n, equal parts, and the points of division be joined by chords, 
 a regular inscribed n sided figure is formed; if tangents be 
 drawn to all the points of division, a regular circumscribed 
 n sided figure is formed. For in both cases the figure will 
 
 coincide with itself, by being turned - - revolution round the 
 centre. 
 
 EXAMPLES. 
 
 59. In a square ABCD a diagonal AC is drawn, and on this 
 marked off AE = = AB, at E a perpendicular is raised 
 cutting BC in F. Prove that BF = = FE == EC. 
 
 60. Prove that the diagonals of a rectangle are equal. 
 
 61. Prove that the diagonals of a rhombus are perpendicular 
 to each other. 
 
 62. Prove that a rectangle is formed by joining the middle 
 . points of the sides of a rhombus. 
 
 63. Prove that a rhombus is formed by joining the middle 
 points of the sides of a rectangle. 
 
64. The distance between two parallels is a; any line AB 
 is drawn between the parallels, and the two supplement- 
 ary angles A and B are bisected; how great is the alti- 
 tude of the right-angled triangle, formed thereby? 
 
 65. In a right-angled triangle, the right angle is bisected, from 
 the point where the bisecting line cuts the hypothenuse, 
 lines are drawn parallel to the two sides containing the 
 right angle. Prove that the figure contained by these 
 two lines and the two sides is a square. 
 
 66. The point C bisects the line AB. The three points are 
 projected on to any line in A^ B l and C l ' shew that C l 
 bisects A l B l . 
 
 67. Two points A and B lie on the same side of a straight 
 line. From A a line is drawn to the point 0, which lies 
 symmetrically with B. This line cuts the given line in 
 a point D. Prove that AD and ED make equal angles 
 with the given line. 
 
 68. A line is parallel to the diagonal of a parallelogram; 
 shew that the one pair of opposite sides cut off a part 
 of the line, equal to the part cut off by the other pair. 
 
 69. Prove that the opposite sides of a square cut off a part 
 of any line, equal to the part cut off a line perpendicular 
 to this, by the other two opposite sides. 
 
 70. In a right-angled triangle squares are described on the 
 sides containing the right angle, and from the outermost 
 angular points of these, perpendiculars are dropped on 
 the hypothenuse; prove that the sum of the perpen- 
 diculars equals the hypothenuse. 
 
 71. From a point in the base of an isosceles triangle, per- 
 pendiculars are drawn to both the sides. Prove that 
 the sum of these perpendiculars equals the altitude from 
 an extremity of the base. 
 
 72. Which parallelograms can be inscribed in a circle, and 
 which can be circumscribed? 
 
 73. Prove that a polygon can be constructed, when all its 
 parts are known, except three, which follow after each 
 
33 
 
 other (i side and 2 angles or 2 sides and i angle). 
 Which proposition on the congruence of polygons follows 
 from this) 
 
 74. In a sector of 60 a circle is inscribed ; prove that its 
 radius is a third of the radius of the circle to which the 
 sector belongs. 
 
 75. Which is the longest line that can be drawn between 
 the circumferences of two circles through their one point 
 of intersection. 
 
 76. Two given lines intersect in 0. In each of the lines 
 any point, A and 5, is chosen, and a triangle ABC with 
 given angles is constructed. The angle at C is equal to 
 the angle at and lies to the same side of AB. Shew 
 that C falls in the same line through 0, wherever A and 
 B are chosen. 
 
 77. From a point a line is drawn to the centre of a circle, 
 and on this line as diameter, another circle is described. 
 Prove that this circle bisects all chords passing through 
 the given point. 
 
 78. In a right-angled triangle a circle touches one of the sides 
 containing the right angle and also the other side at its 
 extremity; prove that the part which the circle cuts off 
 from the hypothenuse is equal to twice the altitude of 
 the triangle. 
 
 79. In a parallelogram a diagonal is drawn, and through a 
 point in this, lines parallel to the sides of the parallelo- 
 gram. The parallelogram is thereby divided into four 
 smaller parallelograms. Prove that the two of these, 
 through which the diagonal does not pass, are equal. 
 
 80. Prove that the one part of the altitude of a triangle, 
 reckoned from the point of intersection of the altitudes 
 to the foot, is equal to the prolongation of the altitude 
 to the circumference of the circumscribed circle. 
 
 8 1. A triangle is divided into two other triangles by a line 
 from an angular point to the point of contact of the 
 opposite side with the inscribed circle. Shew that the 
 
34 
 
 circles inscribed in the smaller triangles touch the side, 
 which the triangles have in common, at the same point. 
 
 82. Through the vertex of an angle and a given point in 
 the line bisecting the angle any circle is described; prove 
 that the sum of the two chords, which the circle cuts 
 off from the legs of the angle, is constant. 
 
 83. On the three sides of a triangle arcs are described 
 inwardly, containing angles of 120; shew that the three 
 arcs pass through the same point. 
 
 84. On the three sides of a triangle equilateral triangles are 
 described outwardly. Shew that the three lines joining 
 the outermost angular points of the equilateral triangles 
 with the opposite angular points of the given triangle 
 i) are equal, 2) cut each other at angles of 120, and. 
 3) pass through the same point. 
 
 85. Prove that a quadrilateral can be inscribed in a circle^ 
 when its opposite angles are supplementary. 
 
 86. On the three sides of a triangle squares are described 
 outwardly. Prove that the three lines, joining the ex- 
 tremities of the outermost sides of the three squares, are 
 twice as great as the median lines of the triangle, and 
 perpendicular to these. 
 
 87. Shew that the altitudes of a triangle bisect the angles 
 of another triangle, having its angular points at the 
 feet of the altitudes. (Find in the figure systems of four 
 points lying in the circumference of the same circle). 
 Prove by the help of this proposition, that the three 
 altitudes intersect in one point. 
 
 88. Any tangent is drawn to a circle with centre 0. Let it 
 cut a fixed line through in M and mark off on the tan- 
 gent MP = MO. What is the locus of P? 
 
 89. In a given circle a triangle is inscribed, the two angular 
 points of which are fixed, whilst the third moves on the 
 arc. Which are the loci of the point of intersection of 
 the altitudes of the triangle, and of the centre of the in- 
 scribed circle? 
 
35 
 
 90. AB is a fixed chord, C a moveable point of the circum- 
 ference. A is produced to D, so that CD = CB. Find 
 the locus of D. 
 
 9 1 . From any point in the circumference of a circle lines are 
 drawn to the angular points of an inscribed equilateral 
 triangle; shew that one of the three lines equals the 
 sum of the other two. 
 
 92. Two men, both living in the neighbourhood of a lake, 
 jointly own a boat. Where should this be placed on the 
 shore, so as to be equally distant from them both? 
 
 93. Construct a triangle, having given an angle, the oppo- 
 site side, and the median line to one of the other sides. 
 (Seek the middle point of this side, when the given side 
 is fixed). 
 
 94. Describe a circle with given radius, passing through two 
 given points or touching two given straight lines. 
 
 95. Construct a triangle, having given a side, the altitude 
 and median line to this. 
 
 96. Construct a triangle, having given a side, the altitude to 
 it, and the opposite angle. 
 
 97. Construct a quadrilateral ABCD, having given AB, BO, 
 AC, BD and L D. 
 
 98. Construct a quadrilateral A B CD, which can be inscribed 
 in a circle, having given A, AB, AC and BD. 
 
 99. Given a line and in it the point A, and outside the line 
 the point P. Find in the given line a point X, such 
 that AX-\- XP equals a given line. 
 
 TOO. In a given line determine a point, such that lines from 
 it to two given points on the same side of the line make 
 equal angles with it. (Ex. 67). 
 
 101. In a given sector to inscribe a circle. 
 
 102. In a given circle to' draw a chord equal and parallel 
 to a given line. 
 
 103. Construct a right-angled triangle, having given a point 
 in each of the sides containing the right angle, two 
 points in the hypothenuse, and the length of the altitude 
 to the hypothenuse. 
 
 3* 
 
36 
 
 104. Construct a triangle, having given the centres of its 
 escribed circles (Ex. 87). 
 
 105. Each side of a square is divided into two parts m and 
 w, so that two equal parts nowhere meet together. Prove 
 that the quadrilateral, having its angular points in each 
 of the four points of division, is a square. (Turn \ re- 
 volution). 
 
 106. In a regular pentagon all the diagonals are drawn. Prove 
 that a new regular pentagon is formed thereby. 
 
 107. Prove that two diagonals in a regular hexagon are par- 
 allel, and that a third diagonal is perpendicular to the 
 two preceding ones, and parallel to two of the sides of 
 the hexagon, and that three of the diagonals form an 
 equilateral triangle. 
 
 III. 
 
 I. SIMILAR FIGURES. 
 
 80, When several parallels ctit off eqttal parts of a straight 
 line , the parts which they cut of from any other straight 
 lines, will also be eqzial. 
 
 If AB == BC and if DO and EH 
 V be drawn parallel to AC, we have 
 
 Af. \D DO = = AB and EH = BC (69). 
 
 therefore 
 
 y /v DG = -^ 
 
 f n 
 
 CfL ~F Further 
 
 and Z G = /.. H . . . . (35), 
 
 therefore A DGE m EHF, 
 
 and consequently 
 
 DE = EF. 
 
37 
 
 81. To divide a straight line AB into a given number, 
 for example, 5, equal parts. 
 
 From A draw any line, and along it ^ . B 
 
 mark off 5 equal parts. Join the extremity 
 C of the last part with B, and through 
 the other points draw lines parallel to CB. 
 The parts of AB will then be equal (80). 
 
 82. Two triangles containing the same angles are said to 
 be similar. ,,Similar to" is denoted thus cvi. 
 
 The letters at the vertices of equal angles are put in the 
 same place. Corresponding sides are such which subtend 
 the equal angles; the letters by which they are denoted are 
 placed in the same order, for example, when 
 
 A ABC cv> abc, 
 
 then Z. A = a; B = b\ C = c, and AB, BC, and AC corre- 
 spond respectively to ab, be and ac. 
 
 83. In similar triangles the ratios between two pair of 
 corresponding sides are equal. (The sides are proportional). 
 
 Place the triangle abc on 
 ABC, so that /. b coincides 
 with /L B, a falls in M, and c 
 in N\ we then have MN ^ AC, 
 as Z. M = /- a = Z. A. Now 
 BM and BA must either have 
 a common measure or not. 
 a) If they have a common 
 
 measure (are commensu- A C 
 
 rable), that is, if there is 
 
 a small line which is contained in each of them a certain 
 
 number of times exactly, let this be marked off along 
 
 them, and be contained, for example, p times in BA and 
 
 q times in BM', we then have 
 
 BA p_ 
 BM = q ' 
 
 Now if lines be drawn through the points of division 
 parallel to AC, these will divide BC and BN respectively 
 into p and q equal parts (80); therefore we have 
 
38 
 
 BC p_ 
 
 BN " q' 
 and consequently 
 
 BA BC 
 
 BM ~ ~ 
 
 ba be 
 
 ft) If the lines have no common measure (are incommensu- 
 rable), divide BM into an indefinite number for example q 
 equal parts, and mark off these further; the point A will 
 then fall between two points of division for example the 
 p ih and (p + i) th ; we then have 
 
 p_ , BA , p+i 
 
 q ' - BM ' q 
 
 If we draw parallels as before, we also get 
 
 P ^ BO p + i 
 q ' ' BN ' q 
 
 73 A 
 
 The two ratios ,. and therefore both lie be- 
 
 v t) I i 
 
 tween the fractions and , the difference of which 
 
 2 9 
 
 is . The difference between the ratios must therefore 
 $ 
 
 be less than ; but this fraction can be made as small 
 
 
 as we please, for q can be taken as great as we please; 
 
 but when the difference between the ratios is less than 
 any ever so small quantity, it must be zero, and the ratios 
 therefore be equal*). 
 
 *) This proof does not hold good here alone, but shews in general, that 
 when two kinds of quantities are proportional when the ratios are com- 
 mensurable, they will also be proportional when the ratios are incom- 
 mensurable. 
 
39 
 
 In the same way we get, by placing c in 6\ 
 BC CA 
 
 be ca 
 
 AB BC CA 
 
 therefore 
 
 ab 
 
 ca 
 
 84. We have also shewn by this, that a line parallel to 
 one side of a triangle cuts off proportional parts of the two 
 other sides. 
 
 Conversely : When a line cuts off parts of two sides of 
 a triangle, which are proportional to the sides, then the line 
 will be parallel to the third side. 
 AD AE 
 
 -AB AC 
 
 Th. DE =fc BC. 
 
 For if DE were not parallel to BC, 
 we would be able to draw another line 
 for example, DF-=f=. BC\ but from that would 
 follow AD AF 
 
 which is in opposition to what was given. 
 
 According to a proposition in proportion, the proportion 
 
 AD 
 AB 
 
 AE 
 
 AC 
 
 can also be written 
 
 AD 
 
 AE 
 
 or 
 
 AB AD 
 AD 
 DB 
 
 ACAE 
 AE 
 
 so that this proportion holds, when DE zf=. BC, and conversely. 
 
 85. To cons trite t a fourth proportional 
 to three given lines, that is, a line which 
 is the fourth term in a proportion, in which 
 the three other terms are the given lines. 
 
 On the legs of any angle B mark off 
 BM, BA, and BN equal to the given lines; 
 if thereupon MN be joined, and A C drawn 
 parallel to MN^ then BC will be the re- 
 
40 
 
 quired line. If both the mean terms are equal, the required 
 line is called the third proportional to the two given lines 
 and is constructed in the same way. If the given lines are a, 
 
 ft C* 
 
 b, and c, the required line a?, -we must have -=- = . Here 
 
 o x 
 
 it is of no consequence, whether the letters represent concrete 
 or abstract numbers, expressing the lines measured by the 
 
 / 5 feet 5 \ 
 same unit I as, -^-j - = J ; but if the equation is written 
 
 ~bc 
 x = , it can only be understood in the latter way, as 
 
 there would be no meaning in multiplying two concrete numbers. 
 
 abc 
 Hereby we can again construct x = 3 , by first con- 
 
 ab 
 
 vc 
 
 
 
 A F 
 
 G B 
 
 structmg y = -j- , and thereupon x = ; it is apparent how 
 
 abed .... 
 this may be expanded to x = 7 - , when there is one 
 
 J J 
 
 line more in the numerator than in the denominator. 
 
 86. To divide a given line into parts, which are to one 
 another as given lines or mtmbers. 
 
 AB is to be divided into parts, which 
 are to eacn other as the given lines AD, 
 DE and EC, which are marked off on a 
 line from A\ join EC, and draw EG and 
 parallel to BC] we then have 
 BG GA GF FA 
 CE EA ED DA 
 
 If numbers are given, any line must be chosen as unit, 
 and this must be marked off as many times as the numbers 
 indicate. 
 
 87. A line bisecting an angle of a triangle divides the 
 opposite side into two parts, which are to each other as the 
 sides containing the angle. 
 
 If we draw CE ^ AB, we have 
 
 A ABD cv> A CED, 
 
 AB AD 
 
 therefore = - , 
 
but CE = = CB. . . (21); 
 
 AB AD 
 therefore - = _. 
 
 The line, bisecting the 
 angle adjacent to B, cuts AC 
 (produced) in a point Z? 1 . By 
 putting Z)j instead of Z), all Z ?; - 
 
 over in the demonstration 
 
 above, we prove that AB : CB == ^Z^ : D 1 C. We say that 
 D l divides AC (externally) in the same ratio as D divides AC 
 (internally), or that AC is divided harmonically by D and D l 
 in the ratio AB: BC. 
 
 88. Two figures (systems of points) are said to be simi- 
 lar in the ratio m, when there to every point in the one figure 
 is a corresponding point in the other figure, and the distance 
 between two points in the one figure all over is m times the 
 distance between the corresponding points in the other figure. 
 When m = = i, the figures are congruent. 
 
 That there always is a figure abed .... similar to a given 
 ' figure A BCD .... in a given ratio is shewn thus: 
 
 From any point draw lines 
 to the angular points of the given 
 figure, and on these mark off 
 Oa == m . OA ; Ob = m . OB, &c. 
 a, b, c, d will then be the angu- 
 lar points of the required figure. 
 is called the centre of simili- 
 tude, the lines through rays 
 of similitude* Tl\e figures are 
 said to be similarly situated. Of such figures it holds that: 
 
 Corresponding points lie in the same ray by construction. 
 
 Corresponding lines (lines joining corresponding points) 
 
 are parallel, for example, ab ^ AB (84) 
 
 Oa Ob 
 
 tor -OA "OB = m ' 
 
 Corresponding lines are proportional in the ratio m : i , for 
 example, ab Oa 
 
 AB ~OA 
 
42 
 
 This also holds for broken lines, for when each separate 
 line becomes m times greater, their sum must also become 
 m times greater. 
 
 Corresponding angles are eqzial (35). 
 
 The figiire we get, will be the same wherever we choose the 
 centre of similitude, for if we had, by choosing it somewhere 
 else, constructed a figure a l b l c l d 1 . . . , then this and abed . . . 
 must have all the corresponding lines and angles respectively 
 equal, and consequently be congruent. 
 
 When certain points in the one Jigtire lie in a straight 
 line, the corresponding points in the other figtire also lie in 
 a straight line. 
 
 This will be apparent, when we imagine the centre of 
 similitude chosen in the straight line. 
 
 The points in the corresponding lines are therefore jointly 
 corresponding. 
 
 When certain points in the one figiire lie in a circle, 
 the same must be the case with the corresponding points in 
 the other figitre. 
 
 This is apparent, when we imagine the centre of simili- 
 tude chosen in the centre of the circle. 
 
 We have termed triangles, containing the same angles, 
 similar; as such triangles can be placed so that they are sim- 
 ilarly situated (see fig. to 83), they are also similar in the 
 general sense of the word. 
 
 89. When the cases in which two figures are congruent 
 are known, we can therefrom deduct the cases in which they 
 are similar; we can namely in the demonstrated propositions 
 on congruence read ,,similar" instead of congruent", when 
 we simultaneously instead of sides equal" read Asides pro- 
 portional" \ we will apply this to triangles. 
 
 90. Of the proposition: 
 
 Two triangles are congruent, when they have all three 
 sides respectively equal, 
 the new proposition is formed: 
 
 Two triangles are similar, when they have all three 
 sides proportional. 
 
43 
 
 Hyp. 
 
 ab 
 
 be 
 BC 
 
 nc 
 
 = m. 
 
 AB BC AC 
 
 We choose any centre of similitude 0, and construct a l b l c l 
 cv: ABC in the ratio m, when m is the value of the equal 
 ratios; we then have 
 
 These ratios are equal to 
 the given ones, and as the conse- 
 quents respectively are the 
 same, the same must be the 
 case with the antecedents, there- 
 fore 
 
 The small triangles are there- 
 fore congruent, and as one of them is similar to the large 
 one, the other must also be. 
 
 91. Of the proposition: 
 
 Two triangles are congruent, when they have two sides 
 and the angle contained by them respectively equal, 
 the new proposition is formed: 
 
 Two triangles are similar, when they have two sides 
 proportional and the angle contained by them eqiial. 
 
 db 
 Hyp. /LA = Z.; 
 
 ac 
 
 By the same construction as before, we have 
 
 CLiO* (I , C , 
 
 AB AC 
 
 wherefrom, by comparison with what is given 
 
 ^a == G t ; ctb = a l b l \ ac = a l c l ^ 
 therefore A abc 25 A a l b l c l ;v A ABC. 
 
 92. Of the proposition: 
 
 Two triangles, having one angle, an adjacent side, and 
 the opposite side respectively equal, are congruent, if the 
 opposite side be greater than or equal to the adjacent side, 
 the new proposition is formed: 
 
44 
 
 Two triangles, having one angle equal, and an adjacent 
 side and the opposite side proportional, are similar, if the 
 opposite side be greater than or equal to the adjacent side. 
 
 The demonstration is precisely the same as for the pre- 
 ceding" proposition. 
 
 93, In the same way it is demonstrated, that two figures, 
 having all their corresponding angles eqztal and sides pro- 
 portional, can be placed so as to be similarly situated \ such 
 figures can therefore, by corresponding diagonals, be divided 
 into similar triangles. If the condition for the congruence of 
 polygons only involves the equality of one side in each, this 
 condition is omitted in the condition of similarity. Thus all 
 regular polygons with the same number of sides are similar. 
 
 94. To draw a figure similar to a given figure , so that 
 one of its lines has a given length. 
 
 Let ABCDEF be the given 
 figure and ab the given line, which 
 in the required figure is to corre- 
 spond to AB\ mark off Ab = ab, 
 from A draw lines to the angular 
 points of the figure, therupon draw 
 be ^ EC, cd^ CD, &c. Abcdef is 
 then the required figure, as it is 
 similarly situated with ABCDEF, 
 
 A being the centre of similitude. 
 
 Hereby we can draw a regular n sided figure with a given 
 
 side, if we can draw any regular n sided figure. (Compare 76). 
 
 S 
 
 EXAMPLES. 
 
 108. From the centre of a circle perpendiculars are drawn 
 on two of the sides of an inscribed triangle. Prove that 
 the line, joining the feet of the perpendiculars, is half 
 as great as the third side of the triangle. 
 
 109. In a parallelogram ABCD, E is the middle point of AB, 
 and F of CD. Shew that DE and BF divide AC into 
 three equal parts. 
 
45 
 
 1 10. The middle points of the four sides of a quadrilateral 
 are joined. Prove that the quadrilateral thus formed is 
 a parallelogram. A new proposition can be formed from 
 this, by considering the diagonals as sides and a pair 
 of opposite sides as diagonals. 
 
 in. Prove that the altitudes of a triangle are reciprocally 
 proportional to the . corresponding bases. 
 
 112. In a triangle ABC two altitudes Act and Bb, cut each 
 other in 0; prove that Cb . CA = = Ca . CB. 
 
 113. Prove that BO . Ob = = AO . Oa. 
 
 114. Prove that Bb . Ob = - Ab . bC. 
 
 A OB can be regarded as the given triangle; C then 
 becomes the point of intersection of the altitudes; thereby 
 new propositions are formed of 112, 113, and 114. 
 
 115. A diameter AB is drawn in a circle, and also tangents 
 AC and BD. E is a point of the circumference, in which 
 BC and AD intersect. Shew that the diameter is a mean 
 proportional between BD and AC. 
 
 1 1 6. Shew that a chord is a mean proportional between the 
 diameter and the perpendicular from one extremity of 
 the chord to the tangent at its other extremity. 
 
 117. In an inscribed triangle ABC, AD is drawn parallel to 
 
 o 
 
 the tangent at B. Shew that AB == BD . BC. 
 
 1 1 8. A square is inscribed in a right-angled triangle, so that 
 one of its sides lies in the hypothenuse. Shew that the 
 one part of the hypothenuse is a mean proportional 
 between the two others. 
 
 119. In an isosceles triangle each of the equal sides is divided 
 into three equal parts, and a line is drawn through 'the 
 upper point of division of the one side and through the 
 lower point of division of the other side, till it cuts the 
 base produced. Prove that the two parts of the line 
 are equal, and that the prolongation of the base is a 
 third part of the whole base. 
 
 120. In a triangle ABC mark off on AC two points D and E, 
 so that AE = AB, and so that AE is a mean proportional 
 between AD and AC. Shew that BE bisects the angle DBC. 
 
4 6 
 
 121. Prove that the median lines of a triangle divide each 
 other into parts, which are to each other as i : 2. 
 
 122. Prove that the line in a trapeziitm, joining the middle 
 points of the sides which are not parallel, eqztals half 
 the sum of the parallel sides. 
 
 123. The parallel sides of a trapezium are produced in oppo- 
 site directions, so that each of the prolongations equals 
 the opposite side. Shew that the line, joining the ex- 
 tremities of the prolongations, bisects one of the diago- 
 nals of the trapezium, and cuts the other in a point lying 
 equally distant from one extremity of the diagonal as 
 the point of intersection of the diagonals from the other 
 extremity. 
 
 124. From a given point to a given line, any line OA is 
 drawn, and on this a point a is marked off, the product 
 OA . Oa being given. Shew that a describes a circle 
 through 0, when A traverses the given line. 
 
 125. In a triangle BAG the angle A is a right angle, and CD 
 bisects the angle C. Prove that AB.AD = AC. (BCAC). 
 
 126. Two circles touch each other externally, shew that the 
 part of their common tangent between the points of 
 contact is a mean proportional between the diameters. 
 
 127. In a triangle ABC the median line AD is drawn, and 
 through A a line AE =. BC. Any line through D cuts 
 AE in E, AB in f\ and AC in G\ prove that 
 
 DF-.DQ = EF-.EO. 
 
 128. Through the point of intersection of the diagonals of a 
 trapezium, a line is drawn parallel to the parallel sides. 
 Prove that these sides are to each other as the parts 
 in which the diagonal divides the not parallel sides. 
 
 129. AD is a diameter of a circle and AB, BC, and CD tan- 
 gents. Shew that both diagonals in ABCD bisect the 
 perpendicular from the point of contact of BC to the 
 diameter. 
 
 130. Two given circles touch each other in 0. Through 
 draw any line, cutting the circles in A and B. Shew 
 
47 
 
 that the radius of a circle, touching the one circle in A 
 and passing through 5, is constant. 
 
 131. A circle touches another internally at A. BC is a chord 
 in the large circle, touching the lesser one at D. Shew 
 that AD bisects the angle BAG. 
 
 132. From a point in the circumference of a circle, perpen- 
 diculars are drawn to two tangents, and to the chord, 
 joining their points of contact; prove that the last per- 
 pendicular is a mean proportional between the two first. 
 
 133. /// a triangle ABC a line is drawn, cutting AB in c, 
 AC in b, and BC produced in a. Prove that 
 
 Ba.Ac. Cb == Ab . Ca.Bc. 
 
 Draw Cm^=AB, and employ bmCrobcA, 
 and eliminate Cm from the two equations obtained thus. 
 
 134. Front the angular points of a triangle ABC, the lines 
 Aa, Bb , and Cc are drawn to the opposite sides and 
 cutting each other in the same point ; prove that 
 
 Ba.Ac.Cb == Ab. Ca.Bc. 
 
 Apply ex. 133 to A ABb, cut by Cc, and to BbC, cut 
 by Aa\ thereupon eliminate AC from the two equations. 
 
 135. Demonstrate the proposition in 87, by the help of per- 
 pendiculars drawn from the extremities of the divided 
 side to the bisecting line. 
 
 136. In tzuo given circles with centres C and c draw any 
 two parallel radii CA and ca. Shew that Aa cuts the 
 line of centres in a fixed point (the exterior centre of 
 similitude), when the two radii go in the same direction, 
 and in another fixed point (the interior centre of simil- 
 itude)^ 10 hen the radii go in opposite directions. What 
 constriction of the common tangents to the circles can 
 be deducted from this? 
 
 137. The points a and b divide the line AB harmonically; 
 prove that A and B also divide ab harmonically. 
 
 When A and B are fixed, and a moves from A to J5, 
 how does b move simultaneously? When AB = = k, 
 Aa = r n Ab = r 2 , then prove that 
 
4 8 
 
 = _L _. 
 
 ^ k r l r 2 
 
 138. Shew that OA (Ex. 137) is a mean proportional between 
 Oa and Ob, when is the middle point of AB. 
 
 II. THE RIGHT-ANGLED TRIANGLE. 
 
 95. The alt^tde on the hypothenuse of a right-angled 
 triangle, divides it into two triangles, which both are sim- 
 ilar to the whole triangle, and therefore similar to one an- 
 other; for we have 
 
 Z A = A, 
 D = C == B. 
 A ADC :%j A AGE. 
 ^ In the same way we get 
 
 P~&~ B A CDB ** A ACB 
 
 ~~lH and therefore also 
 
 A ADC A CDB. 
 
 From this different proportions are deducted, of which 
 we will mention the most important. 
 From ADC ^ CDB 
 
 AD CD 
 
 we get ______ or p- . aft 
 
 which shews, that 
 
 a) the altitude is a mean proportional between the parts of 
 the hypothenuse. 
 
 From ADC ^ ACB 
 
 we get 
 
 AD AC DC 
 
 AC AB CB 
 
 or b 2 = ph and ab = ph, 
 
 from which is seen, that 
 
 b) one of the sides containing the right angle is a mean pro- 
 portional between its projection on the hypothenuse and 
 the ivhole hypothenuse, and that 
 
 c) the prodztct of the sides equals the product of the alti- 
 tude and the hypothenuse. 
 
49 
 
 The most important proposition of the right-angled tri- 
 angle is the one discovered by Pythagoras in the 6 th cen- 
 tury B.C. 
 
 d) The square on the hypothenuse eqitals the sum of the 
 squares on the sides. 
 This is found by adding 
 
 2 == fth 
 
 and the analogous a 2 = a/z, 
 
 from which a 2 + 6 2 = (a + ft)h = 7* 2 . 
 
 This proposition shews that the one side of a right- 
 angled triangle can be calculated, when the two others are 
 known; we find 
 
 a == Vh'* 6 2 ; b = V/i 2 a*; li = Va 2 -h6 2 . 
 By the help of the propositions concerning the right-angled 
 triangle, when two of the parts o, , 7, a, /?, and p are known, 
 the others can be found; some of these problems, however, 
 lead to equations of the second degree. 
 
 If Z C is obtuse, and the opposite side is c, and the sides 
 
 if 
 
 containing it a and b, then c > I/a 2 -f- 
 
 is acute, then 
 
 c < I/a 2 -f- ' 2 ; for if the triangle be compared to the right- 
 angled triangle with the sides a and b, then in the first case 
 c>7^, in the second c < h. (46). 
 
 96, To construct a mean proportional between two given 
 lines, a and b. 
 
 a) On the greater line b mark off a(AB\ and on b as 
 diameter describe a semicircle; at the extremity of a raise 
 a perpendicular , and 
 
 draw the chord x(AD)] 
 x will be the required 
 line (95, b). 
 
 b) On a -j- b as di- 
 ameter describe a semi- 
 circle; where a and b 
 
 meet together raise a perpendicular x(BD], which then will 
 be the required line (95, a). 
 
 4 
 
 A B 
 
 C A 
 
 B 
 
50 
 
 // rv* 
 
 As = -7- , then x = Vab . A line which can be expres- 
 sed by known lines in this form is therefore constructed in one 
 of the methods indicated; if, for example, x = a 1/5 = 1/50 . #, 
 then x will be a mean proportional between a and $a. We 
 could also put x = \/(2a)* -\- a 2 , and therefore construct x as 
 hypothenuse of a right-angled triangle, in which the one side 
 is a and the other 2a. 
 
 z = Vab-\-cd is constructed by putting ab = x 2 , cd = y^\ 
 x and y are then found by 96; thereupon z is found as hy- 
 pothenuse of a right-angled triangle with sides x and y. 
 
 III. THE POTENSE OF A POINT WITH REGARD TO 
 
 A CIRCLE. 
 
 97. a) When two chords of a circle intersect, the prod- 
 ^lct of the two parts of the one chord equals the product 
 of the parts of the other chord. 
 
 Draw the lines Ab and Ba\ then 
 
 24 
 
 /- b = a 
 therefore A AbO ^ BaO, 
 from which it follows that 
 OA Ob 
 OB U~a 
 or OA.Oa = OB. Ob. 
 
 b) When two secants cut each other, 
 the product of the one secant and the part 
 outside the circle equals the proditct of 
 the other secant and the part of it out- 
 side the circle. 
 
 The former demonstration also applies 
 to this case. As the proposition holds 
 good for every position of the secants , it 
 also holds good in the case in which b 
 
5 1 
 
 and B coincide; here both Ob and OB are equal to the tan- 
 gent; so that the proposition will be: 
 
 The tangent is a mean proportional between the whole 
 secant and the part outside the circle. 
 
 These three propositions may be expressed under one, 
 thus: 
 
 When a circle cuts several straight lines, meeting at the 
 same point, the prod^lct of the two parts cut off from each line 
 (both reckoned from the fixed point to one of the points of 
 intersection of the circle with the line) is the same for all 
 the lines. This product is called the potense of the point 
 with regard to the circle; by drawing the line through the 
 centre, we see that the potense of a point, of which the dis- 
 tance from the centre is a (radius r), is a 2 r 2 when a > r, 
 and r 2 a 2 when r > a. 
 
 IV. PTOLEMY'S PROPOSITION. 
 
 98. In an inscribed qiiadri lateral the product of the 
 diagonals equate the sum of the products of the opposite 
 sides. 
 
 Make Z CBE = /_ DBA ; 
 
 we then have CBE ~ DBA , 
 
 whence EC: b l = = b : d\ 
 
 EBA CBD, 
 
 whence AE : a l = a:d\ 
 
 but from b . b l = d . EC, 
 
 a . a l = d . AE 
 we get by addition 
 
 aa l -\- bb l = dd l . 
 
 If the inscribed quadrilateral is a rectangle, the proposi- 
 tion gives Pythagoras' proposition (95) as a particular case 
 of Ptolemy's. 
 
 4* 
 
EXAMPLES. 
 
 139. In a right-angled triangle the sides containing the right 
 angle are a and , their projections on the hypothenuse 
 and /9, the altitude /?, and the hypothenuse h\ how great 
 are the other parts, when 
 
 0. . 545 ft = 9-6? 
 
 2) a = 3.6; a = 6? 
 
 3) a 0.5; b = 1.2? 
 
 4) A == 6; a == 4.8? 
 
 140. Construct 
 
 l/a 3 H-6 3 , . . N . 
 
 V - ~r -> when a, 5, and c are given lines. 
 
 r -2Q 2O 
 
 141. In a triangle, the sides of which are 
 
 a transversal, d = i .68, is drawn parallel to the side c ; 
 how great are the parts in which it divides the sides a 
 and bl 
 
 142. In a rectangle the one side is 6 feet, and the diagonal 
 is 2 feet longer than the other side. How great is this? 
 
 143. In a triangle the one altitude is 15, and the parts of the 
 base 6 and 10. At what distance over the base is the 
 point of intersection of the altitudes? 
 
 144. How great is the diagonal of a square with the side a? 
 
 145. In a right-angled triangle the hypothenuse is m 2 -f- n 2 and 
 the one side 2mn. Find the other side. 
 
 146. How great is the altitude of an equilateral triangle with 
 the side a? 
 
 147. In a quadrilateral the diagonals are at right angles to 
 each other. Prove that the sum of the squares on the 
 one pair of opposite sides is equal to the sum of the 
 squares on the other pair. 
 
 148. A circle passes through the centre C of another circle 
 and touches it at A. A line perpendicular to AC cuts 
 the first circle in Z>, the second in E. 
 
 Shew that AE = 2 AD. 
 
53 
 
 149- From a point J^*in the circumference of a circle a per- 
 pendicular EP is dropped on a diameter. D is the 
 middle point of the semicircle with radius r. Shew that 
 
 150. In a circle with radius r an angle of 45 is drawn at the 
 centre. On the one leg of the angle a perpendicular CE 
 
 is raised, cutting the other leg in D and the circle in E. 
 
 2 2 
 Shew 1 that CD + CE = r 2 . 
 
 151. From a point to a circle a secant is drawn, both parts 
 of which equal a. How great is the tangent from the 
 same point? 
 
 152. In a right-angled triangle with the sides containing the 
 right angle equal to 1.41 and 1.88, a perpendicular is 
 raised on the middle of the hypothenuse. How great 
 are the parts in which it divides the one side? 
 
 153. In a semicircle there is over each half r of the diameter 
 described other semicircles. How great is the radius of 
 the circle touching the three semicircles? 
 
 154. In a triangle with sides a, b, and c, a perpendicular 
 is dropped on c from the opposite vertex. Prove that 
 02 2 == 2c i^ w h en l i s the distance from the foot of 
 the perpendicular to the middle point of c , and when 
 a > b. Prove that a 2 + b 2 = ic 2 + 2m*. fm , the me- 
 dian line). 
 
 155. A and B are two given points in a diameter, equidistant 
 
 from the centre, P is any point in the circumference. 
 
 2 2 
 
 Shew that AP -f- BP is constant. 
 
 156. Through A (Ex. 155) any chord is drawn, the extrem- 
 ities of which are joined to B. Shew that the sum of 
 the squares on the sides of the triangle thus formed is 
 constant. 
 
 157. In a triangle with sides a, b, and c, the angle between 
 the two first is bisected. How great are the parts in 
 which the bisecting line divides the third side? 
 
54 
 
 158. EG is a diameter and A and F two points in the cir- 
 cumference. FT) -L- BC cuts BA in E and CM in G. 
 Shew that DF is a mean proportional between DE and Z)6r. 
 
 159. Shew that any point in the circumference of one of two 
 concentric circles has the same potense with regard to 
 the other circle, as any point in the circumference of 
 the second circle has with regard to the first. 
 
 1 60. Prove that the line bisecting the angle between two 
 opposite sides of an inscribed quadrilateral divides the 
 two other sides into parts which form a proportion. 
 
 161. In a triangle with sides 13, 20, and 21, the altitude is 
 drawn to the latter side; how great are the parts of 
 the base, and how great is the altitude? 
 
 162. Given two sides of a triangle and one of the parts in 
 which the altitude divides the third side; how great is 
 the third side? 
 
 163. a and b are two parallel chords, and the distance be- 
 tween them d\ how great is the radius of the circle? 
 
 164. A tangent at cuts two parallel tangents in A and B. 
 Shew that OA . OB is constant. 
 
 165. A right-angled triangle is divided by the altitude into 
 two other triangles. Find an equation between the radii 
 of the circles inscribed in the three triangles. 
 
 1 66. How long a shadow is cast by an object 2 inches high, 
 when a candle 8 inches high is placed at a distance ot 
 6 inches? 
 
 167. Two circles have radii E and r and the line of centres c. 
 At what distance from the centres is the line of centres 
 cut by a line touching both the circles? 
 
 1 68. In a quadrilateral with diagonals d and D a rhombus 
 is inscribed, having its sides parallel to the diagonals; 
 how great is the side of the rhombus? 
 
 169. A and B are the extremities of a line, C and D two 
 points in a line perpendicular to AB. Prove that 
 
 2 2 2 2 
 CA CB = DA DB. 
 
55 
 
 170. In a right-angled triangle the one angle is 30, the oppo- 
 site side a. Find the other sides, the perpendicular, and 
 the parts of the hypothenuse. 
 
 171. Prove that the sum of the squares on the four parts of 
 two chords, which are perpendicular to each other, 
 equals the square on the diameter. 
 
 172. o, b, and c are three chords, the arcs of which are to- 
 gether 1 80. By what equation is the diameter of the 
 circle determined? 
 
 173. The four sides of a trapezium are given; find the sum 
 of the squares on the diagonals. 
 
 174. A circle has with regard to two points A and B respec- 
 tively the potenses p l and p 2 , whilst P is its potense 
 with regard to a point which divides AB into the parts 
 a and /9. Shew that 
 
 175. M is the middle point of the one side of a triangle, the 
 line bisecting the opposite angle cuts the side in K, the 
 altitude cuts it in H, and the inscribed circle touches it 
 at N. Shew that 
 
 MN.HN = MH.NK. 
 
 176. On a line AD as diameter a semicircle is described, and 
 from a point B in this is drawn BF -*- AD. From D 
 draw the chord DC == DB AB and draw CE -- AD. 
 Shew that AE = 2BF. 
 
 V. THE DIVISION OF THE CIRCUMFERENCE OF 
 THE CIRCLE. 
 
 99. A circle is divided into 4 equal parts 
 
 by tiuo diameters perpendicular to each other. 
 
 If the radius be denoted by r and the 
 
 chord of - - of the circumference by & (the 
 side of an w sided figure), then 
 
56 
 
 therefore & 4 = r]/2. 
 
 By continued bisection of the arcs, the circle can be di- 
 vided into 8, 1 6, 32 .... 2 n equal parts. 
 
 100, kg or the chord of an arc of 60 equals the radius. 
 When the arc is 60, the angle at the centre is also 60, 
 and as the triangle is isosceles, the two other angles will also 
 each be 60, the triangle therefore is equi- 
 lateral, and consequently the chord equals 
 r / \ r the radius. The circumference, therefore, 
 
 is divided into 6 equal parts by marking 
 off the radius 6 times as chord; it can be 
 divided into 3 equal parts by missing out 
 every other point of division; by continued bisection of the 
 arcs, it can be divided into 12, 24, 48 ... 3.2" equal parts. 
 
 101. k 10 or the chord of 36 equals (Vs *) 
 
 The required chord is base of an isosceles triangle, the 
 sides of which are radii, and whereof the vertical angle is 36 
 
 and the angles at the base therefore 
 each 72. If one of these be bi- 
 sected by the line AC, then A CAB is 
 isosceles , for Z B = /. C = = 72 ; 
 therefore AC=k\ also A A CO is isos- 
 celes, therefore OC = k and CB = 
 r k. Now the angles shew that 
 AOABcoACB, therefore (see also 87) 
 
 k r k 
 Then k * rk 
 
 or 
 
 but as fci + r4 
 
 we get by extraction of the square root 
 
57 
 
 or 
 
 k = 
 
 The constriction of k 10 by dividing the radius in extreme 
 and mean ratio. 
 
 The first expression for k above 
 shews how the side of a decagon may 
 
 be constructed. |/ r 2 -)- ( ) is namely 
 the hypothenuse of a right-angled tri- 
 angle, the sides of which are r and 
 
 The figure shews the construction; bisect the radius OZ), make 
 OB -L- OD and join AB\ thereupon make AC = AB, then OC 
 will be the side of the decagon, for we have OC = AC AO 
 
 A line is said to be divided into extreme and mean ratio, 
 when it is divided into two parts, so that the greatest of 
 these is a mean proportional between the least and the whole 
 line; the proportion we got for the determination of k there- 
 fore shews that the side of the decagon is the greatest part 
 of the radi^is thus divided. 
 
 When the circumference is divided into 10 equal parts, 
 we can, by doubling or continually bisecting the arcs, divide 
 it into 5, 20, 40, 80 .... 5.2" equal parts. 
 
 102. If an arc of 36 be subtracted from an arc of 60, 
 we get an arc of 24, and thus, therefore, a circle can be di- 
 vided into 15 and, by continued bisection of the arcs, into 30, 
 60 .... 15.2" equal parts. The expression for k^ will be 
 found below (Ex. 183). 
 
 103. When a chord is calculated, its distance from the 
 centre p is found by 95, d ; thus if the chord is k , radius r, 
 we get 
 
104. When the radius is given, and the chord of a cer- 
 tain arc has been calculated, we can from that again calcu- 
 late the chord of half and the chord of double the arc. 
 
 Let AB = K, AE == k, OD JL AB, 
 c therefore ^^5 = 2^. 
 
 we then have (103) 
 
 OD =- I/A 
 
 and from this 
 
 DE = 
 
 As now Z CAE === 90, then AE or 
 k is a mean proportional between DE and the diameter CE, 
 therefore 
 
 & l/2r . DE 
 
 or 
 
 Example. From k$ = r we hereby find &i 2 = r 2 1/3 ; 
 
 from k = H/2 we find k 8 = rV 2 1/2 . 
 
 From k to calculate K, we remark that 
 2r . AD = k . AC (95, c) 
 
 /-\t* M j\ Z I/ A i*2 7 *2 
 
 UI / xx ft, y tLf K * 
 
 from which J5T 
 
 Example. From &i 
 
 (1/5 i) we find 
 
 105. When the side and the least radius of an inscribed 
 regular polygon are calculated, we can calculate the side and 
 
59 
 
 the greatest radius of the circumscribed regular polygon with 
 the same number of sides. 
 
 Let AB be the side of the inscribed 
 polygon, from the centre draw a 
 perpendicular OF to AB] if thereupon 
 the tangent CD be drawn, it becomes 
 the side, and 00 the greatest radius of 
 the circumscribed polygon, for we have 
 AB =. CD, as they both are perpendic- 
 ular to OF, and therefore the triangles AOB and COD are 
 similar. From this it follows 
 
 CD CO FO 
 AO 
 
 AB 
 
 EO 
 
 but if we put CD 
 
 t_ R 
 
 k r 
 
 *, 
 
 then 
 
 AB = fc, 00 = R, OE = p, 
 
 rk_ 
 P P P 
 
 = =_ or t = - 
 
 EXAMPLES. 
 
 and p. s , t 6 and 
 
 p 8 
 
 and 
 
 177. How great are k. A 
 
 178. What is the product It & put 
 
 179. What is 3 , and what is the least radius of an equilateral 
 triangle with the side O. 
 
 180. Prove that the line joining BO on the fig. to 101 is the 
 side of the inscribed pentagon. 
 
 181. Find the mean proportional between the chord of 30 
 and the chord of 150. 
 
 182. Find & 64 , 7c 96 , ^ 64 , and p 96 . 
 
 183. a, , and c are sides of an inscribed triangle; find c, 
 when a, 6, and r are given. 
 
 Draw the diameter from the point of intersection of 
 a and &, and lines from the extremities of the diameter 
 to the extremities of a and b. These lines become 
 
 l/4r 2 a 2 and l/4r 2 b 2 , and we then have (98) 
 
 2r c = l/4r 2 b 2 + bV^r' 2 a 2 , 
 the upper sign when c is the chord of the sum of the 
 
6o 
 
 arcs, and the lower sign when c is the chord of the 
 difference of the arcs , of which a and b are the chords 
 
 Hereby we get & 45 , when a = r, b= (1/5 i) (102), 
 
 * 15 = -f (V 10 + 2V 5 -1/3(1/5-!)). 
 4 
 
 VI. THE LENGTH OF THE CIRCUMFERENCE OF 
 THE CIRCLE. 
 
 106, A curved line cannot be measured by a straight line, 
 for the unit must be of the same kind as that which is to be 
 measured; we must therefore more closely explain what is 
 meant when we speak of the length of the circumference. 
 
 It is evident that an inscribed regular 2n sided figure has 
 greater perimeter than an inscribed n sided figure, whilst the 
 opposite is the case with the circumscribed figures. If we 
 therefore continue to double the number of sides of an in- 
 scribed and a circumscribed regular figure, the perimeter will 
 in the first case continually increase and in the second con- 
 tinually decrease; as now the latter always is greater than the 
 former, they must approach nearer and nearer to each other, 
 and we can prove that the difference between them can be 
 made less than any assigned quantity. The perimeter of the 
 inscribed n sided figure with the side k is nk and of the cir- 
 cumscribed figure nt\ the difference is therefore, as k = 
 
 ("5), 
 
 but r 2 p 2 = 
 
 therefore the difference 
 
 f * 
 
 *' 
 
 but here the first factor nt is the perimeter of the circumscri- 
 bed figure and is accordingly less than the perimeter we had 
 
6i 
 
 before doubling the number of sides, whilst the second factor 
 can be made as small as we please, by doubling the number 
 of sides till k is sufficiently small. 
 
 The lengths of the perimeters of the inscribed and cir- 
 cumscribed figures therefore approach nearer and nearer to 
 each other, so that they have a common limit, a value to 
 which they both approach. // z's this limit that is meant, 
 ro/ien we speak of the length of the circumference of a circle. 
 
 The limit will be the same, whichever regular polygon 
 we take. 
 
 Let the circumscribed and inscribed polygons with several 
 sides have the perimeters P and p, whilst G is the limit. B J 
 continuing to double the number of sides, we can make P G 
 and G p less than any ever so small quantity. Let P n p t , 
 and G l be the perimeters and the limit, when we take another 
 regular polygon. As p l lies within P and p within P l AVC 
 can (Ex. 56 expanded) put 
 
 P- Pl - ; P,-p -/9 
 where a and /? are certain positive quantities; therefore 
 
 p t pi+p-p"+ft- 
 
 As here P 1 p 1 and P p by continued doubling become 
 less than any ever so small given quantity, the same must 
 be the case with the positive quantities a and /?. P l must 
 therefore have the same limit as p, or G x must be equal to G. 
 
 107. The lengths of the circumferences of two circles 
 are to each other as the radii. 
 
 For the perimeters of two regular figures with the same 
 number of sides are to each other as the greatest radii, and 
 this proposition continues to hold, how ever often the num- 
 ber of sides of the polygon is doubled; therefore it must also 
 hold good for circles. 
 
 If we denote the circumferences by P and p, the radii by 
 R and r, and the diameters by D and d, we therefore have 
 
 P__ D__ R_ P_ _ V_ 
 
 ~~ " " " ^ = ~' 
 
62 
 
 which shews that the ratio between the circumference and 
 diameter is the same for all circles (is constant); this ratio, 
 which is an abstract number, is denoted by the letter TT, so that 
 
 P = n D = 27tE. 
 
 It therefore depends upon, once for all, to calculate this num- 
 ber TT; when it is known, we can always of the three quan- 
 tities P, />, and R calculate the two, when we know the third. 
 As the ratio is the same for all circles, we take radius equal 
 to i , therefore the diameter equal to 2 , and calculate the 
 perimeter of an inscribed polygon with several sides; if the 
 number of sides be w, we have 
 
 nk n <p < nt n 
 
 nk n nt n 
 
 therefore - < TT < - - 
 
 2 2 
 
 We have k e = i and found from this (104) k i2 = V 2 1/3 
 = 0.517638090. . .; from this we find by degrees & 2 4, & 48 
 for example, to k 1Q8 ] by multiplying the value found thus by 
 768, we get the perimeter of the 768 sided figure (nk n ) equal 
 to 6.283160...; after division by 2, we therefore have 
 
 3. 141580 <TT. 
 
 From &768 we thereupon calculate p 1S8 (103) and from this 
 again 768 ; if this be multiplied by 768, we get the perimeter 
 of the circumscribed 768 sided figure (nt n ); for this we find 
 the numbet 6.283212 . . ., and get by dividing by 2 
 
 71 < 3.141606. 
 
 As n lies between the values found thus, we have with 
 4 correct decimals 
 
 7T = 3.I4I6. 
 
 By easier methods, which cannot be shewn here, n has 
 been found with several hundred decimals; the first are 
 
 TT - 3.1415926535 .... 
 The number is incommensurable, but is nearly expressed 
 
 by the fractions and 25 o f wn i c h the first already was 
 
 7 IJ 3 
 
 found by Archimedes, 
 
108. As an arc of i is ^- of the circumference, its length 
 360 
 
 will be -^r , and consequently the length b of an arc of y 
 
 b = = I8o ' 
 
 By the help of this equation, when two of the three 
 quantities 6, r, and g are known, we can calculate the third. 
 If g be expressed in minutes or seconds, the denominator must 
 be multiplied by 60 or 6o 2 respectively. 
 
 EXAMPLES. 
 
 184. How great is the circumference of a circle, when the 
 radius is 2' 4", and how great is the radius, when the 
 circumference is 3' 6" ? 
 
 185. A circle touches another circle internally in A and passes 
 through its centre. Through this centre a line is drawn 
 cutting the circles in B and 0; shew that ^AB = *^>AC. 
 
 O ' 
 
 1 86. Two circles with radii R and r touch a circle with ra- 
 dius R -(- r internally respectively in A and B, whilst C 
 is one of their points of intersection. Prove that ^ AB 
 
 187. How long is an arc of 12 13' 20", when the radius is 
 i' 8"? 
 
 1 88. How many degrees are contained in an arc, the length 
 of which is 3", when the radius is 2"? 
 
 189. A row of semicircles is so placed that the diameters 
 are prolongations of each other. Prove that the sum of 
 the lengths of the circumferences of all the semicircles 
 is equal to the length of a semicircle , the diameter of 
 which is the sum of the given diameters. 
 
 190. Find the number of degrees contained in an arc, the 
 length of which equals the radius. 
 
 191. How great is the radius of the earth, when one degree 
 of latitude is 60 geographical miles? 
 
 192. is the centre of a circle and A a point, the distance 
 of which from is n times the radius. On OA as di- 
 
193' 
 
 64 
 
 ameter a circle is described. Two lines from A intercept 
 arcs on the first circle, having the lengths b l and b 2 ; 
 how long is the arc intercepted by the two lines on the 
 second circle? 
 
 An undulating line is formed by dividing a line a into 
 n equal parts, and on each of the parts as chord descri- 
 bing an arc of 60, alternately upwards and downwards. 
 How long is the undulating line? What does this length 
 become, and what becomes of the undulating line, when 
 n grows infinitely, whilst a remains unchanged? 
 
 IV. 
 
 AREA. 
 
 109. The area of a figure can be measured by the area 
 of a square, the side of which is the unit of length. 
 
 According as this is foot, inch, &c., the unit of surface 
 is called square foot, square inch, &c. ; the designations are 
 the same as for lineal measure, but with the addition of the 
 sign Q. 
 
 110. Rectangles with equal bases are to each other as 
 their altitudes. 
 
 If a certain line m is con- 
 tained exactly in both alti- 
 tudes, for example, H = pm\ 
 li qm , then 
 
 JL P. 
 
 k ' ' q ' 
 
 Parallels through the points 
 of division divide the rectangles respectively into p and q 
 equal rectangles, therefore 
 
65 
 
 B_ _p_ 
 
 r q 
 
 . R Jf 
 
 and consequently = - . 
 
 If the altitudes are incommensurable, the proposition also 
 holds good and is proved as /? in 83. 
 
 111. Rectangles with equal altitudes are to each other 
 as their bases. 
 
 This proposition is the same as the preceding one, for 
 any of the sides we please can be taken as base. 
 
 112. The ratio between any two rectangles is the prodzict 
 of the ratio between the altitudes and the ratio between the 
 bases. 
 
 Let the rectangles be R, with altitude // and base G, 
 and r, with altitude h and base g\ let us then imagine a third 
 rectangle P, with altitude H and base g. 
 
 R G PR 
 
 Then - == - and - = , 
 
 P g r k 
 
 from which by multiplication 
 
 R_ B_.G_. 
 r h g 
 
 If r denote the unit of surface, therefore 7^ and g the units 
 of length, this equation shews that : 
 
 The number of units of surf ace in a rectangle is found 
 by multiplying the number of units of length in the altitude 
 by the number of units of length in the base ; this is written 
 
 R == H.G, 
 
 where R, /7, and G denote abstract numbers, and we ourselves 
 must remember the denominations; to prevent this, we have 
 adopted the phrase, that foot multiplied by foot gives square 
 foot &c., by which we are enabled to calculate with concrete 
 numbers; we therefore say, that 
 
 The area of a rectangle eqtials the prod^tct of the alti- 
 tude and the base. 
 
 The area of a sq2iare is therefore the square of the side; 
 we therefore have i' Q = = 144" Q and i" D = = 144'" D Duo- 
 decimal measure, i" Q = 100'" D Decimal measure. 
 
66 
 
 113. A parallelogram is equal to a rectangle with the 
 same altitude and the same base. 
 
 For we have 
 
 A EAB & A FDC. 
 By subtracting these triangles one at 
 a time from the whole figure, the par- 
 allelogram AC and the rectangle AF will 
 D be left respectively; they are therefore 
 
 equal. 
 
 From this it follows that the area of a parallelogram is 
 the product of the altitude and the base. 
 
 114. The area of a triangle is half the product of the 
 
 altititde and the base. 
 
 For the triangle ABC is half of the par- 
 allelogram AB , which has the same base 
 and the same altitude. The area of a right- 
 angled triangle is half the product of the 
 sides containing the right angle. 
 In an equilateral triangle with the side s the altitude is 
 
 Vs , and the area therefore - . 
 
 2 4 
 
 115. The area of a figttre in which a circle can be in- 
 scribed is half the prodztct of the radius of the circle and 
 
 the perimeter of the figure. 
 
 From the centre draw lines to the 
 angular points; thereby the figure is di- 
 vided into triangles, which all have the 
 radius r of the circle for their altitude, 
 and the bases of which are the sides of 
 the figure o, b, c . . . , we therefore have 
 the area of the figure 
 
 . . .) = J rp , 
 
 A == 
 
 --rc ---- = = r 
 when p is the perimeter. 
 
 The radius of a circle inscribed in a triangle is there- 
 fore equal to the area divided by half the perimeter. 
 
116. The area of a circle with radizts r is Trr 2 . 
 
 According to the former proposition the area of a cir- 
 cumscribed polygon is \ rp ; if the number of sides in this be 
 made infinitely great, it will coincide with the circle, and p 
 becomes 27rr, therefore the area of the circle C=7tr 2 . 
 
 117. The area of a sector of a circle of g z's 
 
 i 
 
 360 
 
 For a sector of i is _ 
 .2 36o 
 
 of the circle and therefore its 
 
 area is 
 
 TIT' 
 
 a sector of g therefore 
 
 118. The area of a segment A is the difference between 
 the areas of the corresponding sector 
 
 and the triangle ; for the calculation of the 
 latter, one of the radii is usually taken 
 as base; the altitude h is then half the 
 chord AB of double the arc of the seg- 
 ment, and this is often known. 
 
 119. The ratio between two triangles, having one angle 
 eqzial, is the product of the ratios of the sides containing 
 the angle. 
 
 Let the triangles be 
 denoted by T and t, the 
 
 altitudes by H and /?, then / 1 \ / 
 
 we have (114) 
 
 B 
 
 but A and H, a and h are corresponding sides of similar tri- 
 
 H A .TAB 
 
 angles, so that -=- = , and therefore = -j- 
 
 ha tab 
 
 120. The ratio between two 
 similar triangles is the duplicate 
 of the ratio between a pair of corre- 
 sponding sides. 
 
 ,,, , ' , T A B 
 
 We have (119) - T 
 
 ^ yj t a b 
 
 but 
 
 TD 
 
 , therefore 
 b 
 
68 
 
 121. The ratio between the areas of any two similar fig- 
 ures is the duplicate of the ratio between a pair of corre- 
 sponding lines or of the linear ratio. 
 
 For two similar figures are divided by corresponding diag- 
 onals into similar triangles (93) ; if these be respectively de- 
 noted be T n T 2 , T 3 . . ., n 2 , t 3 . . ., and the ratio beween 
 any two corresponding lines or the linear ratio by /, we have 
 T T 7 1 
 
 * 
 
 from which, by a well-known proposition in proportion, 
 
 rri I fry i np 
 
 ^1~"^2~T" 2 3 ' ' __ r z 
 j I -/ I * 1/1 
 
 . 
 
 which was to be proved. 
 
 As the proposition holds good independently of the 
 number of sides, it also holds when this is infinite, therefore 
 also of figures contained by curved lines. 
 
 122. If similar figures be constructed on the sides of a 
 right-angled triangle, so that these are corresponding lines 
 in the figtwes, then the figure on the hypothenuse eq^tals the 
 sum of the figures on the sides containing the right angle. 
 
 If the sides be denoted by a, b, /?, the corresponding sim- 
 ilar figures by A, B, Z7, we have by 121: 
 
 J?_ A. JL - A + B 
 
 h* ~~ a* = ~ ^ ~~ a* + l* ' 
 
 but, as 7* 2 == a 2 + 6 2 , then also H = A + B. 
 
 By the help of this proposition and of 94, it is easy to 
 solve this problem, to draw a figiire, the area of which shall 
 be equal to the sum or difference of 'two given similar figures, 
 and which shall be similar to these. 
 
 Specially must be noticed the problem, to draw a square 
 equal to the sum or difference of two given squares. 
 
 123. To change a given polygon to a sqziare. 
 
 From the polygon cut off a triangle ACB by the diago- 
 nal A B\ if CD be drawn parallel to AB, every triangle, having 
 
69 
 
 its vertex in this line, and the base of which 
 is AB, will have the same area as ACB, 
 and therefore could be placed instead of 
 this; if the vertex be taken in D, which is 
 found by producing BE, then the new poly- 
 gon has one side less than the given one. 
 We continue in this way till the poly- 
 gon is changed to a triangle; if we call the altitude of 
 this /*, its base g, the side of the required square x, we must 
 have 
 
 * 2 = P<7, 
 
 which shews that x is a mean proportional between h and g 
 or between \g and h, it can therefore be constructed by one 
 of the methods in 96. 
 
 124. Triangles can be changed to other triangles satisfying 
 certain given conditions; if thereby one of the sides is to 
 remain unchanged, the altitude on this must also remain 
 unchanged^ and the vertex therefore fall in a known line par- 
 allel to the base; one more condition must then be given, 
 which can give one more locus of the vertex. If one of the 
 angles is to remain unchanged, the product of the sides con- 
 taining it imtst also remain unchanged (119); now if one of 
 the sides be given, the other is easily determined as fourth 
 proportional to this side and the sides containing the angle 
 in the given triangle. 
 
 125. To draw a figure similar to a given figiire and n 
 times as great. 
 
 If one side of the given figure be a, the corresponding 
 side in the required figure x, 
 
 x 2 n . 
 
 we have = (121) 
 
 a 2 i 
 
 or x 2 = na z 
 
 x is then constructed as mean proportional between a and na, 
 
 and thereupon apply 94. 
 
 126. The area of the triangle and the radii of its in- 
 scribed and escribed circles. 
 
7 o 
 
 Let the area of the triangle be T, its angles A, B, C, 
 and the sides opposite to these a, &, c; let r be the radius 
 of the inscribed circle, and r a the radius of the escribed circle, 
 
 touching the side a and the prolon- 
 gations of b and c\ we then have 
 T = A0 1 BA -f- A0 1 CA ^0 1 BG 
 or T = lr a c + r a b r a a 
 = r a (s a) (64). 
 We have thus (115) 
 
 T = rs = r a (s a) = r b (s b) 
 
 = r c (s c). 
 
 From the two first of these equa- 
 tions we get 
 
 T 2 = rr a s(s a), 
 but fromA^Oc CXD AO l Bc l we get, as 
 
 Be = s b (64), 
 Bc l = Ac i c = s c (64), 
 
 rr a = (s b) (s c) , 
 
 therefore T = Vs(s a)(s b)(s c). 
 
 127. The diameter 2R of the circumscribed circle equals 
 the product of two sides, divided by the altitude on the third 
 side. 
 
 Draw the diameter AD and join DC; 
 from A BAE^ A DAG we get 
 c h a be 
 
 ~2R ~ ~b~ ' ~2h~ a ' 
 
 When the numerator and denominator 
 of the fraction are multiplied by a, the 
 equation assumes the form * 
 abc 
 
 R = 
 
 or abc = 
 
 EXAMPLES. 
 
 194. Prove that the area of a rhombus equals half the prod- 
 uct of the diagonals. 
 
 195. How great is the area of a right-angled, isosceles triangle, 
 the hypothenuse of which is a? 
 
7 1 
 
 196. Prove that the area of a trapezium is measured by the 
 altitzide, imtltiplied by half the s^tm of the parallel 
 sides. 
 
 197. From a point within a regular polygon perpendiculars 
 are dropped on all the sides ; prove that the sum of the 
 perpendiculars is constant. 
 
 198. In a triangle one angle is 30 and the sides containing 
 it a and b\ how great is the area? 
 
 199. The area of an equilateral triangle is i D' 8 D"; how 
 great is the radius of the circumscribed circle? 
 
 200. Prove that a triangle is divided into three equal parts 
 by lines from the point of intersection of the median 
 lines to the angular points. 
 
 201. By a line, parallel to the side of a triangle, cut off an- 
 other triangle equal to i of the given one (120 and 96). 
 
 202. To divide a triangle into two equal parts by a line 
 issuing from a point in the one side (119). 
 
 203. What is the ratio between the side of a regular hexagon 
 and of an equally great equilateral triangle? 
 
 204. A triangle is divided by two transversals, parallel to 
 the one side, into three parts, the areas of which are 
 A) B, and C. Find the ratio between the distances be- 
 tween the parallels. 
 
 205. Prove that a quadrilateral is twice as great as the 
 parallelogram, formed by joining the middle points of 
 the sides. 
 
 206. The area of a quadrilateral is known and also the seg- 
 ments of the diagonals; find the areas of the four tri- 
 angles, into which the diagonals divide the quadrilateral. 
 
 207. How large is the greatest triangle that can be cut out 
 of a square piece of paper with the side a? 
 
 208. Prove that the area of a regular 2/1 sided figure inscribed 
 in a circle is a mean proportional between the areas of 
 the inscribed and circumscribed n sided figure. 
 
 209. How great is the radius of a circle, the area of which 
 is 
 
210. How great is the radius of a circle, in which a sector 
 of 7 12' has an area of 2 Q"? 
 
 211. How great is the area of a segment of a circle, the arc 
 of which is 90, when the radius is r? 
 
 212. To divide a circle into 3 equal parts, by the help of 
 other circles, concentric with the given one. 
 
 213. How great is the area of a segment of a circle, the arc 
 of which is 30, when the radius is r? 
 
 214. In a circle with radius r two parallel chords are drawn, 
 of which the arcs are 60 and 120; how great is the 
 area between them? 
 
 215. Two equal circles with radius r pass through the centre 
 of each other; how great is the area of the figure they 
 have in common. 
 
 216. Three equal circles with radius r touch each other. How 
 great is the area between them? 
 
 217. A circle is inscribed in a sector of 90; what is the ratio 
 between the areas of the two figures? 
 
 218. In a circle two radii AB and AC are drawn perpendicular 
 to each other; on each of these as diameter a semicircle 
 is described; the two semicircles cut one another in D. 
 Find the ratio between the areas of the figures AD and 
 DBG. 
 
 219. On the sides of a right-angled triangle any similar curves 
 are described, so that the one, corresponding to the 
 hypothenuse, turns to the same side as those, correspond- 
 ing to the other sides of the triangle. Prove that the 
 area of the figure bounded by tlie three curves equals 
 the area of the triangle ; as a special case may be men- 
 tioned that, where the similar figures are semicircles. 
 
 220. A figure is bounded by three straight lines and two arcs. 
 Draw another figure similar to this and three times less. 
 
 221. Find R, r, r a , and r b for an isosceles triangle with the 
 base a and the sides Z>. 
 
 222. Find the area of a triangle with sides 6, 7, and 9; there- 
 upon find 7?, r, r, r 6 , and r c for the same triangle. 
 
 223. Prove that T 2 = - rr a r h r c . 
 
73 
 
 224. Shew that 
 
 r a + r t + r e = = 4# + r. 
 
 225. Prove that the proposition in 119 also holds good, when 
 the two angles, instead of being equal, are supplementary. 
 
 226. Prove that two quadrilaterals, the diagonals of which 
 make the same angle with each other, are to each other 
 as the products of the diagonals. 
 
 227. A circle has its centre on the circumference of another 
 circle. Draw any tangent to the first circle cutting the 
 second circle in A and B. Shew that OA . OB is con- 
 stant (127). 
 
 228. An inscribed q^ladrilateral has the sides a, &, a, and /?, 
 the diagonals d and d (d joins the point of intersection 
 of a and b to the point of intersection of a and /?). 
 Prove that 
 
 d o/9 + ab 
 
 Employ the formula in 127 for expressing the area in 
 a double manner. 
 
UNIVERSITY OF CALIFORNIA LIBRARY 
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