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Del he Sey | VEN aype } AP buseyprie’ Fay ey sing tae pay peers Leah ja hag Pu abae abe eh + ALE Ware AEP Ae ae! 1a) E) I Apanmper tly G art Uy PTT RY BLED Ue suid Fil! ene ACM ARNE ET ae wa eeek irk baie Gale? ” re ue Pp mA “ ; ie fal a bieryaits as a THE UNIVERSITY” ae OF ILLINOIS LIBRARY [O42 ® Return this book on or before the "| Latest Date stamped below. A oe charge is made on all overdue | books, m 6=©=6—SCS—*é less than any other two lines ap, 3 drawn from a and 8 to any other int p in the line pe. 'For Aap = pc and pp is common and the angles at p are izht angles, .. AP=pc. In the same manner, if pe be jined, it may be shown that ap=pc. Hence ap and BP tgether are equal to Bc, and ap, pB are equal to cp, pB. low (Eucl. i. 20.) Bc is less than Bp, pc, and therefore ap, 13 are less than ap, pB; therefore, &c. (7.) Of all straight lines which can be drawn from a given gint to an indefinite straight line, that which is nearer to the arpendicular is less than the more remote. And from the same fint there cannot be drawn more than two straight lines equal to tch other, viz. one on each side of the perpendicular. B2 Sect. 1.] GEOMETRICAL PROBLEMS. 3 4 GEOMETRICAL PROBLEMS. [ Sect. 1. Let a be the given point, and Bc the given indefinite straight line. From A let fall the perpendicular vA Ap, and draw any other lines ar, AG, AH, &c. of which aF is nearer to AD than AG is, and ag than AH; then AF will be less than ac and ag than AH. | For since the angle at p is a right angle, the angle ara if greater than a right angle (Eucl. i. 16.), and therefore create than aGF, hence (Eucl. i. 19.) ag is greater than ar. In thi same manner it may be shown that aun is greater than AG. | And from a there can only be drawn to sc two straigh’ lines equal to each other, viz. one on each side of aD. Mak: DE= pF, and jon az. Then AzE= AF (i.2.). And beside} Af¥ no other line can be drawn equal to ar. For, if possible let AI = AF. Then Seca AT = AF and AF = AB, therefor| ~- eee b-l-kD)COFSCG He to AF. In the same manner it — be shown that no oti but AE can be equal to AF, therefore, &c. (8.) Through a given point, to draw a straight line, so that th parts of it intercepted between that point and perpendiculas, drawn from two other given points may have a given ratio. ij Let a and B be the points from which the perpendiculars al to be drawn, and c the point through which the line is to F drawn. Join ac, and produce it to p, making AC :° cD in tl given ratio; join BD, and through c draw A | ECF per qaantioaltn to BD. ECFis the line required. Draw AE parallel to Bp, and .*. perpen- dicular to uF. The triangles ACE, DFC, having each a right angle, and the angles at c equal, are equiangular, whence CE: CF:!AC : CD, i.e. in the given ratio. l~ or Sect. 1.| GEOMETRICAL PROBLEMS. — (9.) From a given point between two indefinite right lines given in position, to draw a line which shall be terminated by the given lines, and bisected in the given point. _ Let as, Ac be the given lines, meeting in B ‘a. From p the given point draw pp parallel E to Ac one of the lines, and make DE = DA. Join ep, and produce it tor; then willerbe 2% Us pisected in P. For since pp is parallel to ar, (Eucl. vi. 2.) 4 ae EP; PF‘: ED; DA,1.e. in a ratio of equality. | Cor. Ifit be required to draw a line through p which shall ye terminated by the given lines, and divided in any given ratio in P, draw PD parallel to ac, and take ap : Des in the given ratio, and draw EPF, it will be the line required. | (10.) From a given point without two indefinite right lines given ‘n position ; to draw a line such that the parts intercepted by the noint and the lines may have a given ratio. Let as, ac be the given lines, and Pp the given point. Draw >D parallel to ac, and take AD : DE in the given ratio. Join ha, and produce it to F. Then PF: PE vill be in the given ratio. For the triangles PDE and AEF are similar, laving the angles at © equal, as also the mgles PDE, EAF, (Kucl. 1. 39.) | She eer.) A Bi ED nd comp. PF : PE::AD : DE, 1. e. in the Banding os Feed atio. G Pr _(11.) From a given point to draw a straight line, which shall ut off from lines containing a given angle, segments that shall ave a gwen ratio. Let anc be the given angle, and p the given point, either nthout or within. In Ba take any point a, and take aB: 6 GEOMETRICAL PROBLEMS. | Sect. i BC in the given ratio. Join Ac, and from p draw PDs parallel to ac. PDE is the line required. For since DE is parallel to ac, (Kucl. vi. 2.) DB: BE}: AB: BC, 1. e. in_the given ratio. (12.) If from a given point any number of straight lines b drawn to a straight line given in position ; to determine the locu of the points of section which divide them in a given ratio. Let a be the given point, and Bc the line given in position. From a draw any line AB, and divide it at © in the given ratio; through E draw EF parallel to BD; it is the locus required. From a draw any see line AD pane EF in F; then (Kucl. vi. 2.) AF: FD:: AE: EB,1.e. in the given ratio. In ane same manner any other 7 line sha from A to BD will be divided in the given ratio, b EF, which therefore is the locus required. (13.) A straight line being drawn parallel to one of the lin containing a given angle and produced to meet the other ; throug a given point within the angle, to draw a line cutting the ot three, so that the part intercepted between the two parallel lime may have a given ratio to- the part intercepted between the give point and the other line. : Let asc be the given angle, DE parallel to as, and Pp the given point. From Pp draw pc parallel to DE or AB, and take BE: CF in the eiven ratio. Join Fp and produce it to A; APF is the lt required. For since DE and cp are parallel to aB, é AD! EB:: DF: EF?! PF: CF s alt. AD: PF?! BB: CF i.e. in the given ratio. i NI Sect. 1. | GEOMETRICAL PROBLEMS. (14.) Two parallel lines being given in position; to draw a hird, such that, if from any point in it lines be drawn at given ngles to the parallel lines, the intercepted parts may have a given atio. Let AB, cD be the given parallel lines; a AB take any point £; and draw EF, G making angles equal to the given an- les; produce mr, and take nn; EG in he given ratio; produce Fx to 1 so that I;1E:: HE: EF; through 1 draw u1 arallel to AB; it is the line required. Draw 1k parallel to EG; then the triangles 1mK, FFG are quiangular, “IE? 1K?) EF‘: EG ) but FI: 1E?: HE: EF; |.ev equ. FI: 1K !: HE: EG, i. e. in the given ratio; and KE = EGF which is one of the given angles, and by construc- on 1FG is equal to the other. Also lines drawn from any dint in LI, making with aB and cp angles equal to the given igles will be parallel and equal to F1, 1K, and... in the given itio. (15.) If three straight lines drawn from the same point and in ie same direction be in continued proportion, and from that point so a line equal to the mean proportional be inclined at any wgle ; the lines joining the extremity of this line and of the pro- rtionals will contain equal angles. Let aB ; ac:: Ac: AD, and from A let a E be drawn equal to ac, inclined at any igle to AB; join EB, EC, ED; the angle (EC is equal to the angle cep. B cD A ee mcesn SAC 32 AC2AD, and AE= Ac; ~AB;AE:: AE: AD, 1. e. the sides about the gle A are Sproportional, and .*, the triangles AEB, AED are nilar, and the angle arp is equal toEBA. Also since cA = 8, the angle AEc=ECA; but ECA is equal to the two BEC, AJ haa saad 8 GEOMETRICAL PROBLEMS.. [ Sect. 1. | HBO, (Eucl. i. 32.) .*. also anc is equal to the two BEC, EBC; of which pra is equal to EBC; .*. the remainder DEC is equal to the remainder BEC. : i (16.) To trisect a right angle. Let acs be aright angle. In ca take any B, point A, and on ca describe an equilateral triangle acD, and bisect the angle pca by the straight line ce; the angles BCD, DCE, ECA are te! to one sawn For the angle pca being one of the angles of an equilateral triangle, is one third of two right angles, an therefore equal to two thirds of a right angle BCA; conse quently Bcp is one third of pca; and since the angle DCA Is bisected by cr, the angles pc#, ECA are each of them equal te one third of a right angle, and are therefore equal to BcD an¢ to each other. c= (17.) To trisect a given finite straight line. © Let aB-be the given straight line. On it describe an equilateral triangle ABC; bisect the angles cAB, CBA by the lines AD, BD meeting in Db, and draw DE, DF parallel to ca and cB respectively. AB will be trisected in E and F. Because ED is parallel to ac, the ( angle EDA=DAC=DAE and therefore ; 3 AE=ED. For the same reason pF=FB. But vx being paral lel to ca and pF to cs, the angle DEF is equal to the angl CAB, and pFrE to cBA, and therefore EDF = acB; and hence the triangle EDF is equiangular, and consequently equilatera therefore DE=EF=FD, and hence AE = EF = FB, and AB™ trisected. ¥ E (18.) To divide a given finite straight line into any number { equal parts. ‘ » Sect. 1.] GEOMETRICAL PROBLEMS. 9 Let aB be the given straight line. Let ac ge any other indefinite straight line making my angle with a8, and in it take any point p, and take as many lines Dn, EF, FC, &c. 2ach equal to AD as the number of parts into; which AB is to be divided. Join cs, and Jraw DG, EH, FI, &c. parallel to pc; and 2% therefore parallel to each other; and draw px parallel to as. Then because GD is parallel to HE one of the sides of the Inangle AHE, AG: GH:: AD: DE; hence ag=Gu. For the same reason DL=LM. But pM being parallel to a1, and pe, ‘ea to mi, the figures DH, HM are parallelograms; therefore DL=GuH and LM= HI, consequently ¢Ho=un1. In the same jmanner it may be shown that 41 = 1B; and+so on, if there be any other parts; therefore AG, Gu, u1, 1B, &c. are all equal, and AB is divided as was required. Cor. If it be required to divide the line into parts which shall have a given ratio; take AD, DE, EF, &c. in the given ratio, and proceed as in the proposition. (19.) To divide a given finite straight line harmonically. Let aB be the given straight line. From Bs draw any straight line Bo, and join Ac; and from any point E in Ac draw ED parallel to cB, and make FD=FRH, join DC cutting AB in G AB is harmonically divided in G and F, ‘Since sc is parallel to rp, the angle BcG is equal to GpF and the vertically opposite angles at G are equal; therefore the triangles DGF, BGC are similar, and BC : BG :: FD: FG. But Fx being parallel to Bc, (Eucl. vi. 2.) AB: BCO!: AF: FE=FD. ”. ex equal, AB: BG 3: AF: FG or AB: AF :: BG: FG. Bis: 10 GEOMETRICAL PROBLEMS. [ Sect. 1, (20.) If a given finite straight line be harmonically divided, ana from its extremities and the points of division lines be drawn to meet in any point, so that those from the extremities of the second proportional may be perpendicular to each other, the line drawn from the extremity of this proportional will bisect the angle formed by the lines drawn from the extremities of the other two. Let the straight line aB be divided harmonically in the points G and Ff, and let the lines Ac, Bc, Gc, Fc be drawn to any point c, so that Gc may be perpendicular to ca; the angle BOF will be bisected by ca. Through c draw eG@p parallel to ca, meeting cr in p; then £G being parallel to ac, the triangles EGB, ACB are similar; also the triangles acF, DFG; hence AF: AC i! FG: DG but AB: AF ?: GB: GF, "er @quo AB: AC 3: BG: GD. But AB: AC!: GB: GE *, (Eucl. v. 15.) BG: GD :: BG: GE, and therefore GD = GE, and Gc is common, and the angles at G are right angles, therefore the angle pcG = Gcr, and FCB is bisected by ca. (21.) If a straight line be drawn through any point in the line bisecting a given angle, and produced to cut the sides containing that angle, as also a line drawn from the angle PTR Cn aCe eae to the bisecting line ; tt will be harmonically divided. Let the angle asc be bisected by the line Bp, and through any point p in this line draw GDFE meeting the sides in G and F, and BE a perpendicular to BD in E; then will EG: EF ?: GD : FD. For through p draw ac par- allel to BE and therefore perpen- dicular to Bp; then the angles Sect, 1.] GEOMETRICAL PROBLEMS. 11 .DB, CDB being right angles are equal, and anp = cBD, and sD is common to the triangles ApB, cpB, .. AD=pc. But yc being parallel to un, EG:GD:} EB: DC:: EB! AD!: EF: FD, lince the triangles EFB and AFD are similar, “. HG > EF ?:.GD : FD. _ (22.) If from a given point there be drawn three straight lines wrming angles less than right angles, and from another given point ‘vithout them a line be drawn intersecting the others so as to be armonically divided ; then will all lines drawn from that point neeting the three lines be harmonically divided. | From a let AB, AC, AD be f£ A rawn making each of the an- I\\ les BAC, CAD less than a right ngle, and from a given point E +t EBD be drawn so as to be armonically divided in c and .; then will any other line EF e harmonically divided in @ and nu. Through e draw 1x parallel to Bp, then mc jiCB tiaK.G 3 G4; But pco:cBp:: DE: EB foi Huchiy. 1Lb.Ephei eB: K G+) GX and alt. DE: KG! EB: GI | and since DE is parallel to ax, (Eucl. vi. 2.) DE: KG:: EF: FG and xB being parallel to e1, HB; GI ?; EH : HG, hence (Eucl. v. 15.) EF; FG‘: EH: HG and ali, EF : EH ?: FG: HG. (23.) If a straight line be divided into two equal, and also into pi unequal parts, and be produced, so that the part produced may 12 GEOMETRICAL PROBLEMS. [ Sect. 1 have to the whole line so produced the same ratio that the unequal. segments of the line have to each other ; then shall the distances of the point of unequal section from one extremity of the given line, from its middle point, from the extremity of the part produced, and from the other extremity of the given line, be proportionals. ~ Let AB be divided into two equal parts & BD oc. in c¢ and into ok setae parts in p, and produced to 8, so that BE: EA‘: BD: DA; then will AD: pC :: ED: DB. For since BE: EA??? BD: DA Nv. AB: EB:: AD; DB div. AB: BE 3: 20D: DB ONG. A OB ee On) se Give Oe OD oe) Bist ob .. comp. AD: DC i: ED: DB. Cor. The converse may easily be proved to be true. (24.) Three points being given; to determine another, through which if any straight line be drawn, perpendiculars upon it from two of the former, shall together be equal to the perpendicular from the third. Let a, B, c be the three given points. Join AB, and bisect it in p. Join cp, from which cut off DE equal to a third part of it. £ is the point required. Through 5, let any line re be drawn, and let fall on it the perpendiculars a1, BG, DH, CF; then the angles at F and u being right angles, am the vertical anak: at EB equal, the triangles CFE, DHE are equi; angular, ry 3; POs DHA eCE iE Dic. aes als | *. FC = 2p'H; but since Al, BG, DH are parallel, and ap= DB .AI+ BG = 2DH=FC. | q (25.) From a given point in one of two straight lines given in position, to draw a line to cut the other, so that if from the poi tS es e Sect. 1.] GEOMETRICAL PROBLEMS. 13 of intersection a perpendicular be let fall upon the former, the seg- ment intercepted between it and the given point, together with the irst drawn line may be equal to a given line. Let as, Bc be the lines riven in position, and a she given point. Draw 4D perpendicular to aB, and meeting BC in D; lraw DE parallel to ax, ind equal to the given ine. And draw EF paral- el to AD, meeting cB in *, Join FA, and produce it, and from p draw pa = pg, meet- ng FG in G, and draw au parallel to pa, and let fall the yerpendicular H1; Au and Ax together are equal to the given ine. _ Through u draw Kx parallel to pm; then since ep is parallel OAH and HL to DE, oD.G A Hea WEP CD ii Has batpG — nr ex =H ny, | “, AH + AI= KL = DE = the given line. | _ (26.) One of the lines which contain a given angle, is also gwen. lo determine a point in it such, that of from thence to the indefinite ine there be drawn a line having a given ratio to that segment of t which is adjacent to the given angle ; the line so drawn, and the ther segment of the given line, may together be equal to another wen line. _ Let aB be the given line, and Bac the ven angle. From ps draw sp to ac, uch that it may be to AB in the given ‘atio *; produce it till B& = the other iven line. Through r draw xc parallel OAB, meeting Acin c. Join Bc, and * That is, the given ratio must be less than that of as to the perpendicular 14 GEOMETRICAL PROBLEMS. [ Sect. + parallel to Fp, EB; H is the point required. For produce 1G to meet CE in K; then (Kucl. vi. 2.) ED: KG 3: CD: CG‘: DF: BG, but ED = DF, .. KG = BG, and HG + GB = HG + GK = BE = the given line, and HG: HA !: BD: ABi.e. in the given ratio. (27.) Two straight lines and a point in each being given in posi tion; to determine the position of another point in each, so the the straight line joining these latter points may be equal to a give line, and their respective distances from the former points in given ratio. | Let a and B be the given points in the lines Ac, BD which are given in position, and produced to meet in c. Take BD : AC in the given ratio, and from B draw BE parallel and equal to Ac. Join DE, and produce it to meet cr drawn at any angle from c, equal to the given line; dra} FG parallel to BB, and from e draw eu parallel to rc; G and | are the points required. For BE being parallel to Gr, | DG:GF::DB: BE, OL. DG 7 ACs. DB AC, .”.. (Hucl. y. 49. Cor.) BG: AH :: DB: AC in the given ratio, and HG = CF = the given line. (28.) If a straight line be divided into any two parts, and pr¢ duced so that the segments may have the same ratio that the who line produced has to the part produced, and from the extremitt of the given line perpendiculars be erected ; then any line draw, through the point of section, meeting these perpendiculars, will ¢ divided at that point into parts, which have the same ratio th those lines have, which are drawn from the extremity of the pré duced line to the points of intersection with the perpendiculars. 8 | * Sect. 1.] I. | GEOMETRICAL PROBLEMS. 15 Let AB be divided into any two E yarts in c and produced to p so that k@:CB:: AD: DB, and from 4 and slet Az, BF be drawn perpendicu- ars to AB, and through c: let any ime ECG be drawn meeting them in 4 . Sect. 1.] GEOMETRICAL PROBLEMS. 17 3F ; BE in the given ratio of the parts o be cut off. Join Az, FE; and draw dH and BC parallel to er, and uc yarallel to DB meeting BC in co, and hp in 1. hen (Kucl. vi. 2.) Al: 1H 2: AB: Ein the given ratio of the remainders; nd the triangles Bc1, BFE having the angle cui = the alter- late angle BFE, and CIB = FBR, are equiangular, ée Bl LC .. BEBE, a the ratio of the parts to be cut off; and | AB, HC (= DB) are the given lines. (32.) Three lines being given in position; to determine a point h one of them, from which if two lines be drawn at given angles ) the other two, the two lines so drawn may together be equal to a ‘wen line. | Let as, Ac, Bc be the p Dr a aree lines given in posi- on, take ap =the given ne, and making with as ‘a angle equal to one of he given angles. Through | draw pda parallel to an, id meeting ac and Bc in and 6. Draw a& to meet Bin £ making the angle EC = the given angle to > made by the line to be awn, with Bc. In az ike ad = ap, and join ad cutting Bc in Fr. Draw re parallel * EA meeting Ac in G, which is the point required. For through ¢ draw 1@ parallel to pa, then the triangles (3 I, @AD are similar, and CA AD Ad, aG:>G1% But GAT Ad 20 G5: GF, | | | | | | 18 GEOMETRICAL PROBLEMS. [ Sect. 1 and’, GI=GF, ..GH+GF =GH+GI = AD=the given line) and the angle GuB = DAB, and GFC = AEC, .. GHB, GFC ar equal to the given angles. (33.) If from a given point two straight lines be drawn includ ing a given angle, and having a given ratio, and one of them b always terminated by a straight line, given in position ; to deter mine the locus of the extremity of the other. Let a be the given point, and Bc the line given in position. From a draw any line ap, and make the angle DAE equal to the given angle, and take ax such that AD! AE may be in the given ratio; and through = draw er making the angle AEF = ADB; EF is the locus required. Draw any other line a8, and make the angle BAF =DAkE. Then the angle BAD=FAE and ADB= AEF, .*. the triangles ABD, AEF are equiangular, whence A’ - AF‘: AD! AB, in the given ratio. The same may be prove of any other lines drawn from a and containing an angle equi to the given angle, and one of them terminated in Be. (34.) If from two given points straight lines be drawn, col taining a given angle, and from each of them segments be cut 0] having a given ratio; and the extremities of the segments of lines drawn from one of the points be in a straight line given position; to determine the locus of the extremities of the segmen of lines drawn from the other. ‘ Let a and B be the given points, and cp the line given in position. From A to cp draw any line AE. Make the angle EAF =the given angle, and AE : AF in the given ratio, and let re be © the locus of the points F (i. 33.).. Draw BH equal and parallel to ar, and through u draw u1 parallel toer. It is the locus required. ect. 11. | GEOMETRICAL PROBLEMS. CM i, ' Draw any lines ax, BK containing the angle at x = the iven angle. Make the angle LAm = the given angle; AL : M in the given ratio, and m is in the linear. And since AF parallel to Bu, and FM to HN, and BK to Am (since the agles BKA, LAM are equal) and Ar=Bu, .*, the triangles HIN, AFM are similar and equal, . am = By; but AL : Am is equal to the given ratio, “. also AL : BN is equal to the given ratio. nd the same may be proved of any other lines drawn in the me manner. SECTION II. | (1.) [fa straight line be drawn to touch a circle, and be parallel a chord ; the point of contact will be the middle point of the ¢ cut off by that chord. ‘Let cp be drawn touching the circle ABE ¢€ E D the point ©; and parallel to the chord AB; is the middle point of the arc arn. vs b Jom An, EB. The angle Bax is equal to ‘e alternate angle cE A, and therefore to the gle HBA in the alternate segment, whence t==EB, and (Kucl. iii. 28.) the arc ax is equal to the arc En. Cor. 1. Parallel lines placed in a circle cut off equal parts of 2 circumference. If ra be parallel to a; the arc EF=EG, whence AG=BrF. Cor. 2. The two straight lines in a circle, which join the tremities of two parallel chords are equal to each other. For AB, FG be parallel, the arcs ac, BF are equal, therefore ucl, i, 29.) the straight lines AG, BF are also equal. G a (2.) If from a point without a circle, two straight lines be drawn the concave part of the circumference, making equal angles with | line joining the same point and the centre, the parts of the es which are intercepted within the circle are equal. From the point P without the circle a Bc let two lines PB, PD c 2 20 GEOMETRICAL PROBLEMS. [Sect. 11. be drawn making equal angles with po, the line joining p and the cen- tre; AB shall be equal to cp. Let fall the perpendiculars of, or; then since the angle at & is equal to the angle at r, and EPO —rpo, and the side po, opposite to one of the equal angles in each is common, .*. OF=OF, and consequently (Hucl. i. 14) AB=CD. (3.) Of all straight lines which can be drawn from two gwen points to meet on the convex circumference of a given circle ; the sum of those two will be the least, which make equal angles wit. the tangent at the point of concourse. | Let A and B be two given points, cE a tangent to the circle at c, where the lines ac, Bc make equal angles with it; and let lines ap, BD be drawn from A and B to any other point D on the convex circumference; Ac and cB together are less than aD, DB together. | Let ap meet the tangent in &. Join £8; then (1. 6.) At and cB together are less than An and EB; but AE, EB are les than ap, DB (Nucl. i. 19.), .. @ fortiori ac, cB are less tha Ap, ps. And the same may be proved of lines drawn to ever, other point in the convex circumference. (4.) Ifa circle be described on the radius of another circle any straight line drawn from the point where they meet, to th outer circumference, is bisected by the interior one. | Let aps be acircle described on the radius AB of the circle Acz. Draw any line Ac meeting the circle ABD in D; AD is equal to Dc. Join ps. Then the angle apB being in a semicircle is a right angle; and there- fore Bp being drawn from the centre B of the circle AcE bisects Ac (Kucl. in. 3.). Sect. 11.] GEOMETRICAL PROBLEMS. 21 5.) Lf two circles cut each other, and from either point of inter- section diameters be drawn; the extremities of these diameters and the other point of intersection shall be in the same straight hne. ‘Let the two circles ABc, ABD cut each other in A and B; draw the diameters ac, ap, and ‘oin BC, BD; CB and BD are in the same straight ine. Join as; the angles anc, aBp being angles nm semicircles are right angles, and therefore Eucl.i. 13.) cB and sp are in the same straight ine. (6.) If two circles cut each other, the straight line joining their ioints of intersection, is bisected at right angles by the straight me joining their centres. Let the two circles whose centres re c and p cut each other in A and B; 9m AB, Dc. vc bisects aB at right Ze ngles. Nea (Jom BD, DA, AC, CB. Since ap= B 'B, and Dc is common to the triangles ‘DC, BDC, and the base ac=cB, .*. (Eucl. i. 8.) the angle \DE=BDE. Hence the two sides ap, pEare equal to the two D, DE, and the included angles are equal, .:, (Kucl. i. 4.) Az = EB, and the angle DEA=DEB, and being adjacent, they are ight angles, i. e. pc bisects an at right angles. A (7.) To draw a straight line which shall touch two given reles. 1, If the circles be equal. Let a and B be the centres, Cc D in AB; and from a and B ‘aw AC, BD at right angles it; joncp. Then ac being _ wallel and equal tops; cp ™ A 22 GEOMETRICAL PROBLEMS. [ Sect. 11 - is parallel to AB, ... CABD is a rectangular parallelogram ; anc the angles at c andl D being right angles, cp is a tangent t both circles (Kucl. iti. 16. Cor.). | 2. If the circles be unequal, and the line be required to touel them on the same side of the line joining the centres. Let A and B be the cen- tres; join AB; and with the centre B, and distance C equal to the difference of the given radii, describe a sey: circle; and from a draw Ak touching it. Join BE, and produce it top; draw AC parallel to BD, and join cD. Then ac poug parallel and equal to Dr, cD is equal an parallel to AE .*. ACDE is a parallelogram; and the angle az being a right angle, AED is also a right angle; hence the angle at c and p are right angles, and therefore cp touches bot circles. 3. If the line be required to touch them on opposite sidesll ( the line joining the centres. With the centre B and radius equal to the sum of the given radii describe a circle, to which from A draw a tangent AE. Join BE, and let it cut the given circle in D. Draw Ac pa- rallel to BE; join CD. Then ac being equal and parallel to ED, ACDE is a parall ogram ; and the angle aED being a right angle, the angles at and p are right angles, and therefore cp touches both circles, | K (8.) If a line touching two circles cut another line joining the centres, the segments of the latter will be to each other ast diameters of the circles. . - e Sect. 11. | GEOMETRICAL PROBLEMS, 93 - Let the line a8 touch the circles, lyhose centres are Cc and D, in A and » and cut cp in the point £; CE vill be to ED in the ratio of the dia- aeters of the circles. Joinca, BD. ‘Then the angles at a and B are right angles, nd the angles at & are vertically opposite, therefore the triangles \EC, BED are equiangular, and consequently CE.cAED.: | CATS BD DCA * Deh. (9.) Ifa straight line touch the interior of two concentric circles, nd be placed in the outer ; it will be bisected at the point of jontact. Let as touch the interior of two circles, D apse common centre is 0, in the point c; aB 4s bisected in c. | Join oc; then (Eucl. iii. 18.) the angles at c a ‘re right angles; and oc drawn from the centre NS Ge B f the circle apps at right angles to aB, bisects + (Buel. iii. 3.). (10.) If any number of equal straight lines be placed in a circle ; ) determine the locus of their points of bisection. _Let there be any number of lines ax, cp, laced in the circle whose centre is 0, and let 1em be bisected in 5, F; jo O8, oF; then fucl. iii. 14.) these lines are equal, and there- ore the locus will be a circle whose centre is 0, ad radius equal to the distance of the points of isection from o. | (11.) If from a point in the circumference of a circle any num- Ee of chords be drawn ; the locus of their points of bisection will 2 a circle. 24 GEOMETRICAL PROBLEMS. From the given point a let any chord | AB be drawn in the circle, whose centre 1s ww o; bisect itin p. Join ao, Bo, and draw (7 \ DE parallel to Bo. APR 2 Then pz being parallel to Bo, the tri- | angles ADE, ABO are similar, and Bo is equal to AO, .. DE =| EA; but AE: A0:: AD: AB (Eucl. vi. 2.), whence AE = 4AC - ED = EA = 1,0, and the locus will be a circle described ry AO as a diameter. | (12.) If on the radius of a gwen semicircle, another semicire. be described, and from the extremity of the diameters any lines i drawn cutting the circumferences, and produced so that the pa produced may always have a given ratio to the part intercepte between the two circumferences; to determine the locus of ti extremities of these lines. | On as the radius of the semicircle arc let a semicire’ ApB be described; and from a draw ~~ | any line ADE, which produce till BF: | ED in the given ratio. [/ Tall a BR i \e Produce ac to G, making CG: CB, in the given ratio, and jo DB, EC, ‘¢ ts a FG3 then since FE : ED}: GC: CB, pe eVGC ED. CR DAK” ARN en fo Ay whence (Eucl. vi. 2.) re is parallel to cz and ps, and the ang AFG is aright angle, and is in a semicircle whose diameter | AG; hence the locus required is a semicircle. (13.) If from a given point without a given circle, straight l : be drawn, and terminated by the circumference ; to determine H locus of the points which divide them tn a given ratio. | Let a be the given point and c BCD the given circle. Find o S its centre and join Ao, anddi-_ 8B D EF i vide it in £, so that Ao : AE in H | the given ratio; and find a ect. 11. | GEOMETRICAL PROBLEMS. 25 oimt F, so that EF may be to op in the given ratio; and with ie centre E and radius EF describe a circle ; it will be the locus quired. ' Draw any line AGC; join oc, EG. Since ao: AE ina given tio, as also OD : EF; fe OC SE GIITALON) AB, hence 0c is parallel to ne, and AC ; AG :: OC : EG, #. e. in the given ratio. _In the same manner it may be shown that every line drawn om a to Bev will be divided by the circumference of the rcle Gru in the same ratio, i. e. Gru will be the locus iquired. \(14.) Having given the radius of a circle; to determine its intre, when the circle touches two given lines which are not prallel. Let BA, AC be the two lines \aich touch the circle, whose radius i given. 'Bisect the angle Bac by the line if, the centre of the circle will be in lis line (Kucl. iv. 4.). From a draw ,D at right angles to aB, and make jequal to the given radius; through p draw Do parallel to as eeting AE in O; then the centre of the circle being in this line 30, must be at the point of intersection o. R (15.) Through three given points which are not in the same ‘aight line, a circle may be described; but no other circle can Ass through the same points. ‘Let a, B, c be the three given points. Join 43, BC, and bisect them in p and E; from | } Cc \uch points draw po, £o at right angles to /\ tem; these lines will meet in some point 0; IN f - if not, they are parallel, and therefore an, i ee » must be parallel, which is contrary to the pposition. Join AO, BO, CoO. | > io | | =| 26 GEOMETRICAL PROBLEMS. [ Sect. 1 t Since AD = DB, and DO is common, and the angles at equal, ... Ao = BO. In the same manner it may be shown thi BO = co; and the three lines 0A, 08, oc being equal, a cire| described from the centre o at the distance of any one of the) will pass through the extremities of the other two. And besides this, no other circle can pass pee Ay B, © for if it could, its centre would be in pF and EH, and .. in the intersection; but two right lines cut each other only in aa point, .*. ae one circle can be described. (16.) From two given points on the same side of a line given: position, to draw two straight lines which shall contain a 7 angle, and be terminated in that line. Let a and B be the given points, and cp the given line. Join AB, and on it describe a segment of a circle containing an angle equal to the given angle, and (if the problem be possible) meeting CD in P; P is the point required. For join PA, PB; the angle aArB Demet in the segment. equal to the given satay , (17.) If from the extremities of any chord in a circle perpe: diculars be drawn, meeting a diameter ; the points of intersecti) are equally distant from the centre. | | At c and p the extremities of the chord cp, let perpendiculars to it be drawn meeting a diameter AB in E and F; E and F are equally distant from the centre o. Draw oG perpendicular to cp, and therefore bisecting it; then oc is parallel to pF; whence GD: OF :! HG: 8 fect. 11. | GEOMETRICAL PROBLEMS. 9” ince the triangles 1G0, HEC are equiangular ; . (Hucl. v. 18,15.) pe: or:: 60:08 but GD = GC, .. OF = OR. | (18.) If from the extremities of the diameter of a semicircle erpendiculars be let fall on any line cutting the semicircle; the arts intercepted between those perpendiculars and the circumfer- nce are equal. _ From a and ps, the extremities of the dia- D 1eter AB, let Ac, BD be drawn perpendicu- Gr w to any line cp cutting the semicircle in 6 \ \ ‘and F; CE is equal to FD. A O _ From o the centre draw og perpendicu- w to cD, it will be parallel to ac and sp, _ whence CG : GD !: AO: OB, @.¢. in a ratio of equality. tut (Kucl. ii. 3.) ee = Gr, and .. cE = FD. _(19.) Ina given circle to place a straight line parallel to a wen straight line, and having a given ratio to it. _ Let as be the given line in the circle anc hose centre is 0. Draw the diameter cp at ght angles to AB: and taking a line uF (hich has to aB the given ratio (Eucl. vi. 12.), lace it in the circle asc; bisect it in @ and moc; make ou = o«, and through 4, raw 1K parallel to AaB; 1K is the line -quired, _ For since og = on, .-. (Eucl. iii. 14.) 1k = EF, and EF: AB : the given ratio; .. 1K : AB in the given ratio. 20.) Through a given point, either without or within a given rele, to draw a straight line, the part of which intercepted by the rele, shall be equal to a given line, not greater than the diameter “the circle. 28 GEOMETRICAL PROBLEMS. [ Sect. | Let p be the given point without -the circle asc, whose centre is 0. E- In the circle place a straight line x ) AB equal to the given straight line ; | which bisect in E; and join OE. 5 1 With the centre o and radius o£ ae describe a circle; this will touch as | in £, since the angles at & are right angles (Eucl. iii. 3.); fra p draw pcp touching the circle in F. PCD is the line requial Join or. Then oF being equal to o£, cp will be equal AB (Eucl. iii. 14.), 2. e..to the given line. 1 cutting each other, may have a given ratio. Let c be the given point in the a diameter BA produced. Make Bc : cD in the given ratio; and from the 3EZ i points B and p, inwhich cp cuts @ x ee | the semicircle, draw EB, AD to the extremities of the diameter. cp is the line required. Since the angles EDA, EBA in the same segment are equ and the angle at c common to the two triangles acD, CEB, {f triangles are equiangular, whence BE: AD !: BC: CD, i. e. in the given ratio. (22.) From the circumference of a given circle to draw ti straight line given in position, a line which shall be equal at parallel to a given straight line. | Let aB be the given circle whose centre is 0, and DE the line given in position. From o draw oF par- allel and equal to the given line ; and with the centre F, and radius equal to os, the radius of the given et. 11.) GEOMETRICAL PROBLEMS. 29 | cele, describe a circle cutting DE In G: join FG, and draw 0a jrallel to it; jom AG; AG is the line required. Since FG = OB = on and is parallel to it, Ag is equal and jrallel to oF, and .°, equal and parallel to the given line. \(23.) The bases of two given circular segments being in the same ‘aight line; to determine a point in it such, that a line being awn through it making a given angle, the part intercepted ca the circumferences of the circles may be equal to a given 1. Let ap, cp, the bases of the gments be in the same line. arough o the centre of the circle 31, draw EoG making with as angle equal to the given angle, ed make o£ equal to the given line. From & draw &F, to the ccle CFD, equal to the ie oB; draw o1 parallel to EF; jn IF cutting AD in H; H is the ne required. For 01 being equal cane parallel to EF, OF is equal and irallel to 1F, .. 1F is equal to the given line; and 1F being jrallel to uG, the angle ruc is equal to EGB, ii e. to the given cole. If the distance of = from the centre of the circle crp be less ‘e the sum of the radii, there are two points in the circum- ‘ence CFD, and two corresponding points in aD, which will eswer the conditions. (24.) Iftwo chords of a given circle intersect each other, the gle of their inclination is equal to half the angle at the centre wich stands on an arc equal to the sum or difference of the arcs L ercepted between them, according as they meet within or without } circle. Let as, cp cut one other in the point 5; ad first within the circle anc; the angle a zlination is equal to half the facie at thecen- A t: standing on an arc equal to the sum ofca * D ed DB. Cc _ 30 GEOMETRICAL PROBLEMS. [ Sect. . - Through p draw pF parallel to Ba. Find o the centre of t circle ; jom co, ro. Then aB being parallel to rp, (i. 1.) 4 is equal to Bp; and the angle cEA is equal to cpF, 7. e. to hi the angle cor, which stands on the arc cr equal to ca andi together. 2. Next, let AB, cp intersect in E, without the circle. The same construction being made, i the angle cEA is equal to the angle oY cpF, t.e. to half cor, z.e. to half the B — angle standing on cF which is the dif- Ce 7 ference between ca and AF, or cA and BD. (25.) If from a point without two circles which do not meet ea other, two lines be drawn to their centres, which have the sa, ratio that ther radu have ; the angle contained by tangents drain from that point towards the same parts will be equal to the anj contained by lines drawn to the centres. From the point A let the lines A AB, AE be drawn to the centres of two circles, and let them have the same ratio that the radii Bo, DE, have; from A draw the tan- gents AC, AD; as also AF, AG; fe each of the angles cap, FAG will D i be equal to BAE. Since AB : BC !! AE! ED, and the lesa at c and D are right angles, . .. the triangles Af ADE are equiangular (Kucl. vi. 7.), and the angle cAB = DA to each of these add the angle BAD; and CAD = BAE. C In the same manner FAG may be shown to be | to BAD. (26.) To determine the Arithmetic, Geometric, and Harmo means between two given straight lines. Chat. | GEOMETRICAL PROBLEMS. 81 ‘Let as, Bc be the two given lines. Let D 2m be placed in the same straight line, and | Ac describe a semicircle apc. Through s Me Pheng ww BD at right angles to ac, join op, and & oO BC jon it let’ fall the perpendicular pr. Then ) being half of the sum of a8, Bc is the arithmetic mean; jd since (Kucel. vi. 8.) AB: BD :: BD: Be, .. BD is the geo- rae mean. And DE is the harmonic mean, for (Eucl. vi. 8.) (0 =) AO: DB: DB: DE, ze. it is a third proportional to fh arithmetic and geometric means, and .-. is the harmonic yan. 27.) If on each side of any point in a circle any number of Wal arcs be taken, and the extremities of each pair joined; the i of the chords so drawn will be equal to the last chord produced. qneet a line drawn from the given point through the extremity of first are. twee AB, BC, CD, &c., \, EF, FG, &c. be equal is and let their extremi- i BE, CF, DG be joined ; im ; the part of the line between the points of contact, is a mean i portional between the diameters of the circles. uet AEB, CED be two circles touching A : AC : CB. Take o the centre of the circle, join 0; then (Kucl. ii. 18.) the angle cpo is a right angle, and .°. duel. vi. 8.) CO:0D:: 0D: 08, or CO: 0A :: 0A: OF, .. div. and comp. AC: CB! AE: EB. Cor. The converse may easily be proved. | © eteoneer 48.) If from the extremity of the diameter of a given semi- ele a straight line be drawn in it, equal to the radius, and Strom ie centre a perpendicular let fall upon it and produced to the ircumference ; it will be a mean proportional between the lines vawn from the point of intersection with the circumference to the étremities of the diameter. From B the extremity of the diameter an let a C he Bc be drawn, equal to the radius Bo; and uh « it let fall a perpendicular o p meeting the cir- cmference in D; join DB, DA; DO is a mean Joportional between pa and ps. Jom pc. Then the angles BaD, BCD on the same base are cual. Also since op bisects BC, it bisects the arc BDC, .°*. tio the straight line Bp=pc and the angle pDBC=DcB, but OA=OAD, .. the triangles ODA, DBC are similar, .. AD: 1) :: (BC =) DO: ps. (44) If from the extremity of the diameter of a circle, two les be drawn, one of which cuts a perpendicular to the diameter, » ad the other is drawn to the point where the perpendicular meets ? circumference ; the latter of these lines is a mean proportional hween the cutting line, and that part of it which is intercepted bween the perpendicular and the extremity of the diameter. AQ GEOMETRICAL PROBLEMS. Let cx be at right angles to the diameter AB of the circle ABco, and from a let ap, ac be drawn, of which ap cuts CE in F, then will D ADSM GAO ALR B For since the circumference Ax is equal to the circumference Ac, (Eucl. iii. 27.) the angle ECA is | equal to the angle apc, and the angle at a is common to t. two triangles Apc, ACF, .*. the triangles are similar, and AD) AG IT¥ACG (AF. But if the point of intersection f be without the circle, drs du parallel to ca, then, as before, the angle Huda is equal | acd, and the angle at 4 common to the triangles and, acd, AA ULE AC.ce WA ELS pA nih C PEA Je | (45.) In the diameter of a circle produced, to determine a poi. from which a tangent drawn to the circumference shall be equal) the diameter. | From a the extremity of the diameter AB, draw AD at right angles and equal to aB. Find the centre 0, joining op cutting the circle in c; and through c draw cz at right angles to oD meeting BA produced in £. Then because the angle oAD, is equal to oce, each being a right angle, and the angle at o is common} the two triangles OAD, OCE, and OA=oCc, .. AD=cE. fF AD was made equal to AB, .. CE=AB, and E is the point quired. (46.) To determine a point in the perpendicular at the extrem of the diameter of a semicircle, from which if a line be drawi the other extremity of the diameter, the part without the cn may be equal to a given straight line. From s the extremity of the diameter of the semicircle aps, let a perpendicular Bc be drawn; in which take BE equal to the given line; and on it as a diameter describe a circle; through the centre of which draw ct. 11. | GEOMETRICAL PROBLEMS. Al @F, and with a as centre and radius ar describe a circle cut- ag Bceinc. Join ac; cp is equal to the given line. Join BD. Then sp being perpendicular to ac, (Eucl. vi. 8. Cor.) Ac: AB :: AB: AD, and (Kucl. i. 386.) AB: AF:: AG: AB, | *. €@ @quo, AC; AF 3: AG: AD, ‘hence AG=AD, and ., DC=GF=BE. (47.) Through a given point without a given circle, to draw (straight line to cut the circle, so that the two perpendiculars gawn from the points of intersection to that diameter which passes irough the given point, may together be equal to a given line, not eater than the diameter of the circle. e Let p be the given point without the «cle ABC, whose centre is 0; AB the (ameter which passes through p. On 1) describe a semicircle. From p draw ip at right angles to pB and equal to Jif the given line; through p draw pe parallel to ps meeting te semicircle in ©; join px, and produce it toc; pc is the he required. For, draw FG, EH, CI perpendiculars to as. Join o£; then : angle PEO is a right angle, and .°, (Kucl. iii. 3.) er = EC; 1ence FG and C1 together are equal to 2EH=2pp=the given : (48.) If from each extremity of any number of equal adjacent cs in the circumference of a circle, lines be drawn through two (ven points in the opposite circumference, and produced till they iret; the angles formed by these lines will be equal. Let aB, Bc, be equal arcs, and F, = two jints in the opposite circumference, trough which let the lines ari, BEI, 37H, CEH be drawn, so as to meet; the é gles at 1 and u will be equal. From & draw Ek, EL, respectively parallel to ra, FB. Since 42 GEOMETRICAL PROBLEMS. [ Sect. 1 _ EK is parallel to ra, the angle KEB is equal to the angle at; for the same reason the angle LEC is equal to the angle at} But since the arcs AB, BC, are equal, and AK, BL bemg eac equal to ©F (ii. 1.) are also equal to one another, .*, KB, LO, a also equal, and (Eucl. ii. 27.) the angles KEB, LEC, are equi . also the angles at 1 and w are equal. The same may | proved whatever be the number of equal arcs AB, BC. | (49.) To determine a point in the circumference of a cire from which lines drawn to two other given points, shall have given ratio. ! Let a, B be the two given points; join AB, and divide it in p so that aD : pB may be in the given - ratio; bisect the arc ACB in C; join CD, and pro- duce it to E; £ is the point required. Join AE, EB. Since Ac = cB, the angle AEC is equal to the angle cEB, .*. AB is cut by the line Ep bisecti| the angle AEB, and consequently (Kucl. vi. 3.) AE: EB‘: AD: DB, i.e. in the given ratio. (50.) If any point be taken in the diameter of a circle, which not the centre; of all the chords which can be drawn through t point, that is the least which is at right angles to the diameter, — In aB the diameter of the circle ADB, let any point c be taken which is not the centre, and let rN DE, FG be any chords drawn through it, of which 4 DE is perpendicular to AB; DE is less than FG. Take o the centre and draw ou perpendicular to rg. Now in the triangle ocu, the angle at is a rig angle and .*. greater than the angle oc, .*. Co is greater th) ou, and consequently (Eucl. ii. 15.) D£ is less than FG. | E (51.) If from any point without a circle lines be drawn touchit it ; the angle contained by the tangents, is double the angle ce tained by the line joining the points of contact and the diame drawn through one of them. ect. 11. | GEOMETRICAL PROBLEMS. 43 From the point & without the circle asc let A, ECD be drawn touching the circle in a ad oc, and let Ep meet the diameter a3, drawn om A, in the point p. Join ac; the angle eC is double of cas. Through c draw the diameter cor; then the agle rcp is a right angle, and .*. equal to AD, and EDA is common to the triangles DA, COD, .. the angle cop is equal to arp. But cop is ouble of CAD, .°. AEC is double of cap. (52.) If from the extremities of the diameter of a circle tangents | drawn, and produced to intersect a tangent to any point of the reumference ; the straight lines joining the points of intersection vd the centre of the circle form a right angle. From a and B the extremities of the diameter c is .B let tangents AD, BE be drawn, meeting a tan- ” ‘nt to any other point c of the cir cumterence, 1 In A E. jand ©; and let o be the centre; join Do, £0; ie Sele DOE isa right angle. Join co. Then since ce = EB, CO = OB, and the angles at and 8, being right angles, are equal, .. the angle cEO=oOEB, ‘dcxBs is bisected by Eo. In the same manner it may be own that the angle apc is bisected by po. And since the sgles CEB, CDA are equal to two ae angles, .“. CDO and (0 are equal to one right angle, and .*, (Eucl. i. 32.) pox is eight angle. * i @ If from the extremities of the diameter of a circle tangents drawn ; any other tangent to the circle, terminated by them, is | divided at the point of contact, that the radius of the circle is a ian proportional between its segments. ‘Let ap, BE be two lines touching the circle anc, (see the it Fig.) at a and B the extremities of its diameter, and meet- is DCE any other tangent to the circle; take o the centre, and faego-; then will pc : co :: co: CE. AA GEOMETRICAL PROBLEMS. [ Sect. n Join po, £0; then as in the last proposition, it may b shown that pox is a right angle: and since from the righ angle oc is drawn perpendicular to the base, .*. (Hucl. vi. 8 it is a mean proportional between the segments a the base, or DO: C07: CO: CE. (54.) Two circles being given in magnitude and position; 1 find a point in the circumference of one of them, to which of a tan gent be drawn cutting the circumference of the other, the part ¢ it intercepted between the two circumferences may be equal to given line. Let o and c be the centres of the two given circles. To any point a in the cir- cumference of one of them let a tangent AP be drawn, and make aB equal to the given line. With the centre c and distance cB describe a circle pBD cutting the other in the point p, and from p draw pe touching the former give circle; E will be the point required. _ Join ca, cB, cp, cr. Since cA=cr and cB=cp, and tl angles at A and §£ are right angles, .. DE is equal to BA, : to the given line. If the circle ppp neither cuts nor touches DD, it is evide the problem will be impossible. (55.) To draw a straight line cutting two concentric circles; that the part of it which is intercepted by the circumference of t greater may be double the part intercepted by the crewmen of the less. Let o be the centre of the two circles. Draw any radius 0 of the lesser circle and produce it to B, making AB=Ao. On AB describe a semicircle AcB cutting the greater circumfer- ence in ©; join Ac, and produce it to B; CE is the line required. F Join cB; and let fall the perpendicular op. Then the ang ADO jee a right angle is equal to the angle acx, and tl oct. 11. | GEOMETRICAL PROBLEMS. 45 irtically opposite angles at a are equal, and the side oA=An, | AC=AD, and pc=2ap; but pc is half of ec and ap half (ar, .. 6c is double of ar. Cor. The same construction will apply whatever be the rela- im required between the two chords. Take op: oA in the iquired ratio, and proceed as in the proposition. 56.) If two circles intersect each other, the centré of the one ling in the circumference of the other, and any line be drawn am that centre ; the parts of it, which are cut off by the common cord and the two circumferences, will be in continued proportion. | i | ‘From any point a in the circumference of iecircle ABG, as a centre, and with any radius, iF a circle BDc be described, cutting the for- rin Band c. Join Bc; and from a draw iy line AFE; AF!AD:!AD: AE. From A draw the diameter AG, it will cut Bo | éright anglesini. Joi qn, ac. Theright angle ArF being ual to the right angle ArG, and the angle at a common, the tangles AIF, AEG are similar, | : $jAFeicAT 7RAG 3A Ee But (Eucl. vi. 8. Cor.) at: ac:: ac: aa, | “. €@ @quo, AF; AC :! AC: AE, Or AF: AD!:AD! AE. 167) If a semicircle be described on the side of a quadrant, ud from any point in the quadrantal are a radius be drawn; the rt of this radius intercepted between the quadrant and semi- p cle, is equal to the perpendicular let fall from the same point on ur common tangent. On az the side of a quadrant let the semicircle 1B be described, and from any point c draw the lus cB, and cp perpendicular to ap a tangent ae EC = CD. Join Ar, Ac; then the angle AEB being in a nicircle, its adjacent angle anc is a right angle, and .*, equal 46 GEOMETRICAL PROBLEMS. [ Sect, 1 to apc; and BcA=BAc=Acp the alternate angle; .-. t] two triangles AEC, ACD have two angles in each equal, and o1| side AC common, .°. EC=CD. . Cor. Any chord of the semicircle drawn from the centre| the quadrant, is equal to the perpendicular drawn to the oth side from the point in which the chord produced meets ¢) quadrantal arc. Produce pc to F; then cE being equal to cp, the remaind BE is equal to the remainder cr. TP shar pv Be 7 (58.) If a semicircle be described on the side of a quadrant, a a line be drawn from the centre of the quadrant to a common ta gent ; this line, the parts of it cut off by the circumferences of t quadrant and of the semicircle, and the segment of the diame of the semicircle made by a perpendicular from the point whe the line meets its circumference, are in continued proportion. On the radius AB of the quadrant acB OR, | let the semicircle AEB be described, and at 3 A A draw the tangent ap. From B draw any line BECD meeting the tangent in D, and | the circumferences in C, £; from & let fall the perpendicular EF; then BD, BC, BE, BF are in continued proportion. Since FE is perpendicular to BA, it is parallel to ap, “BEBE 4 BA) BOARD, But (Eucl. vi. 8.) BF: BE :: BE: (BA=) BOC, “. (EKucl. v. 15.) also BE; BC :: BC: BD, and BF ,: BB 0B Hw C. oye Case iehs (59.) If the chord of a quadrant be made the diameter of! semicircle, and from its extremities two straight lines be drawn) any point in the circumference of the semicircle ; the segment the greater line intercepted between the two circumferences shi be equal to the less of the two lines. | Let o be the centre of the quadrant ADB; join AB, and } it let a semicircle AcB be described; from any point c in whi! 4 : | | ict. 11.] GEOMETRICAL PROBLEMS. Ay |; lines cA, CB be drawn to 4 and B, of which cB nthe preater; cD = CA. Join AD, and complete the circle aBE; take @y point E, and join HA, EB. Since ADBE isa ¢adrilateral figure inscribed in a circle, the an- oS AEB, ADB are equal to two right angles, and . equal to ADB, ADC; whence AEB = apc; but AEB is half ¢A0B which is aright angle, .. apc is half a right angle, and ‘1D being aright angle (Eucl. ii. 31.), cap is hati a right angle, ad .*, equal to cDA, consequently ca = cp. (60.) If two circles cut each other so that the circumference of 9? passes through the centre of the other, and from either point of nersection a straight line be drawn cutting both circumferences ; f: part intercepted between the two circumferences will be equal ‘the chord drawn from the other point of intersection to the point were it meets the inner circumference. Through o the centre of the circle ase, let ‘2 circle AoB be described, cutting ABc in A ule. If any line arp be drawn from a, and B jomed; pe will be equal to EB. Draw the diameter aoc; ; Join Bc, BD. Then ‘ice the angle aos is aca to AEB, .*. the angle coB is equal (DEB. Also the angles o oP: EDB, eee in the same segment, i equal to one another, .*. the triangles OCB, EDB are equi- usular, and .*, since on = 00, the anes OCB is equal to the isle OBC, whence EDB = EBD, and .. ED = EB. hy .) If from each extremity of the diameter of a circle lines be liwn to any two points in the circumference ; the sums of the lines clrawn to each point will have to one another the same ratio that lines have, which join those points and the opposite extremity jt diameter perpendicular to the former. ‘rom A and c the extremities of ac the diameter of the ile ABC, let lines Az, EC, AF, Fc be drawn to any points E 8 48 GEOMETRICAL PROBLEMS. and ¥ in the circumference, and draw the diameter Bp perpendicular to Ac; join ED, Fp; then AE+EC ! AF+FC !: ED: FD. Join AB; and with the centre B and dis- tance BA describe a circle AGC; produce AB, AE, AF to the circumference. Join GH, HI, BE, EF, GI, BF. Then since AG and Bp are diameters of the circles, the angles ANG, AIG are equal to DEB, DFB; but BAKE, BAF are equ to BDE, EDF, and the angle 1G being = HAG = BDE = BF. .. the angle 1A = EFD, and the triangles GAN, HAI @ similar to BDE, EDF, and SUA OA Lee c ELI}. But (ii. 60.) pH = EC, and FI = FC, - AEB+EC: AF+FC !: ED: FD. (62.) If from any two points in the circumference of a cir there be drawn two straight lines to a point in a tangent to th circle; they will make the greatest angle when drawn to the poi of contact. ' Let a and B be the two points, and cp the tan- gent at c; join Ac, cB; the angle acB 1s greater than any other angle apB formed by lines drawn to any other point D. Join pre. Then the angles acB, AEB in the same segment are equal; but apB is less than the exter! angle AEB, and .*, is less than ACB. Cor. If two circles touch each other in ¢, it might be shot in a similar manner, that the angle formed by two straight I! drawn from A and B toc the point of contact will be greater tl the angle formed by lines drawn from the same points to {J point in the exterior circle. } (63.) From a given point within a given circle to draw a strai ect. 11. | GEOMETRICAL PROBLEMS. 49 ne which shall make with the circumference an angle less than 1e angle made by any other line drawn Srom that point. Let p be the given point within the circle B ae x Find o the centre, join op, and produce Sx to the circumference. From p draw pp right angles to oa; it is the line re- cured. re _Join 0B, and on it as a diameter describe icircle op B, which will touch the circle anc in B. Then opp ithe greatest angle that can be included between lines drawn ism o and P to the circumference ABC (i025 Cory oe the igle contained by px and the circumference ap will be the ast. ((64.) To determine a point in the are of a quadrant, from which | lines be drawn to the centre and the point of bisection of the dius, they shall contain the greatest possible angle. fy ‘Let Bc be the arch of a quadrant whose centre »A, and let the radius ac be bisected in p. On ‘9 describe an equilateral triangle a px, and pro- ‘ce AE to F; Fis the point required. Join rp. Then ar = AC, and Ap = AR, but L is half of ac, and .*. Ax is half of AF, and .°. ‘nal to EF; and EA, ED, EF are equal; whence a circle de- ibed from the centre © at the distance of any one of them | pass through the extremities of the other two, and touch arc BC in F, because their centres are in the same straight ie; and aFp (ii. 62. Cor.) is greater than any other angle ‘med by lines drawn from any point in Bc to « and p. [ 1 65.) If the radius of a circle be a mean proportional to two dis- (ces from the centre in the same straight line; the lines drawn im their extremities to any point in the circumference will have / same ratio that the distances of these points from the circum- ence have. E 50 GEOMETRICAL PROBLEMS. [ Sect. 11 Let a and B be the two points, such that AO: CO :: CO: BO, and from A and B let lines AD, BD be drawn to any point pin B cl AO the circumference; these have always the same ratio, viZ. AD! BD :: AC: BC. Join op. Then since op = 0C, 0A: 0D:: OD: GB, 4. the sides about the common ACER AOD of the rian AO] BOD are proportional, and .*. the triangles are similar, cor sequently ) DA :DB::0A: (OD=) OC. But since AO : CO ?: CO: BO, div. AO: AC 3: CO; CB, alt. AO: CO 3: AC: CB, -, (Kucl. v. 15.) DA: DB 3: AC; CB. (66.) Two circles being given in position and magnitude 5 draw a straight line cutting them so that the chords in each cut may be equal to a given line, not greater than the diameter of t smaller circle. | Let asc, EFG be the given circles whose cen- tres are o and mM. In each place a line aB, EF, equal to the given line; and from the centres draw the perpendiculars OI, MK; | and with these distances, and centres o and 1, describe cire} which will touch AB, EF in1 and K; draw cpGH (ii. 7.) whil shall touch these circles in L and n; each of the chords cp al Gu will be equal to the given line. | Join OL, MN; these lines are perpendicular to cp and & and being respectively equal to o1 and MK, CD = AB (Euel. 14.), and @H = EF; but aB and EF are each equal to the giv line, .. cD and «@u are also each equal to the given line. Cor. If the intercepted parts are required to have a gi ratio, take a B and EF in that ratio, and make the same consti": tion as in the proposition. & | | Nect. 11. } GEOMETRICAL PROBLEMS. 51 _ (67.) To determine a point in the arc of a quadrant, through hich if a tangent be drawn meeting the sides of the quadrant pro- uced, the intercepted parts may have a given ratio. Let 0, oB be the sides of a quadrant pro- uced ; and take m and n two right lines which ve in the given ratio, and let oc be a mean pro- ortional between the radius of the quadrant ad m, and op a mean proportional between the idius and n. Join cp, and draw the radius E cutting it at right angles; m is the point required. Through & draw the tangent ars, which being perpendicu- r to ox (Eucl. iii. 18.), will be parallel to cp, | Fem OF CO UD: ad since oc and op are mean proportionals between m and the idius, and wn and the radius respectively, M : N in the duplicate ratio of oc : op, | 2. €. in the duplicate ratio of Ao : os. Jat (Kucel. vi. 8. Cor.) | AE : EB in the duplicate ratio of ao: OB, . . . ° | “, AE > EB :: M:N, 2.é. in the given ratio. | (68.) Tf a tangent be drawn to a circle at the extremity of a rd which cuts the diameter at right angles, and from any point (tt a perpendicular be let fall; the segment of the diameter inter- ited between that perpendicular and chord is to the intercepted it of the tangent, as the chord is to the diameter. Let the chord cp be perpendicular to the dia- re = iter AB, and let cx touch the circle at c; from t 7 point = in which let rr be drawn perpen- A B lular to AB; | FG > CE‘: 0D} AB. Draw the diameter cu; join up, and draw c1 perpendicular ‘er. Since Ec touches the circle, the angle ECH (Eucl. iii. 18.) ‘t right angle, and .*, equal to 1¢p; whence, taking away from | E2 : ' 52 GEOMETRICAL PROBLEMS. [ Sect. 1 each 1cH, the angle C1 = HCD, and BIC, HDC are right angle, .. the triangles ECI, HDC are equiangular. whence IC : CE 3: DC: CH, or GF: CE!: CD: AB. (69.) If a straight line be placed in a. circle, and from ‘ extremities perpendiculars be let fall upon any diameter ; the perpendiculars together will have to the part of the diameter inte cepted between them, the same ratio that a line placed in the cir perpendicular to the former line, has to the former line itself. Let the line cp be placed in the circle a BC, and from its extremities let cn, pF be drawn perpendicular to a diameter AB. From p let A pe be drawn perpendicular to pc; then will CE+DF: EF:: GD: DC. Join c@, which is therefore a diameter of the | circle; and produce cx to 1: join D1, and draw DH perpen: cular to ck. Since cr is perpendicular to aB, CE = EI, Ii HE =DF,.. H1=ceE+DF. Now (Eucl. ii. 21.) the angle} G is equal to the angle at 1, and cpg, pur are right angles,, the triangles cGD, H1D are equiangular, | and: Hl: HD?: DG. DG, OMOH +4DF iE 7G5DGi: DC, (70.) In a circle to place a straight line of given length, so Ui perpendiculars drawn to it from two given points in the circum ence may have a given ratio. Let a and B be the given points in the circumference of the circle whose centre is 0. Join BA, and pro- duce it; and take ac: cB in the given ratio. In the circle place a straight line equal to the given straight line, . and from the centre o let fall a perpendicular upon it. Wit as centre, and distance equal to this perpendicular, descril bet. 11. ] GEOMETRICAL PROBLEMS. 53 ele DG, and from c draw crepF a tangent to it; then HF is ie line required. For (Eucl. iii. 14.) it is equal to the given straight line. And | from A and B, AE, BI be drawn perpendicular to cr, they ‘e parallel to each other, and the triangles CAE, CB1 are ‘nilar, °. AE: BI!: CA : CB,72. e. in the given ratio. os et, ‘(71.) If from any point in the arc of a segment of a circle a he be drawn perpendicular to the base; and from the greater BF, the straight line Bc = BF; and Dx being -ual to Dc, and pB common, and at right angles to EC, .. BE -BC = BF, and the angle Bre is equal to the angle BEF. lw since A FBC is a quadrilateral figure inscribed in a circle, 2 angles AFB, ACB are equal to two right angles, and .*. equal AEB, CEB, of which ACB = CEB, .. AFB = AEB; but BFE BER, consequently AFE = AEF; whence AF = AE. | | (72.) If from the point of bisection of any arc of a circle a per- jndicular be drawn to the diameter, which passes through one »! ‘remity ; it will bisect the segment of the chord cut off by the (2 joining the point of bisection of the arc and the other extremity ) the diameter. Let ac be the arc bisected inp. Jomac; € D 11 from p draw De perpendicular to the diame- (= | AB, and meeting AcinG; jon BD; AG = B EA lates AC is bisected in p, the angle CAD Is sae) to the izle DBA, 2. e. to the angle EDA (Eucl. vi. 8.), .. the right- - - .| | A .. 5A GEOMETRICAL PROBLEMS. [Sect 1 Bae triangle ADF is equiangular to the two triangles BEI DEA .. the angle GFD = GpF, and consequently GD = GF also GAD = GDA, .. AG = GD, whence aG = GF. i (73.) In a given circle to draw a chord parallel to a straight li given in position ; so that the chord and perpendicular drawn to from the centre may together be equal to a given line. Let o be the centre of the circle, 0 the straight line given in position; draw oB perpendicular to it, and equal to the given line. Take oA equal to the half of o8; and join aB, cutting the circle in ©; through c draw cp parallel to 0A; cp is the chord required. 3 Because 0 is half of os, and 04, EC are parallel, .. (Huc vi. 2.) Ec is half of EB, and pc = EB; therefore pc and 0 together are equal to BE and of together, i. e. to BO, or to t given line. (74.) Through a given point within a given circle, to draw straight line such that the parts of ut intercepted between that po and the circumference may have a given ratio. Let p be the given point within the circle aBD. Through pe draw the diameter ars, and take ap : Pc in the given ratio. With p as centre, and radius equal to a mean proportional between BP and po, describe a circle cutting ADB in D; join pp, and produce it to E; DE is the chord required. Since BP: PD !! PD: PG, and (Kucl. ii. 35.) BP : PD 3: PE: PA, a ER PASS PDS ay and ali. PE: PD :; AP: PC, 2. in given ratio. Cor. Since one circle cuts another in two points, there be two chords which answer the conditions. If c coincides W A, the ratio is one of equality, and pe will be perpendicy) to AB. : oct. 11.| GEOMETRICAL PROBLEMS. 55 } " | (75.) From two given points in the circumference of a given rele, to draw two lines to a point in the circumference, which call cut a line given in position, so that the part of it intercepted ( them may be equal to a given line. ' Let A, B be the given points in the circum- irence of the circle ABC; DE the line given j position. From s draw BF parallel to zB, and equal to the given line. Join aF; ad on it describe a segment of a circle AGF ntaining an angle equal to the angle in the ‘gment AcB; and let it cut pe ing. Join ag, and produce Se ito c; and join Bc cuttmg DE in. Ac, BCare the lines iquired. Jom ar. Since the angle AGrF=a cB, GF is parallel to cz; lit FB is parallel to Gu, whence Feu is a parallelogram, and (I = FB. | (76.) If a chord and diameter of a circle intersect each other at y angle, and a perpendicular to the chord be drawn Jrom either etremity of it, meeting the circumference and diameter produced ; 2 whole perpendicular has to the part of it without the circle, 12 same ratio that the greater segment of the chord has to the less. Let the diameter aB and D (ord DE intersect each other ¢ ©; and from p draw DG ¢ [rpendicular to pr, meeting /3 produced in «; H | then GD : GF::DC::CE. Through F draw Fr parallel to pz, and meeting the diameter i I. Join rx, cutting the diameter in 0. Since the angle IE is a right angle, rz is a diameter and o is the centre. And sice the angle 1 Fo is equal to the alternate angle orc, and the agles at o are equal, and ro=o8, .*. the triangles OF I, OFC i equal, and cz = rr. And since F1 is parallel to pe, (Kucl. vi. 2.) GD: GF:: DC: (FI=) CE. In a similar manner it may be shown that JH: gE:: DC: CE. 56 GEOMETRICAL PROBLEMS. [ Sect. 1) (77.) If from the extremities of any chord of a circle, perpend culars to it be drawn and produced to cut a diameter ; and fro the points of intersection with the diameter lines be drawn to’ point in the chord, so as to make equal angles with a ; these lin together will be equal to the diameter of the circle. Let aw be any chord of the circle anc; draw G An and BF perpendicular to it, meeting the dia- OW. meter cp in Eand F; from which let the lines EX, © IN ru be drawn making equal angles with AB; EH and uF together are equal to cD. | Take o the centre, and join Bo, and produce it; it will me AE produced in G. Produce EH, FB to meet ini. Then sin} the angle rHB=AHE=FUHB, and HB is perpendicular to } the triangles ruB, BI are equal, and FH=nH1I. And since i is parallel to rB, the angle E@o=oBF, and the vertical ang] ato are equal, and Go=oB, .". EG=FB=BI1; whence EI=6) and .*. EH, HF together are equal to HI, 7.¢. to GB or CD t} diameter of the circle. | (78.) If from a point without a circle two straight lines } drawn, one of which touches and the other cuts the circle; alt drawn from the same point in any direction, equal to the tange, will be parallel to the chord of the arc intercepted by two lis drawn from its other extremity to the former intersections of | circle. From the point a let as, ap be drawn, of D which aB touches the circle BCD, and AD cuts it; anddraw AE=AB, in any direction; Jom CE, g DE, cutting the circle in F and G, the chord re will be parallel to ax. I™Se Because (Kucl. ii. 386.) DA; AB!. AB: AC, \ and AE=AB, .°, DA: AE;:AE: AC, i. e. the sides about the angle a of the triangles ADE, ACEX proportional, .*, (Eucl. vi. 6.) the triangles are equiangular, id the angle Arc is equal to the angle apz. But since cp@ikk \ a quadrilateral figure in the circle, the angles cpG, cre * Let. 11. | GEOMETRICAL PROBLEMS. 57 | ual to two right angles, i.e. to EFG, CFG, .. CDG=EFG, lence AEF=EFG, and FG is therefore parallel to AE. '(79.) If from a point without a circle, two straight lines be cawn touching it, and from one point of contact a perpendicular i drawn to that diameter which passes through the other; this prpendicular will be bisected by the line joining the point without 2 circle and the other extremity of the diameter. k ‘Let pa, pB be drawn from a point p without ¢ fe circle ABC, touching it in a and B; and from “ jlet Br be drawn perpendicular to Ac the dia- iver ter passing through a; joincD; BBE is bisected VAR Cp in the point Fr. AE Cc ‘For produce ap and cB to G; join AB. Then since pa= 3, the angle DaB is equal to the angle ppa. Now the angle 3 G, being a right angle, is equal to BAG, BGA, of which aBD “BAG, .. DBG = DGB, and DG = DB=DA; and since AG is allel to EB, | BE GDIL-OF UD EE AD, and GD = DA, ... BF = FE. } ee ee ee \(80.) Tf any chord in a circle be bisected by another, and pro- iced to meet the tangents drawn from the extremities of the Lecting line ; the parts intercepted between the tangents and the “-cumferences are equal. ‘Let aB be bisected in = by cp; and to c id D let tangents be drawn, meeting AB pro- oced in F and @; AF is equal to BG. Find o the centre; join oc, op, 05, OF, ( Since o£ is drawn from the centre to the Lint of bisection of aB (ucl. 11. 3.) the angle omF is a right igle; and the angle ocF is a right angle (Eucl. iii. 18.) ; .. a scle may be described about ozrc. Also since op@ and (1G are right angles a circle may be described about onpG; id the angle poG is equal to the angle eG in the same seg- rmt; but pEG is equal to FEC, i.e. to FOC, .. DOG=FOC; 08 GEOMETRICAL PROBLEMS. [ Sect, | and opG, ocF are equal, being right angles; and oc=oD, | or=oG, and consequently Fe=EG. But AE=EB,..FA=B) (81.) If one chord in a circle bisect another, and tangei; drawn from the extremities of each be produced to meet ; the li joining their points of intersection will be parallel to the biseci chord. Let aB be bisected by the line cDin gE, . F and let the tangents ar, BF meet each © other in F, and pG@, caing. Join GF; as GF is parallel to aB. K B Y Join AO, CO, GO, FO; then Go bisects EC. (86.) If from a point without a given circle, any two lines be ( awn cutting the circle; to determine a point in the circumfer- éce, such that the sum of the Serrenneugh from it upon these les may be equal to a given line. From the point without the circle ppc let 43, Ac be drawn cutting the circle; draw AF |vpendicular to as, and equal to the given lie; re parallel to as, and meeting ac pro- cced in @; from @ draw Gu bisecting the é gle acr, and (if the problem be possible) meting the circle in n; H is the point required. Through H draw KL pecan cie to AB, and HI perpendi- clar to ac; then the angle k eu being equal to a1, and the egle at K to the angle at 1, and the side HG, opposite to one ¢ the equal angles in each common, H K=HI; whence HI and IF together are equal to HK and HL together, 7. e. to AF, 2. @. t the given line. If eu cuts the circle, there are two points which answer the caditions. (87.) If two circles cut each other, and any two points be taken i the circumference of one of them, through which lines are drawn 7 G ® 62 GEOMETRICAL PROBLEMS. [ Sect. . ‘ [ from the points of intersection and produced to the circumferen: of the other ; the straight lines joining the extremities of the: which are drawn through the same point, are equal. | | Let the two circles ACB, AEB cut each other in A and B, and in ACB let any two points c and D be taken, through which draw ACG, BCE, ADH, BDF; and jom EG, FH; EG = FH. For the angles c AD, CBD being on the same circumference CD, are equal to one | another, .*. the circumference EF is equal to the circumferen Gu. Add to each ra, and the circumference EFG, is equal } FGH, .*. (Eucl. ii. 29.), the straight lne EG=Fu. | a i | (88.) If two circles cut each other; the greatest line that ¢ be drawn through the point of intersection is that which is paral to the line joining their centres. - | Let the two circles ABE, AFD cut each : other in A. Join o, c their centres, and 8 through A let BAD be drawn parallel to OC; BAD is greater than any other line EAF which can be drawn through a. Draw 0G, CH, perpendicular to BD, and 01, cK perpendicu torF. Then aa being half of as, and au of ap, GHish of Bp. For the same reason 1x is half of er. Draw cu par lel to EF, and therefore at right angles to o1, and equal to1 Then since the angle cio is aright angle, it is greater th COL, .*. the side co is greater than cL, and eu than Ik, co sequently Bp is greater than rr. In the same way Bp may | shown to be greater than any other line drawn through a. | | (89.) Having given the radii of two circles which cut each oth and the distance of their centres ; to draw a straight line of giti length through their point of intersection, so as to terminate their circumferences. | 2ct. 1.] GEOMETRICAL PROBLEMS. 63 ‘Let the two circles Arp, BGD cut each net ther in D; on oc, the line joining their (Sr, intres o and co, describe a semicircle A 0 c B (20; and in it, from c place ce equal to half the given line, id through p draw Fpe@ parallel to it; rq will be the line iquired. ‘Through & draw orn, which (Eucl. iii. 31.) will be perpen- (cular to FG; and draw c1 parallel to on, and .*, perpendicular (vq; then (Eucl. iii. 3.) rp and pe are bisected in w and 1, id .. F@ is double of u1; but HEC being a parallelogram, If=EC; .*, FG is double of Ec, and consequently equal to the even line. | = | (90.) If two circles cut each other ; to draw from one of the jints of intersection a straight line meeting the circles, so that the rt of it intercepted between the circumferences may be equal to nywwen line. \ } Let the two circles ABC, ADB cut each other (Aands. Join aB, and draw Bc touching the cle aBp. Join ac; and take a Fa fourth pro- rrtional to Bc, BA and the given line; join BF, id produce it to &; BFE will be the line re- ired. Since the angle ars together with the angle in the segment ‘9B or (Kucl. ii. 32.) its equal ABe, are equal to two right igles, i. e. to the angles AFB, AFE, .. ABC=AFE; and ACB -A EF, being in the same segment, .*, the triangles AcB, AEF i: equiangular, iy | and AB: BC !: AF: FE, | but AB: BC :: AF: the given line, ‘ence FE 1s equal to the given line. 91.) If two circles cut each other ; to draw from the point of rersection two lines, the parts of which intercepted between the 77 st | cumferences may have a given ratio. 64 GEOMETRICAL PROBLEMS. Let the two circles ABC, ABD cut each other in A and B; in the circle ABD place BE, BF, which have to each other the given ratio ; join Ak, AF, and produce AE to G; EG will have to uF the given ratio. Draw the diameters Ac, AD; join GB, BH, BC, BD; then ADBE being a quadrilateral figure inscribed in a circle, the angles AEB, ADB are equal to two right angles, and .°. equal to BEA, BEG, .. BEG = BDA = BFA. And since AGE is a quadrilateral figure inscribed in a circle, AHB, AGB % equal to two right angles, 7.e. to AHB, BHF, .. AGB = BH hence the triangles GBE, FBH are equiangular, \ '.GE: HF !: BE: BF, #. é. in the given ratio. ; (92.) If a semicircle be described on the common chord of | intersecting circles, and a line be drawn from one extremity of chord cutting the two circles ; the part intercepted between the | shall be divided by the semicircle into segments proportionai perpendiculars drawn in those circles from the other extremity the chord. | Let the two circles AcB, ADB cut each other in A and B; and on AB, the line joining the points of intersection, as a dia- meter, describe the semicircle AEB, and draw any line AFrEG cutting the circum- ferences in F, E, G; and from B draw Bo, BD perpendiculars to AB; then will EF : EG :: BD: BC. Draw the diameters ac, AD; and join FB, EB,GB. Then AFBD being a quad- rilateral figure inscribed in a circle, the angles AFB, ADB are equal to two right ls angles, 7. €. to AFB, BFE, ... ADB = BFE, and the angle # | | pet. 11. ] GEOMETRICAL PROBLEMS. 65 / a semicircle is equal to ABD, whence the triangles FEB, ABD .e equiangular, and ;,"). FE) EBS) DBs BA. Again, Besoure the angle AacB = AacB, and BEG is a right gle, and ., equal to ABC, the triangles EBG, ABC are equi- ‘gular, and | BE; EG:: AB: BO, but FE: EB:: DB: BA, °. €# @qu. FE; EG 3: DB: BC. [ (93.) Two circles being given, the circumference of one of which y sses through the centre of the other ; to draw a chord from that catre such, that a perpendicular let fall upon it from a given jint, may bisect that part of it which is intercepted between. the “cumferences, Let o and c be the centres of the two oven circles, the circumference of the former jssing through c; and let p be the given int. Join co, ati produce it both ways [A and. Join Bp, and produce it to x, king DE=pB. Draw uF touching the cle AF in F; join cr, and produce it to 6 and on it let fall the perpendicular pu ; /m CG is the chord required, and ra is bisected in n. Draw EI parallel to c@, meeting Be produced in 1; produce bitok. Then se being perpendicular to CG (Eucl. Tiara els syarallel to puK, SB UU ca k Kuk i, ; BD = DE, .. IK = KE, whence FH = HG, and ., FG is Jected in H. Jor. Ifit be required to draw ce such, that the perpendicu- 4 DH may divide re in any given ratio, take DE : DB in that <0, and proceed as in the proposition. | 94.) If any number of circles cut each other in the same points, u' from one of these points any number of lines be drawn; the F 66 GEOMETRICAL PROBLEMS. [ Sect. 1 parts of these which are intercepted between the several circui. ferences have the same ratio. . y Let any number of circles ABC, ABE, ABH Cut can | each other in the same points a and B; and from tf A draw AGEC, AHFD, meeting the circumfer- (PS ences; then HF: GE‘: FD: EC. A Join BG, and produce it toK; — — : then (Eucl. iii. 35.) AL? LB: LG: LH, Arie A eels a) ee ey and also :: KL: LD, ., (Kucl. v.15.) 1b: LF?) GL: LH, and (Eucl. v.19.)1@: HF I:IL: UF. For the same reason, 1K } FD :: IL: LF, ) IG) WF) IK : FD. In like manner, GE : GI::GB: GA‘!:GC: KG::BC: 1K, | “. ex @Quo0 GE: HF::EC: FD. (95.) In a given circle to place a straight line cutting two vr which are perpendicular to each other, in such a manner that line itself may be trisected. | Let aBc be the given circle, Ao and os being two radii at right angles to each other; bisect the angle aos by oc; at c draw the tangent cp, and make it equal to 3c0; produce oB to EB; join op, and from F draw FGIK parallel to pc; it will be trisected at the points @ and I. >| Since the angle at c is a right angle, and cos is half a ri angle, .*. also cEO is half a right angle, and equal to cc whence co=cr. And since HF is parallel to cp, ¢ , CE: ED‘: HG: GF, | 4 but Ep is double of nc, .. F@ is double of ne. But HE= since H0 bisects the angle 106, and is perpendicular to 16} rG=ci. AlsoHK=HF, ..IK=GF; | whence FG=GI=IK, and FK is trisected. vet. 11. ] GEOMETRICAL PROBLEMS. 67 | (96.) Ifa straight line be divided into any two parts, and upon. ie whole line and one of the paris, as diameters, semicircles be lscribed; to determine a point in the less diameter, from which ia perpendicular be drawn cutting the circumferences, and the jints of intersection and the extremities of the respective diame- is be joined, and these lines produced to meet ; the parts of them ithout the semicircles may have a given ratio. Let ax be divided into any two parts in te point c, and on AB, Ac let semicircles | described. Take ac: ac the ibaa (the given ratio, and make cD: cB :: AG GB; D will be the point required. AGED Cy From D draw the perpendicular pr; join B&, cr, and pro- cce them to H; join AE, cr; and from x draw Ku parallel to Ti. Since CD : CB::AG: GB, comp. and inv. DB : CD::AB: AG, id since C1 is parallel to BE, and KL to BA, | BD) CD:: BE: CK?:BA: CL, whence (Kucl. v.15.) AaB: AGIIAB: CL, “. AG=CL consequently AG: AC::CL: CA, | a. €. in the duplicate ratio of cx : oF, | or (by similar triangles) of HE : HF. ‘But AG: AC is ie duplicate of the given ratio, | . HE: HF is in the given ratio. 07) If a straight line be divided into any two parts, and from , point of section a perpendicular be erected, which is a mean »portional between one of the parts and the whole line, and a “cle described through the extremities of the line and the perpen- lular ; the whole line, the perpendicular, the aforesaid part, and nerpendicular drawn from its extremit y to the circumference will /in continued proportion. Let as be divided into any two parts in c, and from c Law the perpendicular cp equal to a mean proportional be- F 2 : ; 68 GEOMETRICAL PROBLEMS. [ Sect. 11 tween AB and ac; and through a, B, D D let a circle be described, and draw Az per- & eg pendicular to AB; AB, CD, AC, AE are in A B continued proportion. In aB produced take BF=ac. Join FD, meeting the circumference in 6; join AG, AD, GE. Then be cause BF = AC, ... CF=AB, and cD is a mean proportiona between Ac and cr, .. ADG is a right angle, whence (Kuc iii. 21.) AEG is also a right angle, and equal to HAC; .. HG1 parallel and equal to a8, i. e. to cr; whence (Kucl. 1. 33.) B and GF are equal and parallel, and the angle Ach = cFD= ‘apc, and the triangles AEC, ADC, CDF are similar, es ECR vA BtsC Diy AOR) 7 UAn a, © a meres (98.) If the tangents drawn to every two of three unequal en cles be produced till they meet ; the points of intersection will | in a straight line. Let a, B, c be the centres of the three circles; and let D FG, HI be respectively tangents to each of two circles, meetir the lines joining the centres in the points P, @, R; P, Q, RE the points in which two tangents to the circles would interse Join pa, aR; they are in the same straight line. . ect. 11. | GEOMETRICAL PROBLEMS. 69 Join AD, AF, BE, BH, CG, C1, and draw Bx parallel to pa. hen Bx and Ap being perpendicular to pp are parallel, WADE TB E22 AP BPS OFA. BH. ABR: BP’: AQ? Qk But CG—) Cl AF >: CQ: AQ, “. C# equo C1} BH 3: CQ: QK. But C1: BH 3: CR: BR, tee (uucl. v. 15.) cR: BR!:CQ@: QKk, : ; and CR: CB!:CQ!: CK; ‘so the vertically opposite angles at c are equal, .-. the triangles BK, CQR are similar, and the angle car (Eucl. vi. 6.) is equal ) BKC, ..car and cap are together equal to BKo, cap, Bac ih 29.) to two right angles, whence (Eucl. i. 14.) Q snd QR are in the same straight line. '(99.) If from the extremities of the diameter of a circle any umber of chords be drawn, two and two intersecting each other a perpendicular to that diameter ; the lines joining the extremi- vs of every corresponding two will meet the diameter produced in 1é same point. From a and s, the extremities : c ( the diameter aB of a semi- ele, let Ac, BD be drawn in- icsecting each other in FH, = ms He = \uich is perpendicular toAB. Join cp, and produce it to meet 1. in P; Pisa fixed point, or the icig joining the extremities C every ‘dane two chords intersecting each other in Fu will [ss through p. Join Bc; and bisect Be in 0; and with the centre oO, and ius oB, describe a circle u ae which will circumscribe the ‘ adrilateral figure HGcB. Take & the centre of the semi- ie and jon Hc, Ec. The angle pcx is equal to Pca, ACE trether, 7. e. to DBA, CAE together ; and the angle cue is sual to ACH, CAH together, 7. ¢. to DBA, cAn together, .. a= =CHE, fad the angle at © being common, the triangles cs fg CPE are equiangular ; - 70 GEOMETRICAL PROBLEMS. [ Sect. 1 whence EH ; EC :: EC: EP, i. | in which proportion the three first terms being invariable, £ is also, and the point « being fixed, p is also. | (100.) If from a given point in the diameter of a semicire produced, three straight lines be drawn, one of which is inclin at a given angle to the diameter, another touches the semicire| and the third cuts it, in such a manner, that the distance of t, given point from the nearer extremity of the diameter, and t, perpendiculars drawn from that extremity on the three aforesa lines may be proportional ; then will the lines, which join the e tremities of the diameter and of that part of the cutting line whi is within the circle, intersect each other in an angle equal tot given angle. g From a given point c, in 5 a | the diameter AB produced Sods S| of the semicircle AGB, draw SNE q CD inclined at a given angle yon y to Ac, cG touching, and « 7) C1H cutting the circle in such a manner that BD, BE, BF bei drawn respectively perpendicular to them, cB may be to BD3 BE to BF; then if a1, BH be joined, the angle ALH or BI will be equal to BCD. *| Join 0 H,0G; and draw ox perpendicular to n1. Now i angles at E and F being right angles, as also those at G and. BE is parallel to oc, and BF to OK; = | “ (0G) 08. BEY. COT CS 2) OL Ba. epi. OK 11.BE . BP .. BOs ne } also the angle at p is equal to ok 4H, .*. (Eucl. vi. 7.) the tri. gles OHK, BCD are equiangular, and the angle OHK is equalt pop. But ons is equal to oBu, i. e. to Arm (Kucl. iil, “. OHK is equal to AIH, LHI together, 2 e. to ALH (Eue! 82.) ; wherefore ALH is equal to BCD. ‘ | . > l r z —_ ad ". J ect, III. | GEOMETRICAL PROBLEMS. 71 SECTION III. | _(1.) Any side of a triangle is greater than the difference between ve other two sides. Let anc be a triangle; any of it sides B _ greater than the difference of the other vO. Let ac be greater than as; and cut f ap=AB; join BD; then the angle 4 - BDis equalto aps. But the exterior angle ppc is greater nan DBA, 72. e. than BDA, and .°. greater than psc (Eucl. i 5.) ; whence BC is greater than pc, 7. e. than the difference of 1e sides Ac and aB. In the same way it may be shown that 'B is greater than the difference of ac andspc; and ac greater an the difference of AB and Bc. | (2.) In any right-angled triangle, the straight line joining the ght angle and the bisection of the hypothenuse is equal to half e hypothenuse. . Let acs bea right-angled triangle, whose A Vpothenuse AB is bisected in D; join Dc; Dt ‘c 1s equal to the half of an. | From p draw pe parallel to ac, .*. (Eucl. Ls — . 2.) BE=EC, and ED is common and at izht angles to BC, ... DC=BD, i. e. the half of an. .(3.) Tf from any point within an equilateral triangle perpen- \culars be drawn to the sides ; they are together equal to a per- jndicular drawn from any of the angles to the opposite side. [rom any point p within the equilateral tri- B hate ABC, let perpendiculars pr, pF, DG be yawn to the sides; they are together equal to aa perpendicular drawn from B on the oppo- fe side ac. Jom DA, DB,.pc. Since Pian Sia upon the ‘me and equal bases are to one another as their altitudes, "72 GEOMETRICAL PROBLEMS. [ Sect. 11 ABC: ADC {: BH; DE, also ABC : BDC :: BH: DF, and ABC : ADB ?:: BH: DG; whence ABC: ADC + BDC + ADB :: BH : DE+ DF + DG, | which proportion the first term being equal to the second, .DE+DF+ DG=BH. (4.) If the points of bisection of the sides of a given triangle | joined; the triangle so formed will be one fourth of the gin triangle. | Let the sides of the triangle anc be bisected in the points D, E, F; joi DE, EF, FD; the triangle DEF is one fourth of the triangle ABC. Since AB and AC are bisected in p and F, (Eucl. vi. 2.) DF 1s parallel to Bc; and for the same reason FE is parallel to’ AB, and DFEB IS a parallel gram, .*. the triangle prx is equal to Bx. In the same wi it may be shown to be equal to rec and apF; and.*, it iso fourth of ABC. | (5.) The difference of the angles at the base of any triangle double the angle contained by a line drawn from the vertex pe pendicular to the base, and another bisecting the angle at t vertex. From B the vertex of thé triangle aBc let B& be drawn perpendicular to the base, | and pp bisecting the angle asc; the dif- ference of the angles B Ac, BCA 1s double the angle EBD. A * ED | The angle BAC is equal (Kucl. 1. 32.) to the difference of the angles BEC and ABE, 2. e. of a right ang and ABE. Also the anole BCA is equal to the difference of right angle and esc, .*. the difference of the angles BAc ar BCA is equal to the diference of the angles ABE and EBC, t. (since ABD = DBC) to twice the angle EBD. : ct. 111. | GEOMETRICAL PROBLEMS. 73 | (6.) If from one of the equal angles of an isosceles triangle any le be drawn to the opposite side, and from the same point a line / drawn to the opposite side produced, so that the part intercepted i'ween them may be equal to the former ; the angle contained by {> side of the triangle and the first drawn line is double of the agle contained by the base and the latter. Let asc be an isosceles triangle, hav- A iz the side aB equal to ac. From B draw ay line BD, and also BE cutting off px e1al to DB; the angle ABD is double of apicae ete CE. , rd For the angle pcB is equal to the two Di B, CBE, 2. e. to the two DBE, CBE, or to DBC and twicec BE; bt DCB is equal to ABC, .. ABC is equal to pBc and twice > E, and taking away the angle psc, which is common to both, hk angle ABD is equal to twice CBE. 7.) If from the extremity of the base of an isosceles triangle, a i? equal to one of the sides be drawn to meet the opposite side ; / angle formed by this line and the base produced, is equal to jee times either of the equal angles of the triangle. uet ABC be an isosceles triangle having ni I side AB equal to ac. From c to aB joduced if necessary) draw cp equal to 1, and let Bc be produced; the angle )E is equal to three times the angle B CE \ Ce | . Since CA is equal to cp, the angle cap is equal to cpa, .. } A and twice a Bc are together equal to two right angles, and ‘are equal to cDA, cDB; whence cpB is double of anc. Now licl. i, 32.) the angle pcx ‘is equal to the two angles cpB, ND and consequently is equal to three times the angle asc. ; ; 3.) The sum of the sides of an isosceles triangle is less than the 74: GEOMETRICAL PROBLEMS. [ Sect. 1 sum of the sides of any other triangle on the same base and betw the same parallels. 7 Let acs be an isosceles triangle, and £ C ApB any other triangle on the same base, and between the same parallels aB, ED; Ac and cB together will be less than ap and DB. Since Ec is parallel to as, the angle ECA is equal to cAB; and for the same reason DCB is equal cBA; but cAB being equal to cBA, ECA is equal to DCB; ac and sc drawn from two given points A and B on the sag side of the line EcD given in position make equal angles wh the line, .*. (i. 6.) they are together less than any other to lines AD, DB, draw from the same points to that line. | A (9.) If from one of the equal angles of an isosceles triangit perpendicular be drawn to the opposite side; the part of it ami cepted by a perpendicular from the vertex will have to one of equal sides, the same ratio that the segment of the base has to perpendicular upon the base. | Let asc be an isosceles triangle, having the side AB equal toac. From B and a let fall perpendiculars Bp, AE; then will BF: ac:: BE: EA. Since the angles BDA, AEC are right angles, and the angle p A F common to the two triangles FAD, EAC, .. the triangles are similar. But the triangle Fi is similar to AFD, and .. to EAC; ‘ whence BF: BE $: AC: AE, and BF: AC !! BE: EBA. (10.) If from any point in the base of an isosceles triangle f be drawn to the opposite sides, making equal angles with the bé the triangles Jormed by these lines, the segments of the base, i the lines joining the intersections of the sides and the angles 00 site, will be equal. etm] GEOMETRICAL PROBLEMS. 7D ‘From any point p, in ac the base of the isos- cles triangle A BC, let DE, DF be drawn, making e angles CDE, ADF equal to one another; join 48, CF ;, the triangles AED, CDF are equal. ‘Since the angle ADF is equal to the angle 9c, and FAD = ECD, the triangles ECD, FAD g2 equiangular, and Ap :.pc:: FD: DE. Also sice the angle FDA is equal to EDC, add to each the angle rp z£, the angle ADE = CDF; hence the sides about the equal angles ¢2 reciprocally proportional, and .*, (Eucl. vi. 15.) the triangles 'p E, FCD are equal. (11.) If from any point in the base of an isosceles triangle per- ndiculars be drawn to the sides ; these together shall be equal to cverpendicular drawn from either extremity of the base to the (posite side. Let asc be an isosceles triangle, from any int D in the base of which, let pz, pF be (awn perpendicular to the sides; and from B 1; Be be drawn perpendicular to ac; Be is cual to pg and pF together. Since the angle EBD is equal to the angle at (and the angles at = and F are right angles, the triangles BE D, 176 are equiangular, and Peete) DG* ere DE; DF; comp. BC: DE+DF:: DC: DF. But Be being parallel to pF, pc : DF :: BC : BG, | whence BC: BG !: BC: DE+ DF, | and .. BG = DE + DF. ‘(12.) Of all triangles having the same vertical angle, and whose ises pass through a given point, the least is that whose base is lected in the given point. Let Bac be the vertical angle of any number of triangles, “— bases pass through a given point p; and let Bc be sected in P: ABC is less than any other triangle apr. 76 GEOMETRICAL PROBLEMS. From c draw cF parallel to An; then the angle pBP is equal to pcr, and the vertically opposite angles DPB, CPF are equal, and BP =Ppc, .. the triangle pBp is equal to the triangle pcr, and .. ppB is less than cre; add to each the trapezium ADPC, and asc is less than apxE. In the same manner ABC may be proved to be less than any other triangle whose base passes through pP. (13.) If from the angles of the base of a triangle perpendicuis be let fall on a line which bisects the vertical angle; the pari this line intercepted between these perpendiculars will be bisecd by a perpendicular from the middle of the base. From a and B let perpendiculars ap, BG be drawn to the line cp which bisects the angle at c; the part ap will be bisected by a perpendicular eF from © the middle point of the base aB. Produce pc, FE to Handi. Then IE being parallel to HB, and aE=eEB,.. (Eucl. vi. 2.) At =1H. Also since aD Is parallel to Hae and IF, DF; FG :: AI: 1H, whence DF = FG, and DG 1s bisected in F. | (14.) If from one of the angles at the base of a triangle a be drawn parallel to the opposite side, and from any point tt lines be drawn making any angles with the sides (produce| necessary) ; they will have the same ratio that lines have, wi are drawn parallel to them from the other angles, and termini by the same sides. tl From a one of the angles of the triangle asc, let A1) drawn parallel to Be the opposite side; and from any p! D in it, let DE, pu be drawn making any angles with the sié _ St. m1] GEOMETRICAL PROBLEMS. wy dw BF, CG parallel to them respectively ; D:DH:: BF: CG. jince DE is parallel to Br, and pA to p, the triangles DEA, BFC are equi- avular, | on. DA >; BE : BC; ail in a similar manner it may be shown, tl t Hines Ue B Cra GG 0 aD Hiab Rt G- 15.) To bisect a given triangle by a line drawn from one of its ui les. — set ABC be the given triangle, and a the A ule, from which the bisecting line is to be liwn. SBisect the opposite side Ac in p, - ul join AD; AD bisects the triangle. B = ts ‘or the bases BD, pc being equal, (Eucl. . 8.) the triangles ABD, ADC are also equal. . 16.) To bisect a given triangle by a line drawn from a given nut mn one of its sides. et ABC bé the given triangle, and p A E given point. Bisect Bc in D; join ap, >; and from a draw Ax parallel to pp; 0. PE; Pk bisects the triangle anc. ince Ax is parallel to pp, the triangle #—x——i € \ D is equal to the triangle epp; from 41 of them take away the triangle prp, and Arp = EFD. \o since BD is equal to pe, the triangle aBp is equal to the ragle Apc; parts of which FD, AFP are equal, ., ABEF is gal to prpc; whence ABEF and AFP together, or ABEP "i be equal to prpc and rep together, i, e, to PEC; and ., h triangle a Bc is bisected by px. 73 GEOMETRICAL PROBLEMS. (17.) To determine a point within a given triangle, from whic lines drawn to the several angles, will divide the triangle im three equal parts. ¢ Let asc be the given triangle ; bisect AB, BC, a ; | in E, and D; join AD, CE, BF; Fis the point : | required. ye 4 Since Bp=DC, the triangle BAD is equal to 8 pAc; and for the same reason the triangle BFD is a { pro; .. the triangle BrA is equal to arc. Again, since B| —8A, the triangle BEc is equal to the triangle Arc; parts: which, the triangles BEF, AEF are equal; .*. the tian BE is equal to arc; and .-, the three Brc, BFA, AFC are equal {| one another. (18.) To trisect a given triangle from a given point within it. Let asc be the given triangle, and p A the given point within it. Trisect the a side BC in D and E; join PD, PE; and @ from A draw AF, AG eee eels argh A \\ to them. Join pr, pG,Ap. Those three fi WW lines will divide the triangle into three ®& F > KE G| equal parts. 6 Join AD, AE. Since AF is parallel to pp, tue triangle AY is equal to ApF; to each of these add ABE, .. APFB is equ to apB. In the same manner APGC is equal "he AEC; and} the remainder FPG is equal to DAE. Now the triangles AB ADE, AEC, being on equal bases and of the same altitude, ¢ equal, .. AP FB, PFG, APGC are also equal; and the triang ABC is trisected. (19.) From a given point in the side of a triangle, to draw ta which will divide the triangle into parts which shall have a gin ratio. # | Let aBc be the given triangle, and p the given point the side Bc. Divide so, in the points D, E, F, into pai which shall have the given ratio. Join AD, AB,AF, AP; @! Pree ga et. III. | GEOMETRICAL PROBLEMS. 79 aw DG, EH, FI parallel to ap. Join 4, i3, PH, P1; they will divide the trian- se, as required. For the triangles ABD, ADE, AEF, irc being as their bases will be in < D ip. {e given ratio. And since Dé is parallel to ap, the triangles 3A, DGP are equal, ., DBA, GPB are equal. And since ie triangle ADP =AGP, and AEP = AHnpP, .,. ADE=HPG. JsOAPE=AHP, and APF=AIP, .*. AEF=AHPI, and Arc= 10; .. the parts PBG,GPH, HPIA, IPC are equal to ABD, JE, AEF, AFC, and are .°. in the given ratio. The same may I proved whatever be the number of parts. k 1 (20.) Iftwo exterior angles of a triangle be bisected, and from le point of intersection of the bisecting lines, a line be drawn to 1? opposite angle of the triangle ; it will bisect that angle. | Let the exterior angles EBC, Bor, of the E tangle aBc, be bisected by the lines sp, © meeting inp. Join va; it will bisect ny B t> angle Bac. ie fe fall the perpendiculars, DE, pF, pa. Wm A . . Cc ir en the angles pBE, DB@ being equal, | d the angles at © and ¢ being right angles, and ps common tile triangles DBE, DBG,... DE=pa. In the same manner ta: and ..pe=pDF. Hence in the right-angled_ trian- 48 DAE, DAF, DE is equal to pF and DA is common, .°. the “angles are equiangular, and the angles DAB, DAFare equal, . BAC is bisected by ap. | | | 21.) If in two triangles the vertical angle of the one be equal eg of the other, and one other angle of the former be equal to / exterior angle at the base of the latter; the sides about the ind angle of the former shall be proportional to those about the 4° angle at the base of the latter. Let ABC, DEF be two triangles having the angle Bac equal 8 A 80 GEOMETRICAL PROBLEMS. [ Sect. III, to mpF, and ABC equal to the exterior angle DFG, made by producing the side EF ; then AC : CB :: DE: EF. | At the point p in the line Fp, D -make the angle rp@ equal to the A angle EDF or BAC, and meeting eF produced in c. Since the Le angle Fp«@ is equal to the angle BAC, and DFG is equal to ABC, . the triangles ABC, DFG are equiangular, and | AC: CB; DG? GF. | But since the angle Ge is bisected by pF, .*. (Hucl. vi. 3.) DG: GF?! DE: EF, “ AC: CB :: DE: EF. c B ¢ (22.) In a given triangle to draw a line parallel to one of th sides, so that it may be a mean proportional between the segment of the base. | | Let anc be the given triangle; in the base > a of which take a point E, such that ar may be 7 ' to nc in the duplicate ratio of ac : cB; draw ED parallel to Bc; ED is the line required. ah c Since ED is parallel to BC, AE: ED::AC: | cp. But an: Ec in the duplicate ratio of ac : cB, and there fore also in the duplicate ratio of AE : ED; whence from th definition of the duplicate ratio, | AE! ED: ED: EC. (23.) To draw a line parallel to the common base of two trian gles which have different altitudes, so that the parts of it lei cepted by the sides may have a given ratio. ! Let ABc, DBC be two triangles on the same base BC, th vertex D being in the side ac. Divide Bc in 8, so that Bo : cr may be equal to the given ratio. Join A&, cutting B k in @; and through ¢ draw Fu parallel to Bc; Fu is the li required. a ie 4 ie a Sect. 111. | GEOMETRICAL PROBLEMS. 8] Since Fu is parallel to Bo, FH : GuitBC: CE, i. é. in the given ratio. But if the vertex 1 is not in ac, draw 1p paral- lel to BC; join BD; divide the base Bo, as before ; join AE, and draw rx parallel to Bc. Then it is evident that GH=LK, and .. FH: LK in the , ziven ratio. Cor. If the triangles be upon equal bases, but in the same straight line, the line may be drawn in‘a similar manner. (24.) If the base of a triangle be produced, so that the whole nay be to the part produced in the duplicate ratio of the sides ; he line joining the vertex and the extremity of the part produced vill be a mean proportional between the whole line produced and he part produced. _ Let ac be produced to p, so that B .D may be to pc in the duplicate E atio of AB : BC; join BD; it will be ‘mean proportional between ap and He A c D Draw cx parallel to as; then AB: cE::AD: DC. t.eein 1e duplicate ratio of AB : Bc, whence AB: BC:: BC: CE, 2. @ 1e sides about the equal angles anc, BCE are proportional ; ierefore the triangles ABC, BCE are similar, and the angle at ‘Is equal to the angle csp; .-. the triangles ABD, CBD are yuiangular, and ive lean Bie 1) Ce = '(25.) To determine a point within a gwen triangle, which will vide a line parallel to the base into two segments, such that the cess of each segment above the perpendicular distance between € parallel lines may be to each other in the duplicate ratio of é respective segments. | ‘Let axzc be the given triangle. From c draw cp perpen- cular to aB; and from p draw DE, DF bisecting the angles o¢,BDC. Join BE, cutting cD in P; P is the point required. G 82 GEOMETRICAL PROBLEMS. [Sect. 111. Through p draw cuik parallel to an; then ‘ the angle ppu is equal to the angle HDA, 7. é. / | to the alternate angle PHD; and .. HP, and in Fo | like manner Pr will each be equal to pp the Ne perpendicular distance of gk from AB; and 4 i GH, 1K will be equal to the excess of each seg- | ment above that distance pp. And since GP is parallel to aB, | GP PK) AD gD Bs -GH faa) PE |. hence (Eucl. v. 19. Cor.) Gu: (PI=)PH::PH:1K, and .. GH : 1K in the duplicate ratio of GH : HP, 7 ¢. of GP : PK. j (26.) If perpendiculars be drawn to two sides of a triangle fron any two points therein; the distance of their concourse from tha of the two sides will be to the distance between the two points, a. either side is to the perpendicular drawn from its extremity upoi the other. | From any two points &, F in the sides AB, Ac of the triangle anc, let perpendiculars ED, rp be drawn, meeting in D. Join AD, EF; and from c draw c@ perpendicular to aB; AD: FE‘; AC: CG. | Produce Ep to u. And since the angles AED, AFD are righ angles, a circle described on AD as a diameter will pass throug r and 8, and ., the angles FAD, FED standing in the sam segment are equal; .*. the triangles AHD, HEF are equiangular AMG SAD ee eA Ee AC apUG, | since HE is parallel to ca. Ei (27.) If the three sides of a triangle be bisected, the perpent. culars drawn to the sides at the three points of bisection, will mei in the same point. Let the sides of the triangle a Bc be bisected in the points p, 5, F. Draw the perpendicu- lars EG, FG meeting ina. The perpendicular. at p also passes through G. Join Gp, GA, GB, Gc. Since aF=Fc, and » Sect. 111.] GEOMETRICAL PROBLEMS. 83 FG is common to the triangles ArG, crG, and the angles at r 'are right angles, .. Ac=Gac. In the same way it may be ‘shown that Gc=GB; .. AG=aB; but AD=DBArand, DGiis ‘common to the triangles ana, Boa, .. the angles at D are equal, and .. right angles, or the perpendicular at p passes through «a. Cor. The point of intersection of the perpendiculars is equally distant from the three angles. (28.) If from the three angles of a triangle lines be drawn to the points of bisection of the opposite sides, these lines intersect ‘each other in the same point. | Let the sides of the triangle a Bc be bisected MD,E,F. Join Ax, CD, meeting each other iG. Join BG, GF; BGF isa straight line. _ Join EF, meeting cp in. Then (Eucl. vi. 2.) FE is parallel to AB, and .*, the triangles DAG, GEH are equiangular, sp D AT GH. > His HG, OF DB UeD G20 nH Ber Gs -€.the sides about the equal angles are proportional; .-, the 4 . . . Tiangles BDG, GHF are similar, and the angle pen=nGF; ond .. BG and GF are in the same straight line. | ¥ i q (29.) The three straight lines, which bisect the three angles of a rtangle, meet in the same point. | Let the angles Bac, Boa be bisected by de lines ar, cp, and through @ their point * intersection draw 3B GF; it bisects the i agle at B. | For (Eucl. vi. 3.) Bo: oF :: BG: GF‘: BA: AF, | “BC? BA‘: CF: FA or FB bisects the angle a Bc. G2 5 } + 84. GEOMETRICAL PROBLEMS. [Sect. 111 (30.) If the three angles of a triangle be bisected, and one 0 the bisecting lines be produced to the opposite side; the angle con. tained by this line produced, and one of the others is equal to th angle contained by the third, and a perpendicular drawn from th common point of intersection of the three lines to the aforesau side. | Let the three angles of the triangle aBc be bisected by the lines AD, BD, CD; produce BD to E, and from p draw DF perpendicular to Ac; the angle ADE is equal to CDF. i Since the three angles of the triangle aBc are AS equal to two right angles, .. the angles DAB, | DBA, DoF are together equal to one right angle, i. e. to DOE and cpF; whence the two angles DAB, DBA are together equé to the angle cpF; but ape is equal to the same two angle and .*. ADE is equal to CDF. | (31.) In a right-angled triangle, if a straight line be draw parallel to the hypothenuse, and cutting the perpendicular draw from the right angle ; and through the point of intersection a lin be drawn from one of the acute angles to the opposite side, an the extremity of this line and of the perpendicular be joined ; th locus of its intersection with the line parallel to the hypothent will be a straight line. Let =F be drawn parallel to ac the hypo- thenuse of the right-angled triangle ABC; and from the right angle B let the perpen- dicular psp be drawn, meeting EF in G3; through G draw cau; join uD; the locus of 1, the intersection of nF and HD is a straight line. 1 Because £G is parallel to ac the base of the triangles aH ABD, AK! KD!:EI1:1G::4D: pc. But ap and pe are yj variable, .*. the ratios of AK : KD, and £1: 1G are also. aa the same manner if any other line be drawn parallel to t) hypothenuse, and a similar construction be made, the point: intersection will divide the part intercepted between AB al Sect. 111.] GEOMETRICAL PROBLEMS. 85 ‘BD in the ratio of AD : pc, or AK: KD, and will .°. be in the line BK, which is the locus required. | (32.) If from the angles of a triangle, lines, each equal to a given line, be drawn to the opposite sides (produced if necessary) ; and from any point within, lines be drawn parallel to these, and meeting the sides of the triangle ; these lines shall together be qual to the given line. | From the angles of the triangle ABC let the lines aa, Bd, cc be i Trawn to the opposite sides, each paca! to a given line L; and paral- ‘el to them respectively draw, from _, wy point p, the lines pp, px, PF; B Did ass hese together will be equal to x. _ Join pa, ps, Pc. Then since the triangles ABC, APC are m the same base ac, they are to one another as the perpendi- ‘ulars from B and Pp, i.e. by similar triangles, as Bb: PE, or s L: PE. In the same way, ABC: ABP !!L: PF, and ABC: BPC ::L: PD; »; ABC. APCTABP+BPC!::!L: PE+PF+PD; and since the rst term is equal to the second, the third will be equal to the jurth, or L = pPp+PpE+PrF, i | (83.) If the sides of a triangle be cut proportionally, and lines 2 drawn from the points of section to the opposite angles ; the \tersections of these lines will be in the same line, viz. that drawn Jom the vertex to the middle of the base. ‘Let the sides. of the triangle apc be cut im soportionally, so that AD: AE‘: DF:EG:: /\ a -GL:: HB: Lc. Join BE, BG, BL, CD, } Q *, CH; these lines will intersect each other | the line Ax drawn from a to x the middle ‘the base Bc. | } 4 | 86 GEOMETRICAL PROBLEMS. [ Sect. 111. Join pr. Then since when any number of magnitudes ate proportional, as one antecedent is to its consequent, so are all Ebe antecedents taken together to all the consequents together, .AD!AE!}AB: AC, and DB is parallel to Bc. Jom ko, and let it meet py ini. The triangles BOK, 10E are similar, and therefore, | BK : KO :! EI: 10, and for the same reason, | CK : KO!:DI: 10, whence ae, and DE is bisected by KO; and it is also bisected by Ak, .*. AK passes through 0, In the same manner it may be shown that BG and cF, as alsa BL, CH intersect each other in points which are in the line ak, (34.) If from any point in one side of a triangle, two lines be drawn, one to the opposite angle, and the other parallel to the base, and the former intersect a line drawn from the vertex bisect- ing the base; this point of intersection, that of the line parallel t the base and the third side, and the third angular point are in oe same straight line. From any point p in the side aB of the B | triangle aBc, let DE be drawn parallel to { ! D E Ac, and pc joined; and let pc meet BF drawn from B to the middle of Ac in G; 4, 3 G, E are in the same straight line. Lom | Let pe cut BFin k. The triangles DGK, 4 F ia CGF are equiangular, and P WN GE GO ce Ke CD Ce hence the triangles DGB, AGC, having one angle in each equal | viz. EDG, GCA, and the sides about them proportional, ar - therefore similar; whence the angles AGC, DGE are equal; anc p@c being a straight line, AGE is also. | (35.) If one side of a triangle be divided into any two parts | and from the point of section two straight lines be drawn paralle | to, and terminating at the other sides, and the points of termina tion be joined ; and any other line be drawn parallel to either 0 : the two former lines, so as to intersect the other, and to terminals | | Sect. 111.] GEOMETRICAL PROBLEMS. 87 : vim the sides of the triangle; then the two extreme parts of the ‘three segments into which the line so drawn is divided will always ‘be in the ratio of the segments of the first divided line. ' Let aB be divided into any two parts in p, Metin which draw pr, DF parallel to the other two sides of the triangle; join EF, and draw Gu parallel to pr, meeting pF and uF intand K; GI: KHi!AD: DBS and, if LM 'be parallel to DF, LK : MN:!AD: DB. | Since G1 is parallel to ar, and nx to pr, GI: AF::(ID=) NK! FD!I:!NE: DE!:KH: FC, feet: KH:: AF; FC::AD: DB, since DF. is parallel to se. Again since ML is parallel to Bc, | MN. BE*+.ND: DE‘. KF: FE?! KL: EC, “» MN .KL°: BE: EC:: BD-: DA. (36.) If through the point of bisection of the base of a triangle my line be drawn, intersecting one side of the triangle, and the ither produced, and meeting a parallel to the base Jrom the vertex ; his line will be cut harmonically. | From the vertex B of the tri- ingle ABC, let BE be drawn pa- ‘allel to the base ac, and through he middle point p let any line 26F be drawn meeting AB, BC, 3E, In F, G, E; EG: DG:: FE: FD. _ Since Ap is parallel to BE, jae FES FD?: BE: AD, | but se+ (DC=—)DA :? EG: GD, since the triangles BGE, DGC are equiangular, “. (Kucl. v.15.) ea: GD :: FE: FD, 1 the line is divided harmonically. 87.) If from either angle of a triangle a line be drawn inter- xcting that which joins the vertex and the bisection of the base, ! } | 88 GEOMETRICAL PROBLEMS. [ Sect. it will be cut harmonically. From the vertex a of the triangle © ABC, let Ar be drawn parallel to the base Bc, and aD to its point of bisection p; and from c draw any line CFGE; then will cE : CF:: EG: FG. Draw Gu parallel to Bc. Since AE and sc are parallel, (Eucl. vi. 2.) BA TAG ..CE. EG, and since GH is parallel to BD, Pc SORIA ett ie shalt 6 y Tore ¥G, i since ay triangles DFC, GHF are miles ; CRO ENN Ge Dk ak Gs and CE: CF !: EG: FG. 4 (38.) To draw a line from one of the angles at the base i triangle, so that the part of it cut off by a line drawn from th vertex parallel to the base, may have a given ratio to the part cu off by the opposite side. | From a let Ar be drawn parallel to Bc. E__ A Divide AB in G, so that AB: AG in the = given ratio; join CG, and produce it to ae : meet AEin E. CGE is the line required. a c For the triangles AGE, BGC are equiangular, °, CG: EG:iBG: AG, whence (Kucl. v. 18.) cE; EG::BA: AG, z. €. in the given ratio. (39.) To determine that point in the base produced of a right angled triangle, from which the line drawn to the angle opposit to the base shall have the same ratio to the base produced, eS | the perpendicular has to the base itself. » Let aB be the base, and cs the perpendicular of a right ¥ Sect. 111. | GEOMETRICAL PROBLEMS. 89 jangled triangle. Draw cr at right angles E ‘to Ac, meeting AB produced in z. At the LI point c make the angle EcD=caB. D is Vad J = che point required. Big cs From p draw pF perpendicular to ac, and .*, parallel to 3m. Since the angle rpc is equal to the alternate angle pcx, . €. to CAB, and the angles at F and B are right angles, .*. the mingles DCF, ACB are equiangular; and DAF is also equian- rular to ACB, hence Ko) 2s Ar 2178 GF: CA angeD Co. I has. AC: 2 At, *, e# equo per. CD: DA: CB: BA. (40.) If the base of any triangle be divided into two parts by a me which is a mean proportional between them, and which being - lrawn parallel to the second side is terminated in the third; any me parallel to the base will be divided by the mean proportional produced if necessary) into segments, which will be to each other nwersely as the whole mean proportional to that segment which is ‘erminated in the third side of the triangle. | Let AC the base of the triangle aBc ae ¢ divided into two parts in p, by a line Bee: 16 which is parallel to Bc, and a mean a he roportional between ap and pc; then A ny line r@ parallel to ac, and ene iz e 'E (produced if necessary) in H, willbe 4D ¢ ivided into segments FH, HG, which re to each other inversely as the lines DE, HE. | For since ru is parallel to ap, FH ; AD >: HE: DE, but aD: DE :: DE: (pc =) HG, ,. FH : DE :; HE? HG. (41 (41 .) If from the extremities ve of the base of any triangle, two ‘raight lines be drawn intersecting each other in the perpendicu- ww, and terminating in the opposite sides; straight lines drawn q q 90 GEOMETRICAL PROBLEMS. [ Sect. 4 ; from thence to the intersection of the perpendicular with the base will make equal angles with the base. From A and c, the extremities of Ac, the base of the triangle arc, let AE, CF be drawn intersecting the per- pendicular BD in the same point a. Join FD, ED; these lines make equal angles FD A, EDC with the base. Draw £1, FH_ perpendicular, and KGL parallel to the base; then Fu 1s parallel to Bp, | and". BG BD °° FM 2 FH. | And in the same manner it may be shown that BG: BD:. EN: EI; whence FM : FH !: EN? EI; and .. FM: EN}: FH: El. But FM: EN ?} F@:GN 3! KG: GL:: HD: DI, HDC LeE Hei, {. whence the two triangles DFH, DEI, having the angle at u equa a to the angle at 1, anil the eee about the “equal angles propor: tional, are equiangular; .*. the angle HDF is equal to EDI. : | ; 14 | 1 Pa !a (42.) In every triangle, the intersection of the perpendicular, drawn from the angles to the opposite sides, the intersection of th. lines from the angles to the middle of the opposite sides, and th intersection of the perpendiculars from the middle of the sides, ari all in the same straight line. And the distances of those ~ from one another are in a given ratio. From the angles a and B of the triangle aBc, let AD, BE be drawn perpendicular to the opposite sides, H will be the intersection of the three perpendiculars (vii. 34.). From a and B draw ag, BF to the points of bisection of the teas sides, inter- £ cH z secting in K, which .~, is (iii. 28.) the 4 intersection BE the Bae drawn from the angles to the middla 0 a ‘Sect. 111. | GEOMETRICAL PROBLEMS. 91 the opposite sides; and from Fr and @ draw the perpendiculars FI, GI meeting in 1, which .°, (iii. 27.) is the intersection of the three perpendiculars. Join HK, K1; HKI is a straight line. _ Joiner. (Eucl. vi. 2.) 4B is parallel to, and double of c¥; . by similar triangles ABK, KFG, BK is double of kr, and ak double of ke. And the triangles aus, rie are equiangular, .. Au is double of 1G, and Bu le double of 1F; | HCL eH. [heey : iV Batis fee whence the triangles BHK, KIF having the angles at B and F equal, and the sides about them proportional, are similar, .°. the mgle HKB is equal to IKF, .*. H, K, and 1 are in the same itraight line. And since Bx is double of kF, HK is double of K1, and . heir distances from each other will be in an invariable ratio. (43.) If straight lines be drawn from the angles of a triangle hrough any point, either within or without the triangle, to meet the des, and the lines joining these points of intersection and the sides f the triangle be produced to meet ; the three points of concourse vill be in the same straight line. | | Let asc be a triangle from the ree angles of which let lines .F, BE, CD be drawn through a omt p within the triangle. Join E, DF, EF, and produce them to teet the sides in H, G, 1; these aree points will be in the same aight line. Jon Gu,ur. Then the three é igles of the triangle puG being jual to two right angles, as also the pak EHI, E1H, and (HEI | ) DEF, as also the two pri, GF1; .. the three angles of the jangle pune together with the Aon EHI, E1H, DEF, DFI, FI are equal to six right angles. Now the angles of the tri- gles DEF, FGI are together equal to four right angles, whence HG, DHI are equal to two eH angles; or GH, HI are in the me straight line. - =o Ss —— rr = — A D B 92 GEOMETRICAL PROBLEMS. [ Sect. 1y SECTION IV. | (1.) The diameters of a rhombus bisect each other at R | | angles. | Let aspcp be a rhombus, whose diame- | ters are AC, BD; they bisect each other at | right angles in E. Since AB=AD, and AC is common to the two triangles aBc, ADC, the two Ba, Ac are equal to the two DA, ac, each to each, and Bc=po, .. the angle BAc is equal to the angle pac. Again, since BA, : AE are equal to DA, AB, each to each, and the included a. : are equal, .., BE=ED, and the angles AER, AED are equal, an *, are right angles. For the same reason AE=EC; also th angles BEC, DEC are right angles. | (2.) If the opposite sides or opposite angles of a quadriiae ) figure be equal, the figure will be a parallelogram. © : | | Let ascp be a quadrilateral figure, whose A opposite sides are equal. Join spp. Since el AB=DC, and BD is common, the two as, —— BD are equal to the two cD, DB, each to each, and ap= =BC, . *, the angle ABD=BDC, ‘ d whence (Eucl. 1. 27.) AB is parallel to pc; also the angle AD. =DBC, whence AD is parallel to Bc ; ad the figure is a . q lelogram. Again, let the opposite angles be equal. Then since the ea angles of the quadrilateral figure aBcp are equal to four righ angles, and that BAD, Apc together are equal to pcB, CBA *, BAD, ADC together are cae to two right angles; whence AB is parallel to cp. In the same way it may be shown that A. is parallel to Bo, and .. ABCD isa parallelogram. | ) i | (3.) To bisect a parallelogram by a line drawn from a point i} | | one of its sides. Sect. rv. ] GEOMETRICAL PROBLEMS. 93 | Let ascn be a parallelogram, and p a given point in the side as. Draw the liameter Bp, which bisects the parallel- pgram. Bisect BD in F; join pr, and oroduce it to E. Pps bisects the parallel- ogram. Since the angle pBD is equal to the angle Bpx, and the vertically opposite angles at F are equal, and BF=FD, ., the riangles PBF, DFE are equal. But the triangle app is equal 0 BDC, .. APFD is equal to BFEC; and to niece equals adding he equal andles DFE, PFB, the Grane APED = PECB; and IkG@as .”. Beected by PE. ber in See 1 D 1D) C _ Cor. Any line drawn through the middle point of the diame- er of a parallelogram is bisected in that point. y (4.) If from any point in the diameter (or diameter produced) fa parallelogram straight lines be drawn to the opposite angles ; ey will cut off equal triangles. | From any point &£, in ac the dia- setaagmene < neter of the parallelogram aBcp, LO es t lines EB, ED be drawn; the Ae / D riangles ABE, AED are equal; as “~ lso the triangles BEC, CED. Draw the diameter Bp. The bases Br, Fp being equal, the flangles BFA, DFA (Eucl. i. 38.), as also the triangles Bru, (FE are equal, hence .*, BAH, DAE are equal. And axc being qual to anc, the triangles Bec, DEC are also equal. | (5.) From one of the angles of a parallelogram to draw a line to ‘ve opposite side, which shall beequal to that side together with 4 segment of it which is intercepted between the line and the posite angle. Let ancp be the parallelogram, a the angle from which the heis to be drawn. Produce pe to x, making ce=cp, Join | 8 94: GEOMETRICAL PROBLEMS. Ak, and at the point a make the angle EAF = AEF; AF is the line required. For cx being equal to cD, EF is equal to pc and cr together; and the angles FEA, FAE being equal, FA = FE, and .; AF = Dc and cr together. i Cor. In the same manner if CE i; =— CB, AF —EF=—8C ant sor to- gether. (6.) If from one of the angles of a parallelogram a straight iy be drawn cutting the diameter, a side, and a side produced ; a segment intercepted between the angle and the diameter, is a med proportional between the segments intercepted between the can. . and the sides. From s, one of the angles of B c the parallelogram aBcp, let any line BE be drawn cutting the dia- meter Ac in F, the opposite side { { in G, and AD produced in ES; BF , D : is a mean proportional between | | FG and FE. . Draw pu parallel to Br, and .*. equal to it; .. also AH=FI and AF = cH. Since HD is parallel to BG, FG: DH (:: CF: Qiuse AH oA BR) 25D Hol anas | Or FG: BF ?: BF: FE. (7.) The two triangles, formed by drawing straight lines fro. any point within a parallelogram to the extremities of two oppost sides, are together half of the parallelogram. — Let p be any point within the parallelogram asco, fro which let lines pA, PD, PB, PC be drawn to the extremities ‘ the opposite sides; the triangles PAD, PBC are equal to ha the parallelogram ; as also the triangles a PB, ppc. oe 3 e ‘Sect. Iv. ] _ GEOMETRICAL PROBLEMS. 95 _ Through © draw EprF parallel soap or BC; then (Kucl. i. 41.) whe triangle app is half of arrp, ind src is half of BEE GY . APD, BPC are together half of \Bcp. In the same manner if i line be drawn through P par- illel to AB or DC, it may be shown that ap B, DPC together are aalf of ABCD. A £ B D F Cc (8.) Ifa straight line be drawn parallel to one of the sides of a varallelogram, and one extremity of this line be joined to the pposite one of the parallel side, by a line which also cuts the dia- eter ; the segments of the diameter made by this line will be ‘eciprocally proportional to the segments of that part of it which is Mecented between the side and the parallel line. _ Let =F be drawn parallel to av one of Auber B he sides of the parallelogram aBcp, cut- ing the diameter Bp inc. Join ar, cut- HG ngitalsoinu;thenwillsu:up::up ~——+ c || HG. _ For the angle anu being equal to npr, and AnB to DHF, Qe triangles AuB, DHF are equiangular, and .. BH: HD :: AH HF‘: DH: HG, since the triangles AHD, FHG are also equi- peular. K (9.) If two lines be drawn parallel and equal to the adjacent des of a parallelogram ; the lines joining their extremities, of pro- uced, will meet the diameter in the same point. Let Hi, FG be drawn ‘qual and parallel to the Yacent sides as, Bo of te parallelogram axcop. om HF, Gr; these lines toduced will meet the “ameter DB in the same ont. | 96 GEOMETRICAL PROBLEMS. [ Sect. r Produce AB, cB to K andu. Then the triangles AFH, LB having the vertically opposite angles at F equal, and the alte; nate angles AHF, FLB also equal, are equiangular, whence AF: FB :: (HA=) IB: BL; and in the same manner it may be shown that (GC) B AGIsaR Ret Bah oe APS Cig SB. koe ie But Ar=DG, and CI=DH, .. DG: DH 3: BK : BL, and .”, H) DB, G1 converge to the same point. . (10.) Jf in the sides of a square, at equal distances from ti four angles, four other points be taken, one in each side ; the figui contained by the straight lines which join them shall also be square. Let £, F, G, H be four points at equal dis- =, E tances from the angles'‘of the square ABCD. _,, Join EF, FG, GH, HE; EFGH is also a square. Since AH=EB, and AE=BF, and the an- a D G ( gles at A and B are right angles, .. HE=EF, and the angle AEH is equal to the angle pre. In the san way it may be shown that ne and GF are each of them equal | HE and EF, .*. the figure HEFG is equilateral. It is also rec angular; for since the exterior angle FEA is equal to the inter’ angles EBF, EFB; parts of which AEH and EFB are equal; | the remaining mee FEH is equal to the remaining angle FB) and .*. isa right angle. In the same manner it may be shov that the angles at F, G, are right angles, and .. pre bei! equilateral and rectangular, is a square. | (11.) The sum of the diagonals of a trapezium is less than b sum of any four lines which can be drawn to the four angles fre any point within the figure, except from the intersection of t} diagonals. i Let aBcp be a trapezium, whose diagonals are Ac, BD, CU ting each other in E; theyare less than the sum of any fo ‘ect. tv.] GEOMETRICAL PROBLEMS, 97 jmes which can be drawn to the angles from ‘ny other point within the trapezium. | Take any point p, and join PA, PB, PC, PD. ‘hen (Eucl. i. 20.) ac is less than ap, pc; and \D is less than BP, PD; .*. AC, BD are less A D ‘han AP, PB, PC, PD. | (12.) Every trapezium is divided by its diagonals into four viangles proportional to each other. Let aBcD be a trapezium (see last Fig.) divided by its dia- jonals Ac, BD into the triangles AEB, BEC, AED, DEC; these ire proportional to each other. | For (Kucl. vi. 1.) AEB: BEC!: AE : EC, and AED: DEC ?: AE : EC, ’., AEB: BEC :: AED: DEC. _ (18.) Iftwo opposite angles of a trapezium be right angles ; the ngles subtended by either side at the two opposite angular points all be equal. Let the two angles acs, App of the trape- i um AcBD, be right angles. Join aB, cD; re angles AcD, ABD, subtended by ap, are jual. Bisect Abin E. Join cE, ED, and produce l@to Fr. Then (i. 2.) AE, EB, EC, ED are {ual to one another. Also the angle AED is equal’to the two DB, EBD, 7. e€. to twice EBD; and DEF is equal to the two CE, EDC, 7. e. to twice DCE; and AEF is equal to twice ACE; twice ACE and twice ECD, or twice acD will be equal to ED, 7. e. to twice EBD, ... ACD=ABD. : ‘The same may be proved for the angles standing on any of ‘te other sides. | (14.) To determine the figure formed by joining the points of ./? of the sides of a trapezium ; and its ratio to the trape- alm, H 98 GEOMETRICAL PROBLEMS. [ Sect. Let ABcp be a trapezium, whose sides are bisected in &, F, G,H. Let the points of bisection be joined; and draw the diagonals AC, BD. Since AB, AD are bisected in E and u, (Eucl. vi. 2.) EH is par- allel to Bp; and for the same rea- : son FG is parallel to BD, and .*. to nH. In the same way it mé be shown that EF is parallel to ue, and .. the figure EFGH a parallelogram. | Again (Eucl. vi. 19.) the triangle nBF is to the triangle AB in the duplicate ratio of HB: AB, i.e. in the ratio of 1 34, | EBF is equal to one fourth of anc; for the same reason HDG one fourth of pac, whence EBF and HDG are together equal one fourth of the trapezium. For the same reason HAE a @Fc are together equal to one fourth of the trapezium ; therefo} the four triangles together are equal to half the trapezium ; an consequently HEFG is equal to half of aBep, \ | a Cor. 1. Hence two lines, drawn to bisect the opposite 1 Cor. 2. If the sides of a square be bisected and the poi! of bisection joined, the inscribed figure is a square, and equal) half the original square. a . of a trapezium, will also bisect each other. (15.) To determine the figure formed by joining the points whe the diagonals of the trapezium cut the parallelogram ; and its ra) to the trapezium. | Let 1, K, L, M be the points of intersection; (see last Fi) join IK, KL, LM, M1. And let o be the intersection of the d gonals. Since EK is parallel to 01, BK : KO!:BE: EA, 7%. é. in a ratio of equality. For the same reason at=10. Whence the sides of the t angle AoB being cut proportionally, 1K is parallel to AB. the same manner it may be shown that KL, LM, MI are respr tively parallel to Bc, cD, DA; whence the figure 1K LM will | Eien Ne —— Sect. 1v.] GEOMETRICAL PROBLEMS. 99 similar to Ancp. Also since the triangle m1rx is half the par- ulelogram ux, and MLK half of ex, .. the figure 1K LM is half of HF, and .*, equal to one fourth of the trapezium aBcpD. | (16.) Tf two sides of a trapezium be parallel ; its area is equal ‘0 half that of a parallelogram, whose base is the sum of those two ides, and altitude the perpendicular distance between them. Let aBcD be a trapezium, whose side A BG H 4B is parallel to pc. Produce pc to £, naking cE=azs. Draw Br, and ca, EH garallel to AD, meeting AB produced. $77 7G) E Then ar is a parallelogram of the same utitude with the trapezium, and its base is equal to the sum of he sides AB, pc; and ABcpis half of ax. Since DF=CE, the parallelograms Ar, GE are equal (Kucl. i. 36.) ; and the diameter sc bisects the parallelogram Fre ; whence \BCD=BCEH, and .°, the trapezium is half of the parallelo- yram AE. | (17.) If from any angle of a rectangular parallelogram a line be lrawn to the opposite side, and from the adjacent angle of the tra- iezium thus formed another be drawn perpendicular to the former ; he rectangle contained by these two lines, is equal to the given arallelogram. | From a, one of the angles of the rectangular . C varallelogram ABCD, let any line ax£ be drawn to * he opposite side, cutting off the trapezium ABCE; i rom 8 let fall the perpendicular pr; the rect- heen D mgle contained by az, BF, is equal to the pa- allelogram ABCD. For Ba being parallel to Ep, the angle BAF=AED, and the ngles at F and p are right angles, .. the triangles BAF, AED ‘re similar ; whence BF; BA!!AD: AB, nd the rectangle AE, BF 1s equal to the rectangle BA, AD, i. e. 0 ABCD. ‘I ee * - 100 GEOMETRICAL PROBLEMS. [Sect. rv. (18.) To divide a parallelogram into two parts which shall hil Q a given ratio, by a line drawn parallel to a given line. | Let ascp be the given parallelogram, n and EF the line whose direction is given. Divide ap in G in the given ratio, and make GH=AG. Join BH; bisect it in 1, and through 1 draw KL parallel to EF; KL di- -vides the parallelogram in the given ratio. For (Eucl. vi. 1. and 1. 41.) | ABH: ABCD!: AG ~~ AD. But the triangles Bri, K1H are equal, since BI=1H, and puis parallel to KH; .. ABH=ABLK, and ABLK : ABGCD::AG* AD, 4 “.ABLK } LKDC!:AG : GD, i.e. in the given ratio. angles. i Let ancp be the given trapezium, and A A A the angle from which it is to be bisected. y a Draw the diagonals ac, BD; and bisect Low | BD, which is opposite to the angle a,in 4% G 4 the point E. Join An, cE; and through 4 | E draw FEG parallel to ac. Jom aq; ae bisects the trail zium. a Since DE is equal to EB, the triangles AED, AEB are equal as also DEC, BEC; .. the figure AECD is equal, to the figur AECB. dine (Eucl. i. 38.) the triangles AnG, cEG are equal| take away ., the common part rHG, and annH=euc. To th figure AECD add aEu, and take away its equal Guc; and t AECB add GHC, and take away AEH; and the triangle AGD it equal to the trapezium aGcB, or the eon trapezium is bisectec by ac. | i 7 (20.) To bisect a trapezium by a line drawn from a given poin mm one of its sides. | Sect. rv. | GEOMETRICAL PROBLEMS. 101 Let ascp be the given trapezium, B md p the given point. Join pA, and from the angle Pp bisect he trapezium APrcp (iv. 19.) by the ine PE. On pt make the triangle D‘**SG% Cc EF equal to ABP. Bisect EF in G@; joinpaG. pa bisects the rapezium. Since FG is equal to Gx, the triangle par is equal to the riangle pcr. But pas is equal to half the triangle aBp; nd pec is half the figure papc; whence pGc is half of the rapezium ABCD; which is .*. bisected by Pa. (21.) If two sides of a trapezium be parallel; the triangle con- ained by either of the other sides, and the two straight lines drawn rom its extremities to the bisection of the opposite side, is half he trapezium. ) Let ascp be a trapezium, having the , A * ide AB parallel to pc. Let an be bisected ; as | 1E; join BE, CE; the triangle BEC is alf of the trapezium. Peas F | Through & draw FrEG parallel to Bc, meeting cD in G, and ‘A produced in ¥. The alternate angles FAE, EDG being ‘qual, as also the angles at zr, and AE=ED, .*. the triangles EF, DEG are equal; whence the parallelogram BFGc is equal }) the trapezium ABcD. But Brec and the triangle BEC, eing on the same base Bc, and between the same parallels Bc, i@, the triangle pec is half of prec, and .. also half of (BOD. | Cor. From the demonstration it appears, that a trapezium ‘hich has two sides parallel, may be reduced to a parallelogram ‘yual to it, by drawing through the point of bisection of one of jie sides, which are not parallel, a line parallel to the other of tose sides, and meeting the parallel sides. | (22.) To divide a given trapezium whose opposite sides are wrallel, in a given ratio, by a line drawn through a given point, : : id terminated by the two parallel sides. 102 GEOMETRICAL PROBLEMS. [ Sect. rv Let ascp be a trapezium whose sides a | AB, DC are parallel, and p the given point. ly oder | Bisect AD in B, and draw EF parallel iS to AB, meeting BcinF. Divide EFinG 56 i, “| in the given ratio; and through @ and Pp draw 1PH; 1PH will divide the trapezium e in the given ratio. Draw KGL, MFN parallel to ap; then EA=KG=MF, an ED=GL=FN; but AB=ED, .. KG=GL, and ME=PN whence (iv. 21. Cor.) ADIH = AL, and HICB= KN. Now AL: KN :: EG: GF, | °- ADIH ! HICB:: EG: GF, i. e. in the given ratio. (23.) If a trapezium, which has two of its adjacent angles rigi angles, be bisected by a line drawn from the middle of one of ti sides which are not parallel; the sum of the parallel sides wi have to one of them the same ratio, that the side which is bisected has to that segment of it which is adjacent to the other. | Let asco be a trapezium, having the angles ¢ at A and p right angles, and .*. AB, pc parallel; and let the trapezium be bisected by EF; if AD */S be bisected in &, AB+ DC PAB!) BCU OF; but if Bc be bisected in F, AB+DC:ABi!: AD: DE. Produce pa, cB to meet ing. Join AF, DF, BE, CE, andill fall the Dey ed AH, DI. Since AE=ED, the triangl AFE, DFE are equal, .*. the triangles DFC, AFB are equal; , the rectangles Frc, DI inet BF, AH are equal, 5 and FC : FB ?: AH: DI:: AB: CD, 4 . AB+CD: AB?: (FC+FB=) BC: FC. But if pc be bisected, the triangles EBF, 4 ECF being equal, the triangles AEB, EDC are also equal, E *, (Kucl. vi. 15.) AB: CD?! DE: EA, and AB+CD: AB!:!(DE+EA=) AD: DE. In like manner AB+DC: DC :! AD: AE. a) ‘Sect. 1v.] GEOMETRICAL PROBLEMS. 103 (24.) If the sides of an equilateral and equiangular pentagon be moduced to meet; the angles formed by these lines are together - to two right angles. Let ABcDE bean equilateral and equi- =, F mgular pentagon ; and let the sides be Rese, ye lproduced to meet in F, G, H,.1I, K; the a B 5) 3 Ig angles at these points are together equal E o two right angles. ag For since BcG is the exterior angle of avi G vhe triangle Fc1, it is equal to the angles it Fr andi. For the same reason the mgle cBG is equal to the angles at K and i; and .. the angles at F, G, H,1I, K are equal to the three mgles of the triangle BCG, 2. e. to two right angles. ' (25.) If the sides of an equilateral and equiangular hexagon be wroduced to meet; the angles formed by these lines are together qual to. four right angles. Let ABcDEF be an equilateral and quiangular hexagon; and let the sides )€ produced to meet in G, H, I, K, L, M3 ‘he angles at these points are together :qual to four right angles. _ For Gui being a triangle, the angles ~fG, 1, L are equal to two right angles; ind for the same reason, the angles at H, <, M are equal to two right angles ; .-. the ix angles are equal to four right angles. | (26.) The area of any two parallelograms described on the two “ides of a triangle is equal to that of a parallelogram on the base, vhose side is equal and parallel to the line drawn from the vertex f the triangle to the intersection of the two sides of the former arallelograms produced to meet. 104: GEOMETRICAL PROBLEMS. [ Sect. 1v Let Be and ce be parallelo- grams described on the sides AaB, BC of the triangle aBc; and let EF, HG be produced to meet in D. Join DB; produce it, and make 1K =DB; through a draw au equal and parallel to 1K; and complete the parallelogram am. AM is equal to AF and CG to- gether. Produce LA, Mc to N and oO; ~ since ND is parallel to aB, and AN to BD, NABD is a para’ gram, and equal to EB, which is on the same base, and betwee the same parallels. It is also equal to AK; because they a upon equal bases, DB, 1K, and between the same parallel *..AK=EB. In the same manner IM=BH, .’. AM is equal ¢ AF and ca together. L K M | | (27.) The perimeter of an isosceles triangle is greater than ; perimeter of a rectangular parallelogram, which is of the sam altitude with, and equal to the given triangle. Let asc be an isosceles triangle, whose base A 7 s Bc. Draw ak perpendicular to Bo, and va r *, bisecting it; and draw ap, cp parallel re- | “poarele to BC, AE; then DE is a rectangular / | parallelogram of ihe same altitude with, and * - @ equal to the triangle aBc (Kucl. i. 42.). The perimeter of AB is greater than that of DE. Because AB=AC, and BE=EC, the perimeter of aBc double of ac and Ec together; also the perimeter of DE | double of az and EC feerrcee But since AEC is a right angl Ac is greater than Arn; and .. the perimeter of aBc greats than that of DE. . 5s | (28.) If from one of the acute angles of a right-angled triangl a line be drawn to the opposite side ; the squares of that side an the line so drawn are together equal to the squares of the sequi@ adjacent to the right angle and of the hypothenuse. Sect. 1v.] GEOMETRICAL PROBLEMS. 105 Let asc be a right-angled triangle, and from , let ap be drawn to the opposite side; the quares of AD and BC are together equal to the quares of ac and BD. For the squares of ap and Bc together are qual to the squares of AB, BD and Bo,i.e.to © DathB he squares of Ac and BD; since the squares of aB and BC are qual to the square Ac. > (29.) In any triangle if a line be drawn from the vertex at right ingles to the base ; the difference of the squares of the sides is qual to the difference of the squares of the segments of the base. ' From a the vertex of the triangle anc, let A 1D be drawn perpendicular to the base; the lifference of the squares of aB, ac is equal to he difference of the squares of Bp, Dc. a | For since a Bp is a right-angled triangle, the quare of AB is equal to the squares of Ap, BD; and since apc s a right-angled triangle, the square of ac is equal to the quares of AD, DC; whence the difference of the squares of ac nd AB is wae to the difference of the squares of cp and pp. CG _ (80.) In any triangle, if a line be drawn from the vertex bisect- ng the base ; the sum of the squares of the two sides of the tri- ingle is double the sum of the squares of the bisecting line and of alf the base. From the vertex a of the triangle aBc, let A .D be drawn to the point of bisection of the vase; the squares of aB, Ac, are together ; fouble the squares of AD, DB. y ao From a draw ak perpendicular to Bc; Then (Eucl. ii, 12.) an’=ap?+pB?+2BDx DE, and (Kucl. ii. 13.) ac?-=ap’?+pc’—2cpx DE =AD°+DB’?—2BDxX DE, whence AB’+Ac?=2apd?+2pDB’. 106 GEOMETRICAL PROBLEMS. [ Sect. 1 (31.) If from the three angles of a triangle lines be drawn to tl points of bisection of the opposite sides; the squares of the div tances between the angles and the common intersection are toge ther one third of the squares of the sides of the triangle. From the angles of the triangle axsc, let lines be drawn to the middle points of the opposite sides, intersecting each other in @«; the sum of the squares of AG, GB, GC is one third of the sum of the squares of AB, BC, CA. Join eF. Then ap’?+ac?=2Aa4n’?+2EB’, AB?+BOC°=2ARr+2FB, AC’?+BC°?=2AD?4+2D0’, . AB?+BO0°+CA°7=AER’+BF°+CD' +AF°+EB +AD, Now the sum of the squares of ar, EB, AD is equal to or fourth of the sum of the squares of AB, BC, CA; whence thre fourths of the sum of the squares of AB, BC, CA will be equ to the sum of the squares of an, BF, CD; or three times tl sum of the squares of AB, BC, CA, is aa to four times th sum of the squares of AE, BF, CD. Now BG: GF!: BA! EF?! BC: cE 7:2: 1, , (BG?) BE 37'2.23,.and BG 298 Foe he i whence 48F’=98G*. And the same being true of each of tl rest, three times the sum of the squares of AB, BC, CA, is equ to nine times the sum of the squares of AG, BG, CG; .wetem sum of the squares of AB, BC, CA is three times the sum of t) squares of AG, BG, CG. Cor. If from the angles ofa triangle lines be drawn to th points of bisection of the opposite sides, the squares of tho lines together are to the squares of the sides of the triangle . 3: 4. (32.) If from any point within or without any rectilineal figui ; perpendiculars be let fall on every side; the sum of the squares” d _ the alternate segments made by them will be equal. Let aBcp be any quadrilateral figure (the demonstratic Sect. 1v.| GEOMETRICAL PROBLEMS. 107 yeing the same whatever be the number of ides). From any point 1 let perpendicu- ars IE, IF, 1G, 1H be drawn ; AE’+BF’+ s0+DH’ = EB’+FC’+GpD'+AH’. From 1 draw lines to each of the angles ; A Hi D then AE’+EIr=(ar=) an’+Hr’, BF?+Fl=(BI’=) BE’+EI’, ce’ +@er=(cr’=) cF+FI’, AY DH +HI’=(DIr’=) DG’+er’, yhence, AE’+BF’+CG+DH =EB +FC+GD'+HA’. (83.) If from any point within a rectangular parallelogram ines be drawn-to the angular points ; the sums of the squares of hose which are drawn to the opposite angles are equal. Let ancn bea rectangular parallelogram, A- ind F any point within it; join FA, FB, FC, ‘p; the squares of Fa and Fo are together qual to the squares of FB and FD. Draw the diagonals ac, BD; and jon Fr. Because the riangles ADC, BDC are similar and equal, ac=sp; and .. heir halves, Az and Dk, are equal. ’ | Now (iv. 30.) FD’+FB’?=2DE°+25F’, =2AE’+2EF=AF'+ FC’. | (84.) The squares of the diagonals of a parallelogram are toge- her equal to the squares of the four sides. ' Let ancp be a parallelogram, whose dia- a onals are Ac, BD; the squares of Ac, BD are <— iS ogether equal to the squares of AB, BC, CD, ’ D <* 1A. ' Since pp is bisected by ac, 2 2 2 2 2AE+2ED°’=AD+AB, ‘nd for the same reason, B 108 GEOMETRICAL PROBLEMS. | Sect. 20H’ +2ED°=CD’+cB’, “. 4A 8+ 4ED'=AD?+AB?+CB’+CD’, 1.€ AC +BD?=AD?+AB’?+CB’+CD’, (35.) If two sides of a trapezium be parallel to each other ; th squares of its diagonals are together equal to the squares of a it two sides which are not parallel and twice the rectangle containe by its parallel sides. 4 Let the sides an, pc of the trapezium ABCD yilaeiaen a be parallel; draw the diagonals ac, BD; the VAG Squares of Ac and BD, are together equal to DY the squares of ap and Bc, and twice the rect- 5 i angle AB, DC. - Let fall the perpendiculars cr, pF. a Then (Eucl. ii. 12.), pB’=pa’?+AaB?+2ABX AF, and AC’=CB’+AB’+2ABXBE, anes AC’+DB’=AD?+CB’?+2AB’+2ABX BE+2ABX AF, Now (Eucl. ii. i.), ABX FE=ABX FA+ABX AB+ABX BE, “. AC’+DB’=AD?+CB’?+2ABX DC. (36.) The squares of the diagonals of a trapezium are togelll } double the squares of the two lines joining the bisections of th opposite sides. | Let ascp be a trapezium, whose sides are bisected In E, F,G, H. Join EG, FH; and draw the diagonals ac, Bp. The squares of Ac, BD are together double of the squares of EG, FH. ; } Join EF, FG, GH, HE. Then (iv. 14.) a EFGH is a parallelogram, and Bp is double of EH; a °. BD =4EH’=2EH +2FG’, ie and for the same reason A c?=24EF’+2HG?’, \ hi . AC’ +BD'=2EF'+2Fq’+2GH’+2HE, ‘ =2EG’+2HF’, (iv. 34.) Sect. Iv. ] GEOMETRICAL PROBLEMS. 109 _ (87.) The squares of the diagonals of a trapezium are together less than the squares of the four sides, by four times the square of the line joining the points of bisection of the diagonals. Let ABcD be a trapezium whose diagonals B AC, BD are bisected in 5, F; join EF; the A 7s : squares of AC, BD are less than the squares I of the four sides by four times the square of liam BF. D c Since BE bisects 4c the base of the triangle asc, AB’+BC°=2A5?+2EB’; ind for a similar reason, AD’+DC°=2AH’?+2ED’; “ AB’+BC+CD'+DA°=44 F74+25B'+2ED? = AC’?+2EB?+2ED’ = AC?+4BF°+4FR’ = AC’+ BD’+4FR’, 388. In any trapezium, if two opposite sides be bisected ; the um of the squares of the two other sides, together with the squares f the diagonals, is equal to the sum of the squares of the bisected wdes together with four times the square of the line joining those oints of bisection. Let AB, Dc, two opposite sides of the tra- ezium ABCD, be bisected in x, and F; join 1F; and draw the diagonals ac, Bp. The uares of AD, BC, AC, BD are equal to the uares of AB, DC, and four times. the square ” fEr, Jon AF, BF. Since aF bisects pc the base of the triangle ‘DC, AD’+AC?=2DF°4+2Fa’; din the same manner, BC?+BD?=2DF?+2FB’; Bence AD’+BC?+a0?+BD?=4DF" 4 Ona? + 2B? =DC?+2FA°+2FR?=)0?4+4A874+4EF? =Dc?+AB’+4EF’, \j = I z . FF 110 GEOMETRICAL PROBLEMS. [ Sect. ry! (39.) If squares be described on the sides of a right-angle, triangle ; each of the lines joining the acute angles and the opposi angle of the square, will cut off from the triangle an obtuse-anglei triangle, which will be equal to that cut off from the square by | line drawn from the intersection with the side to that angle of th square which is opposite to it. * From the angles Bp, c of the right-angled triangle Bac, lét lines BG, cD be drawn to the angles of the squares described upon the sides, and from the in- tersections H and 1 let HE, 1F be ‘drawn to the opposite angles of the squares; the triangle Bric= AIF, and CHB=AHE. Ml Join ac, avD. Then (Eucl. i. 37.) the triangle ar1=arg) to each of which add Bi, .*, the triangle BiF = BAG = BCA’ (Eucl. 1. 37.) | From each of these equals take away the triangl) BIA, and BIC=ArF. In the same manner it may be shown th CHB=AHE, ¥ na q (40.) If squares be described on the two sides of a right-ongaa triangle ; the lines joining each of the acute angles of the tri angle and the opposite angle of the square will meet the per pendicular drawn from the right angle upon the hypothenuse, 4 the same point. 7 Let Br, cr be squares described on the sides Ba, Ac containing the right an- gle. Join pc, Be; they intersect aL, which is per- pendicular to Bc, in the same point o. Produce pk, GF, to meet in H. Jom HA, HB, HC. Let Bu, cH respectively meet DC, BGin I and K. Since Sect. 1v. | GEOMETRICAL PROBLEMS. 111 sH=AF=AC, and EA=AB, and the angles HEA, BAC are right ngles, the triangles HEA, BAC are equal, and the angle EH A [=BCA=BAL, 7. e. since EH and BA are parallel, HAL is a traight line, or LA produced passes through u, and HL is per- yendicular to Bc. Again, since AC=cG, AH=BC, and the mgle HAC=BCG, .. the triangles Hac, BCG are equal; .-. feancle CBK = CHL; but BCK=HCL;. .. BKC=HLC, i. e. s aright angle, and BK is perpendicular to nc. In the same aanner it may be shown that c1 is perpendicular to Bu. Hence *, HL, CI, BK are perpendicular to the sides of the triangle 1Bc, and .°. they intersect each other in the same point. (41.) If squares be described on the three sides of a right-angled riangle, and the extremities of the adjacent sides be joined ; the tri- ngles so formed are equal to the given triangle and to each other. | On the sides of the right-angled miangle Asc let squares be de- ‘cribed, and join GH, FD, 1n. The riangles AGH, BFD, ECI are equal 0 ABC, and to each other. Pit is evident that acu = ABc. *roduce FB, and from p draw ps rerpendicular to it. Since aBs and ‘BD are right angles, .. the angles .BC, SBD are equal; and BAC, BSD re also right angles, and Bc = BD, .‘Ds=ac. And the triangles ABC, FBD being upon equal ases AB, FB are as their altitudes ac, ps (Kucl. vi. 1.); and , are equal. Inthe same manner if 1c be produced, and ER Yawn perpendicular to it, it may be shown that ER is equal to .B, and the triangle Ec1 to ABc. And since each of the tri- ngles is equal to A BC, they are equal to one another. D E | | (42.) If the sides of the square described upon the hypothenuse fa right-angled triangle be produced to meet the sides (produced 112 GEOMETRICAL PROBLEMS. [ Sect. 1 if necessary) of the squares described upon the legs; they will ci off triangles equiangular and equal to the given triangle. ] Let ps, EC, the sides of the square described on Be the hypothenuse of the right-angled triangle aBc, be pro- duced to meet the sides of the squares described upon BA, AC, in K and L; the triangles BFK, CIL, cut off by them, are equal and equiangular to ABC. Since FBA and K BC are right angles, the angles rBK, ABC are equal; also | the angles at F and a are right angles, and FB=BA, ... FK; Ac, and the triangles FKB, ABC are equiangular and equal. | In like manner it may be proved that the triangles a Bc, L¢ are equiangular and equal. | D (43.) If from the angular points of the squares described up the sides of a right-angled triangle perpendiculars be let fall upt the hypothenuse produced ; they will cut off equal segments ; a the perpendiculars will together be equal to the hypothenuse. Let rm, IN be drawn from the angles F, 1 of the squares described upon BA, AC, perpendicular to. BC the hypothenuse produced; MB , will be equal to Nc; and FM, IN | together equal to BC. Mie eo D From A draw Ao perpendicular to Bc. Since FBA is a rig] angle, the angles FBM and a Bo together are equal to FBM ar BFM, .*. ABO is equal to BFM; and the angles at m and oa right angles, and AB=BF, .. BM=Ao, and rm=Bo. In same manner it may be shown that cn=aAo, and IN=CC .. MB=NC; and FM and IN together are equal to Bo and ¢ together, 7. e. to BC. | Cc Cor. The triangles FBM, ICN are together equal to ABC, | bd ect. 1V.| GEOMETRICAL PROBLEMS. 1138 (44.) If on the two sides of a right-angled triangle squares be escribed ; the lines joining the acute angles of the triangle and ie opposite angles of the squares will cut off equal segments JSrom ie sides ; and each of these equal segments will be a mean propor- onal between the remaining segments. On as, Ac the sides of the right- qgled triangle Bac, let squares be ascribed, and Bi, cF joined; the seg- ents AP, AQ are equal, and each of lem is a mean proportional between Pp and ca. Since AQ is parallel to n1, and aP to FG, ? Bit: ci te, BAe and (CA=) HI: CG!: AP: (FG=) AB, | “oBH 0G. 4 APs AQ and BH being equal to ce, AP=AQ. Again, the triangles Bpr, acp being similar, as also ABQ, #Q, BES (Be), MpeSS Apo eG and BA: AQ@?: (IC=)AC: CQ, “. 62 @quo BP ; (AQ=) AP 3: AP: CQ. \(45.) If squares be described on the hypothenuse and sides of a tht-angled triangle, and the extremities of the sides of the tmer and the adjacent sides of the others be joined ; the sum of 2 squares of the lines joining them will be equal to five times the ware of the hypothenuse. Let squares be described on the ree sides of the right-angled tri- gle ABC; join DF, EI; the dares of pF and x1 together are lal to five times the square of Bc. Draw FxK, IL perpendicular to 4 EC produced, and am to Be. e angle FBxK is equal to ABC, and -angle at. x to the right angle ~ I 114 GEOMETRICAL PROBLEMS. [ Sect. 1v AMB, and FB=BA,.. BK =BM. In the same way Gr) CM: Now (Eucl. ii. 12.) rp’=pB’?+BF’+2DBX BK =BC’+BA’+2BCX BM, a and E1’=BC’?+CA’?+2BCx CM, r PD + Er =2B0?+BA?+AC?+2BCX BM+2BCX OM | =2p0?+BC?+2B0°=5BC. + | (46.) Ifa line be drawn parallel to the base of a triangle, a terminated in the sides ; to draw a line cutting it, and terminate, also by the sides, so that the rectangles contained by their 9 ments may be equal. Let ED be parallel to cB the base of the triangle aBc; from Dp draw pF, making with ac (produced if necessary) the angle DFE equal to ABC, and draw any line Gu parallel to rp, cutting ED int; the rectan- gle EI, ID is equal to the rectangle G1, 1H. For the angle AGH=AFD= pie Se : and the vertical angles at 1 are equal, .*. the triangles GE1, mr are equiangular ; —- = < ee aaa and HI: 1ID?: IF: 1G, “. the rectangle n1, 1p is equal to the rectangle 41, IG. (47.) If the sides, or sides produced, of a triangle be cut by ai line ; the solids formed by the segments which have not a comm extremity are equal. : ¥ Let ase be a triangle having the sides (produced if necessary) cut by the line peF; thenarxcCDXBE=AE xX DBX CF, 4 Draw BG parallel to ac; the triangles AEF, BEG will, similar, as also CDF, BDG; Sect. 1v.] GEOMETRICAL PROBLEMS. 115 SA haan teB ER BE, and CD.GF 33 BD BG, .AFXCD:AEXCF!: BD: BE, AE. CD BEA E XID B’XICF. (48.) If through any point within a triangle, three lines be drawn parallel to the sides; the solids formed by the alternate segments of these lines are equal. _ Through any point p within the triangle ‘ABC, let HG, EF, IK, be drawn parallel to the sides ; then IDX DGX DF=EDXDKXDH. Since the lines are drawn parallel to the sides, the triangles 1nD, GDK, HDF are simi- ar to ABC, and to one another ; | PL De eA C CS GD UK Alb. A.C DEY DH: BC DAE, whence 1DX DGX DF: DEX DK X DH:!ACX ABX BC ; BCX | AC X AB, . é. in a ratio of equality. (49.) If through any point within a triangle lines be drawn rom the angles to cut the opposite sides ; the segments of any one ide will be to each other in the ratio compounded of the ratios if the segments of the other sides. | Through any point p within the triangle .BC, let lines AE, BF, CG be drawn from Seta Nala sat he angles to the opposite sides; the seg- ‘aents of any one of them as ac, will be in ZN he ratio compounded of the ratios ac: A F Cc /B, and BE : EC. _ Draw 18H parallel to ac, meeting aE and c@ produced in ‘andi. Then the triangles GCA, GBI, and EAC, EBH, as also DF, BDH, and FDC, IDB, are respectively equiangular, 12 . 116 GEOMETRICAL PROBLEMS. [ Sect. whence BH ;: AC 3: BE: CE, and AC ::BI:: AG: BG, °- BH ! BI: AGX BE ! GBX CE. But BH: BI:: AF: FC, 7. ARUsEO 4A GX BEL GBxX CE (50.) If from each of the angles of any triangle, a line be draw through any point within the triangle, to the opposite side; thi solid contained by the segments thereof, intercepted between th angles and the point, will have to the solid contained by the thre remaining segments, the same ratio that the solid contained bh the three sides of the triangle, has to either of the (equal) solid. contained by the alternate segments of the sides. | Let asc be the given triangle, ‘ fe and through any point p within it, Ne] let Ar, BF, CG be drawn from the , eee Si. angles to the opposite sides; then ™ wil ADXDBXDC:}EDXDFXDG:: ig anc ABXBCXCA:AFXCEXBG. Let fall the perpendiculars a4, BI, Hae CRAB LGM FNS we ae rte? EPS re Since EL is parallel to BI, CB : CE :: BI: EL, and em being parallel to AH, BA : BG ::AH : GM, also FN and cx being parallel, ac : AF ?::CK : FN, “ie AC X AB X BC:CE X BGXAF {i BI x AMX CK: BUX [GM x FN. Again since EL is perpendicular to po, and cK to DK, thi triangles DEL, DCK are Stata cane and ... DC } DE 3: CK : EL. In the same manner, DB : DG :: BI : GM, and DA: DF ::AH;: FN, “. DAXDBXDC ; DEXDFXDG:!:BIX AH XCK 7 EL X GM FNIIABXBCXCA ! AFXCEXRG. Sect. v.| GEOMETRICAL PROBLEMS. 117 SECTION V. (1.) A straight line of given length being drawn from the centre ut right angles to the plane of a circle ; to determine that point in t, which is equally distant from the upper end of the line ‘and the ircumference of the circle. A | From o the centre of the circle, let oa be lrawn at right angles to its plane; draw os per- yendicular to OA; join AB, and make the angle ABC equal to BAc. C is the point required. Since the angle ABC=CAB, .*. AC=CB. (2.) Zo determine a point in a line given in position, to which mes drawn from two given points may have the greatest difference: iossible. 4 | Let a and B be the given pointy, and cp the *\ ine given in position. Let fall the perpendi- ular Bc, and produce it, so that cz may be qual to cB; join AB, and produce it to meet DinD. Join BD. D is the point required. | For DE=pB; and .*, AE is equal to the jafference between ap and ps. If then any ther point F be taken, BF==EF; and the dif- erence between AF and BF is equal to the difference between \F and EF, which is less than Az (ii. 1.). The same may be woved for every other point in cp. (3.) A straight line being divided in two given points ; to deter- une a third point such that its distances from the extremities ay be proportional to its distances from the given points. _Let as be the given line, divided in c E jodp. On ap and cB let semicircles be pr \ \ Jescribed intersecting in x. From f let fall «x can eee 1e perpendicular EF; ¥F 1s the point required. t } ito. GEOMETRICAL PROBLEMS. [ Sect. v. For (Eucl. vi. 8. Cor.) AF : FE :: FE: FD, and FE: FB !: FC: FE, aA Rr 2 FE i Oc: side, will contain the greatest angle. Let A and B be the given points, and cp the B given line. Join BA, and produce it to meet cp in p. Take pc a mean proportional be- tween DA and pB. C is the point required. Join Ac, BC; and about the triangle aBc YAS describe a circle; Dc is a tangent at the point c (Eucl. ii. 37.), and .*. the angle is the greatest (ii. 62.). (5.) To determine the position of a point, at which lines draw from three given points, shall wake with each other angles equal tt given angles. Let A, B, c be the three given points; join AaB, =a : and on it describe a segment of a circle containing “4 ny an angle equal to that which the lines from a and | B are to include. Complete the circle, and make Z| the angle ABD equal to that which the lines from ze | A and c are to include. Join pc, and produce it to the circum ference in E. £E 1s the point required. i Jom AE, BE. Then the angle AEc=aBD, and AEB is 0 the given magnitude, by construction. rectangle contained by them may be equal to the square of thet difference. | Let a B be the given line; upon it describe a semicircle ADB From B draw se at right angles and equal to as. Take ¢ the centre, and join oc; and from p draw pz perpendicular t AB; AB is divided in the point ©, as was required. Sect. v.] GEOMETRICAL PROBLEMS. 119 Since Bc is double of Bo, DE is double of o£ c Eucl. vi. 2.), and of being half the difference vetween AE and EB, DE is equal to the differ- mee. Also (Eucl. vi. 13.) the rectangle ax, 2B s equal to the square of DE. rays A Ook B (7.) Ifa straight line be divided into any two parts; to pro- luce it, so that the rectangle contained by the whole line so pro- luced, and the part produced may be equal to the rectangle con- ‘ained by the given line and one segment. | Let a8 be the given line divided into two parts n the point c. On AB as a diameter describe .circle ADB. From B draw Be at right angles ° © AB, and .°. a tangent to the circle; and make BE a mean proportional between aB and ac. Take o the centre; join EO, and produce it to Fr. Pro- luce 4B to G, making BG equal tozD. Then will the rectan- zle AG, GB be equal to the eee BA, AC. | Since DE=B6, the rectangle BG, GA is equal to the rectangle DE, EF, 2. e. to the square of EB, or tts the rectangle AB, AC, oy construction. ' Cor. 1. If it be required to produce the line, so that the ‘ectangle contained by the whole line produced and the part oroduced, may be equal to the rectangle contained by two given ines; find Br a mean proportional between the two given lines, ind. Proceed as in the proposition. | Cor. 2. If it be required to produce the line, so that the sectangle contained by the whole line produced and the part wroduced, may be equal to a given square; take BE equal toa side of the square, and proceed as in the proposition. | (8.) To determine two lines such that the sum of their squares nay be equal to a gwen square, and their rectangle equal to a nwven rectangle. Let aB be equal to a side of the given square. Upon it 120 GEOMETRICAL PROBLEMS. describe a semicircle ADB; and from B draw Dd. BC perpendicular to AB, and equal to a fourth A \ proportional to aB and the sides of the given & rectangle. From c draw cp parallel to Ba. Join AD, DB; they are the lines required. Since cB touches the circle at B, the angle cBp is equal tc DAB, and the angles pcB, ADB are right angles; .*. the tri angles DCB, ADB are equiangular, and AB! AD !: DB: BC, | whence the rectangle aD, DB is equal to the rectangle a8, BC i.e. to the given rectangle. “Also the squares of AD, DB ar equal to the square of 4B, 7. e. to the given square. (9.) To divide a straight line into two parts, so that the rect angle contained by the whole and one of the parts may be equal ti the square of a given line, which is less than the line to be divided Let as be the given line to be divided. Upon it describe a semicircle, in which place the line Ac = to the given line. Join cB; and on it describe a semicircle cpB, cutting 4—p AB in D; D is the point required, Since ne angle ACB is in a sae Na it is a right angle, .” Ac touches the circle cps (Eucl. ii. 16. Cor.) 5 whence th, rectangle BA, AD is equal to the square x A C,.2.é, to the squat) of the given line. {pits $$ ~— (10.) Zo divide a given line into two such parts that the reel angle contained by the whole line and one of the parts may be*(m times the square of the other part, m being whole or fractional. Let aB be the given line, and in it produced, D take Bc = an m™ part of aB. On Ac describe a semicircle, and from B draw BD perpendicular I to ac. Bisect CB in 0; join op, and take o£ =0oD; and aB will be divided in £, as required. On Bc describe a semicircle, cutting oD in F; join FE. The the angle por being common to the triangles Dos, HOF, ani ‘ect. V.| GEOMETRICAL PROBLEMS, 121 0, oB respectively equal to EO, oF, the triangles will be simi- ar and equal, and .*. the angle org equal to OBD, and .. a ight angle; whence FE is a tangent to the circle crs. Hence he rectangle AB, BC is equal to the square of DB, 7. e. to the quare of rx, or the rectangle cz, mB. From each of these quals take away the rectangle cB, BE; and the rectangle ax, 'B is equal to the square of BE, .*. (m) times the rectangle ax, B, 2. é. the rectangle AB, AE 1s equal to (m) times the square fBE. ! (11.) Zo divide a given line into two such parts that the square f the one shall be equal to the rectangle contained by the other nd a given line. . Let az be the given line to be divided, (see last Fig.) and Bc he other given line. Let them be placed so as to be in the ame straight line. On ac describe a semicircle and draw he lines, as in the last proposition; and £ is the point re- vuired. | For the rectangle Ar, cB is equal to the square of BE. | 12.) A straight line being given in magnitude and position ; to raw to it from a given point, two lines, whose rectangle shall be qual to a given rectangle, and which shall cut off equal segments ‘rom the given line. voint. Bisect AB in D, and from p draw po t right angles to a8, and let fall the perpen- licular ce. With the centre c, and radius 4 F\_DEGSB qual to a fourth proportional to 2c and he sides of the given rectangle, describe a circle cutting Do in », Join oc; and with the centre o, and radius oc, describe a ircle CFG, arsine AB in F and G; join cr, ca; they are the ines required. | For (Eucl. vi. C.) the ane CF, CG is equal to the rect- mere contained by 2co and cx, %. e. to the given rectangle. _ Let azn be the given line, and c the given. And since AD=DB, and FD=DG, **. AF=GB. 139 GEOMETRICAL PROBLEMS. (13.) To draw a straight line which shall touch a given + and make with a given line, an angle equal to a given angle. Let aB be the given line, and o the cen- tre of the given circle. From any point a in the given line, draw ac making with it an angle equal to the given angle; from o draw op perpendicular to ac, and through the point = where it meets the circle, draw 4 EF parallel to DA; EF is the line required. For being parallel to ac it is perpendicular to op, and .*, tangent to the circle; and the angle EFB = DAB = the give) angle. Cc EF (14.) Through a given point to draw a line terminating in tw lines given in position, so that the rectangle contained by the tw parts may be equal to a gwen rectangle. —-- = Let aB, cD be the lines given in posi- tion, E the given point; from §& draw EF perpendicular to aB, and produce FE to G, so that the rectangle FE, EG may be equal to the given rectangle. On EG describe a circle, cutting cp in H. Join HE, and produce it to A; AH is the line required. Jon @u. The angle GEH is equal to AnF, and the angle GHE, AFE are right angles, .. the triangles GEH, AEF ar equiangular, and j EH : EG?) EF: EA, whence the rectangle AE, EH 1s equal to the rectangle na, BF i. e. to the given rectangle. (15.) From a given point to draw a line cutting two given ae allel lines, so that the difference of its eons may be equal to given line. 4 Let AaB, cD be the given eae, and p the given point, From p draw any line pB, meeting the given lines in B andE Sect. v. | GEOMETRICAL PROBLEMS. 123 Make EF=EP, and draw FG par- Wel to AB. With any point o as sentre, and radius equal to the siven line, describe a circle cutting 3yFin u. Join on, and draw pga varallel to it. paa will be the ‘ine required. Since PE is equal to EF, .*. (Kucl. vi. 2.) pr=1e@; and AG is qual to the difference of ar and 1p, the segments of pa; and \@ = 08 = the given line. | (16.) From a given point without a circle, to draw a straight ine cutting the circle, so that the rectangle contained by the part af it without and the part within the circle shall be equal to a given square. | Let a be the given point, and pep the riven circle. From a draw a8 touching the circle; and on it as a diameter describe a semicircle AEB, in which place BE equal to aside of the given square. Join AE; and with the centre A and radius az, describe the circle Ec, cutting Bcp inc. Join ac, and produce it to p. AcD is the line required. | For the rectangle ac, aD is equal to the square of AB, 7. e. to ‘he squares of Az and EB or to the squares of ac and EB; take uway from each the square of ac, .*. the rectangle ac, cD 1s 2qual to the square of 5B, 7. e. to the given square. | (17.) From a given point in the circumference of a semicircle, to Traw a straight line meeting the diameter, so that the difference jetween the squares of this line and a perpendicular to the dia- neter from the point of intersection may be equal to a given “ectangle. Let a be the given point in the circumference of the semi- Me . : ; cirele ; from it draw ap perpendicular to the diameter. Take o she centre, and divide po in B, so that the rectangle contained 124. GEOMETRICAL PROBLEMS. [ Sect. 1 by 208, BD may be equal to the given rectangle. , © | Join AB; and draw Bc perpendicular to pB. AB, /N | BC are the lines required. DBO For (Kucl. i. 12.) the square of aB together with twice th rectangle 0B, BD is equal to the difference of the squares ¢ oA and oB, z. e. to the square of Bc; .. the difference betwee, the squares of AB and Bc is equal to twice the rectangle or BD, 2. e. to the given rectangle. | (18.) From a given poini to draw two lines to a third given | position, so that the rectangle contained by those lines may b equal to a gwen rectangle, and the difference of the angles whic, they make with that part of the third which is intercepted betuieah them may be equal to a given angle. | ; 4 Let -a be the given point, and pc the line given in position. From a draw ap perpen- dicular to Bc; make the angle Dax equal to the given angle; and produce ax, till the _ rectangle DA, AE, is equal to the given rectan- “ * = gle. On ar as a diameter describe the circle ara, cutting Bé inFand@. Join AF, AG; they are the lines required. a Draw Gu perpendicular to the diameter ar; then the an HA is equal to the arc ag,and the angle acu to AFG; .*. th angle HGF is equal to the difference of the angles acer, AFG Now the right-angled triangles 1x, KD@ have the angles at equal, .. the angle KAI=K@pD; but KAI was made equal t the given angle; .*. the difference of the angles arG, AGF i equal to the given angle. And (Eucl. vi. c.) the rectangle aF AG is equal to the rectangle pa, au, i. e. to the given reet angle. (19.) Two points being given without a given circle; to deter mine a point in the circumference, from which lines drawn to th two gwen pornts shall contain the greatest possible angle. Let a and B be the given points, and Ep¥F the given cirek whose centre is 0. Describe a circle through a, B, 0. Joir sect. V- | GEOMETRICAL PROBLEMS. 125 e 3F, BA, and produce them to meet in ; From e@ draw ep touching the siven circle ‘in p. Through p, A, B Jescribe another circle ; then since the quare of Gp is equal to the rectangle 1G, GF, 7. e. to the rectangle aG, GB, +, @p touches the circle aBD. Join \D, DB. D is the point required, as is vident from (ii. 62.) | _ (20.) From the bisection of a given arc of a circle to draw a traight line such that the part of it intercepted between the chord f that and the opposite circumference shall be equal to a gwen traight line. Let pat be the given arc of the circle anc, visected in A; Arc the diameter, and u1 the siven straight line. Produce u1 to x, so that the rectangle 1K, K1 may be equal to the rect- ngle ra, ac. From a place in the circle aB = HK; AB is the line required. | Join pc; then the angle are being a right angle is equal to the angle asc, and the angle at a is common, .. the triangles GF, ABC are equiangular, 7 and AF; AG :: AB: AC, *, the rectangle GA, AB is equal to the rectangle FA, AC, 7. e. o the rectangle HK, KI. But AaB=HK, .. AG=K1I, and con- equently GB=HI. ' (21.) To draw a straight line through a given point, so that he sum of the perpendiculars to it from two other given points day be equal to a given line. Let a, B, c be the three given points, a ‘eing that through which the line is to be ‘fer yawn. Join ac, and produce it, making aD Fr ane =Ac. Join Bp, and on it describe a semi- ee arcle; in which place BE equal to the given € ; \ 126 GEOMETRICAL PROBLEMS, [ Sect. line. Join pE; and through a draw Fae parallel to pz; it the line required. | For let fall the perpendicular ca, and draw pu parallel i BE; then the triangles Ace, ann being similar, and ac=Aj .. CG=HD=FE, FD being a parallelogram; .. BF and @ together are equal to BB, i. e. to the given line; and rH bein parallel to ED, BF is perpendicular to ra. ) (22.) To draw a straight line through one of three points give in position ; so that the rectangle contained by the perpendiculai let fall upon it from the other two may be equal to a given squar Let a, B, c be the three given points, c I and A the point through which the line is | to be drawn. Join aB, Ac; and draw cp parallel to Ba, and take cx a third propor- | tional to AB and a side of the given square. P ¥ x | On ac describe a semicircle; and from = draw EF at righ angles to cp, and meeting the semicircle in F. Join ar, an produce it; it is the line required. . Join cr, which will be perpendicular to ap; and from | draw BG perpendicular to ac. Since ce is parallel to Ba and cr to BG, the triangles aBG, cEF will be similar, | “ABS BE?) CF? CH, .. the rectangle BG, CF, is equal to the rectangle aB, CF But since the side of the given square is, by construction, mean proportional between 4B and cx, the rectangle aB, CF is equal to the given square; .*. the rectangle BG, CF is equé to the given square. q | } (23.) A given straight line being divided into two parts; t cut off a part which shall be a mean proportional between the tw remaming segments. Let aB be divided into two parts in the be point c; bisect cB in pD, and draw pr perpen- 4% dicular, and equal to ap; and through the 6 0 points B, c, E describe a circle ; produce ED to F. Jom Ak, and bisect EF in 0; and from o E 8 ect. v.] GEOMETRICAL PROBLEMS. 127 raw 0G parallel to AB, meeting Ar in @; and since AD=pE, , GO=O8, and G is a point in the circumference. From « raw GH perpendicular to ac; uH is the point required. _ For ua, being perpendicular to ap, is perpendicular also to 10, and .*, is a tangent at G; .°. the square of u@ is equal to ne rectangle cu, HB. But since ap=pkxE, .. AH=HG, and msequently the square of an is equal to the rectangle cn, 'B; and Ax is a mean proportional between the two remaining soments CH and HB. (24.) To draw a straight line making a given angle with one of ve sides of a given triangle, so that the triangle cut off may be | the whole in a given ratio. | Let asc be the given triangle; make D ie angle AcD equal to the given angle B hich the cutting line is to make with ac. roduce AB to D; and make AE: AB in ue ratio of the part to be cut off to the < ACD: ACE, and .. AFG=ACE. BULAGE “ACE *°A ER: AT, “AFG: ACB::AE: AB, 7. é. in the given ratio. AD | |(2%5.) Between two given straight lines containing a given angle, | place a straight line of given length, and subtending that angle, | that the segment of the one of them adjacent to the angle may to the segment of the other which is not adjacent, in the ratio two given lines. Let ep, EF be the lines given in length and a Produce one of them rr, till ne : 2 in the given ratio. Join pq@; and with 'e centre E, and radius equal to the given line | be placed, describe a circle cutting pe in \ join Eu, and draw u1 parallel to nF, and | 4. Sa GEOMETRICAL PROBLEMS. [ Sect. > 1c parallel to HE. 1C is the line required. | For (Eucl. vi. 2.) H1:1D:: GE! ED, 3 and uc being a parallelogram, H1=kO, | “. EC! 1D!:GE: ED, i.e. in the given ratio; | and 1c = EH = the given line. | | (26.) From two given points to draw two lines to a point in’ third, such that the difference of their squares may be equal to | given square. Let a and B be the given points; join AB; and from a draw AE perpendicular to it, and equal to a side of the given square. Join BR, and bisect it in F; from F draw the per- pendicular FG, meeting aB in G; and from G draw GD perpendicular to AB, meeting cD in D3; join AD, DB; these are the lines required. Join Gk, it is equal to es. And (iy. 29.) the differenc between the squares of Bp and Ap is equal to the differen between the squares of BG and GA, 7. e. between the squares « EG and G4, or it is equal to the square of Ax, i. e. to the give square. (27.) To divide a given straight line into two such parts, thi the square of the one may be to the excess of a given rectang above the square of the other, in a given ratio. Let as be the given straight line. From B draw se at right angles to as, and make AB’: BC in the given ratio. Join ac. Find a mean proportional between the sides of the given rectangle; and with it as radius, and B ¢< as centre describe a circle cutting ac in D. Join BD, and draw DE parallel to Bc; E is the point required For (Kucl. vi. 2.) az’? : ED’::AB’ : BC’. Now the square | ED is equal to the difference of the squares of BD and BE,%. to the difference of the given rectangle and the square of BE ‘ect. V.| GEOMETRICAL PROBLEMS. 129 , the square of Ar is to the difference between the given ectangle and the square of BE as AB’: BC’, 7. e.in the given tio. _N. B. The given rectangle must not be less than the square * the perpendicular from B upon Ac; and when BD is less ian BO, there are two points E. ' — \ | | (28.) From any angle of a triangle, not isosceles about the angle, draw a line without the triangle to the opposite side produced, hich shall be a mean proportional between the segments of the de. Let asc be the triangle, and B the angle om which the mean proportional is to be ‘awn. About the triangle describe a circle, id to the point B draw a tangent BD mect- g the side ca produced in D. BD is a ean proportional between ap and pe. _(Eucl. iii. 86.) the rectangle ap, Dc is equal to the square of 4 i@epand .. AD: DB::DB : DC. i ee Ee Se ee Sn (29.) From the obtuse angle of any triangle, to draw a line (thin the triangle to the opposite side, which shall be a mean joportional between the segments of the side. Let asc be a triangle having the obtuse ‘gle Apc. Describe a circle about it, and joduce BA to D, making AD=AB. From p (aw DE parallel to ac, meeting the circle in join BE, cutting ac in F; BF will be a 2an proportional between aF and Fc. | For (Eucl. vi. 2.) BF; FE?! BA: AD, D and since BA=AD, .. BF=FE. bw the rectangle ar, Fc is equal to the rectangle BF, FE. 7. é. tthe square of BF; | SARS EEL. EB FC, 130 GEOMETRICAL PROBLEMS. [Sect v (30.) From the common extremity of the diameters of two sem: circles given in magnitude and position; to draw a line | the circumferences, so that the rectangle contained by the cw chords may be equal to a given square. Let as, AC be the diameters drawn from a, and given in magnitude and position. With the. | centre A, and radius equal to a side of the given Z\ | square, describe a circle, cutting the lesser semi- af | circle in p. Draw pe perpendicular to ac, and i" meeting the other semicircle in F, Join ar, and WAY produce it to G@; AG is the line required. For joining Gc, the ak cee AGC, AFE are similar, .AC IAG i: AF: AE, and .*, the Dean cle FA, AG is equal to the rectangle CA, A Ay 4. e. to the square of aD, which is equal to the given square. | (31.) To draw a line parallel to a given line, which shall ( terminated by two others given in position, so as to form wit them a triangle equal to a given rectilineal figure. | 4 " a Let aB, Ac be the lines given in position, C AD the line to which it is required to draw x —P—_F_/ a parallel. Describe a rectangular parallelo- | eram AEFG equal to the given figure. Pro- duce EF to H; and take Ak a mean propor- tional Ris pu and 2rF; draw kc parallel to ap; Ke the line required. , For the angles Dua, CAK being equal, as also pau, >) A G Ky the triangles DAH, AKC are equiangular, and similar; whene AKC; AHD?:AK’; DH’::2EF : DH!:2EFX AE: DHX AB, Now the rectangle pu, Az is double of the triangle AHD, AKC is equal to the rectangle nF, ax, i. e. to the given rectil neal figure. . (82.) To bisect a triangle by a line drawn parallel to one of sides. ; , v Sect. v. | GEOMETRICAL PROBLEMS. 131 Let asc be a triangle to be bisected by a line parallel to its side as. On Be describe a semicircle; bisect Bc in O, and draw the perpendicular op; join cD; and with co as centre, and radius cp, describe a circle cut- ting CBin E; draw BF hess to AB; EF bisects the triangle. (Kucl. vi. 8.) Bo : cD::cD: CO, | mec (CD: EC IDN Sahel. Sith | out the triangles anc, FEC are in the duplicate ratio of Be : on, and .*. in the ratio of 2: 1,i.e. urc is half of ABC, and EF visects the triangle. (33.) To divide a gwen triangle mto any number of parts iaving a given ratio to each other, by lines drawn parallel to one of the sides of the triangle. | Let asc be the given triangle; divide AC into parts AE, EF, FC having the same ‘atio to one another that the parts of the ‘riangle are to have. On ac describe a semicircle, and draw the perpendiculars 1G, FH; and with the centre a, and radii 1G, AH, describe circles meeting ac in I nd x, from which points draw 11, kM parallel to Bc; these will divide the triangle in the ratio required. For the triangles ALI, AKM, ABC are to one another in the luplicate ratio of the sides Ai, AK, AC, #. e. in the ratio of the ectangles AC, AE; AC, AF; and the square of Ac; or in the atio of the lines Az, AF, AC; whence ALI, LIKM, MKCB are 4 the ratio of AE, EF, FC, #. é. in the given ratio. | (34.) To divide a given triangle into any number of equal parts y (nes drawn parallel to a given line. | _ Let anc be the given triangle; from the angle c draw cp arallel to the given line; and let it be required to divide the ‘langle into five equal parts. On ap, BD describe semicircles K 2 132 GEOMETRICAL PROBLEMS. [Sect. AID,BMD3; divide AB into five equal parts in the points &, F, G, H; draw El, FK, GL, HM perpendicular to aB; and make AN, AO, AP respectively equal to AI, AK, AL, andBQ = BM; and draw NR, OS, PT, QV, par- allel to pc; they divide the triangle as required. | 3 (Kucl. vi. 1.) the triangle abc : ADo::AB: AD, (Eucl. vi. If SODA N RAD ANS Da "C2 @QU0, ABC > ANR:°AB. Abs Do. le i. €. ANR Is One fifth of apc. In the same manner ABC : AosS::5:: 2, whence NRSO is also one fifth of ABC. And by a similar manner, opTs and Bay, may each be shan to be one fifth of ABc, .. TP@v will also be one fifth of An Cor. In nearly the same manner the triangle may be divide into any number of parts having a given ratio. 4 if (35.) To divide a trapezium which has two sides parallel in” any number of equal parts, by lines drawn parallel to those sides” Let ascp be the given trapezium having the sides AB, Dc parallel. On as the longer side describe a semicircle Ars; from p draw DE parallel to Bc; with the centre B, and radius B&, describe the arc EF, and from F let fall the perpendicular ra; and divide ac into the given number of equal parts, e. g. three, in H and 1; and draw HK, IL at right angles to AaB. Make BM, BN respectively equal, BL, BK; and draw MoO, NP parallel to Bc; and pa, or parall | to AB; and produce AD, BC tos. bi Since DC=BE=BF, and OR=BM=BL, and P@=BN=B] | Sect. v.] GEOMETRICAL PROBLEMS. 133 he triangle ors is to psc in the duplicate ratio of or to cp, or of BL to BF, 7, e. in the ratio of BI: BG; whence ODCR : DSC::GI : GB In the same manner PDCQ ! DSC:!:GH: GB, *.. ODCR : PDCQ::GI : GH, and ODCR } PORQ!:GI : IH, . é. in a given ratio of equality. _ And in a similar manner aP@B may be shown to be equal to ?ORQ. And so on, whatever be the number of equal parts. Cor. In nearly the same manner, the trapezium might be livided into parts having any given ratio. _ (86.) From one of the angular points of a given square to draw » line meeting one of the opposite sides, and the other produced, in uch a manner, that the exterior triangle formed thereby may have gwen ratio to the square. iven ratio. From a to pc (produced if neces- // ary) draw a line ao, such thatm:M+N::DC: .0. With the centre o and radius oA, describe ‘semicircle meeting pc produced in f and F. Join ar; which vill be the line required. Jon az. Then M:M+N::DC: 40::ABCD: the rectangle .0,AD. Now the triangle apz is similar to ape, and equal » it, since AB=AD; .*, the trapezium AECG is equal to ABCD; nd the rectangle ao, ap is equal to the triangle arr, ' Let ancp be the given square,andm:Nthe «a z G ¢ whence M: M+N::AECG: AEF, “M:NI:AECG : GCF | [; ABCD : GCF. | (87.) From a given point in the side produced, of a given rect- ngular parallelogram, to draw a line which shall cut the perpen- icular sides and the other side produced, so that the trapezium ut off; which stands on the aforesaid side, may be to the triangle “. eee 134 GEOMETRICAL PROBLEMS. [ Sect. | which stands upon the produced part of the opposite side, in given ratio. | | Let AK cD be the given rectangle, and cer i | £ the given point in the side cp pro- hele | duced. On Ec describe a semicircle, and / ; \\ in it place EF=ED; join Fc; and divide EC in G, so that EG: Gc in the given ratio, and draw Gu at right angles to ec. In ax produced tal BK a fourth proportional to EG, Gu and Fc. Join BE; it is tl line required. | For (Kucl. vi. 19.) the triangle Eom : EDI:: EC? : ED’, *, dw. CDIM ! EOM!!EC?—ED’ : EC’, . but ECM : BMK:: EC’: BK’, *, Cx equO CDIM } BMK!:!EC?—EF"’ : BK? | a pa aah 8 ::E@’ ; Gu’, by construction, _ “rae 2.é. in the given ratio. E DG C | . ) (38.) Through a given point, between two straight lines co. taining a gwen angle, to draw a line which shall cut off a tr tang equal to a given figure. | | Let as, Ac be the lines containing the given angle Bac, and p the given point. Through p draw p& parallel to ac, and de- scribe a parallelogram ne equal to the given figure. Draw ex perpendicular to ac, and equal to DE; and make Hc=DF; join cp, and produce it to meet AB in B; CB is the cl line required. For the triangles EBD, D1F, arc being similar, are to 0) another in the duplicate ratio of the sides pg, pr, ac; but tl square of uc is equal to the squares of He, ac; and .* tl | Square of DF is equal to the squares of pE, @c; whence tl triangle p1F is equal to the triangles pBE, ic; .*. the triang) ABC is equal to AEFG, 4, e. to the given figure. ‘ect. v.] GEOMETRICAL PROBLEMS. 135 » (89.) Between two lines given in position, to draw a line equal ; h a given line, so that the triangle thus formed may be equal to a wen rectilineal figure. f Let as, Ac be the lines givenin nu} : vosition, and pe the line whose aagnitude is given. Bisect it in ¥, nd on pF describe a rectangular yarallelogram equal to the given i igure. On DE describe a segment facircle containing an angle equal to the angle at a, and cut- ing HGinI. Join DI,1E; and make AK=ID, and AL=IE. | oin KL; it is the line re Since AK=ID, and AL=1k, and the angle at A=DIE, .. KL =Dk, and the triangle AK L=IDE=HGFD= the given figure. _ (40.) From two given lines to cut off two others, so that the emainder of one may have to the part cut off from the other a wen ratio; and the difference of the squares of the other re- nainder and part cut off from the first may be equal to a given quare. } Perpendicular to AB one of the given lines, 1 lraw Bc equal to a side of the given square ; 5 nd take ap to the other given line in the siven ratio of the part remaining from the first ‘o the part cut off from the second. Join pc; nd with the centre a, and radius equal to the ‘econd given line, describe a circle cutting pc Q E; join Ag, and draw ceF parallel to it, meeting ar, drawn varallel to pc,in F. Then BG and GF are equal to the parts to ve cut off. ' For the difference between the squares of CG, GB is Spins to he square of BC, i.e. to the given square; and AG: GF::AD: Ag, i. é. in the given ratio. i| 1] D BG /* ¥ (41.) From two given lines to cut off two others which shall have ir i 136 GEOMETRICAL PROBLEMS, [ Sect. a gwen ratio, so that the difference of the squares of the vena may be equal to a given square. | ! Let ac be one of the two given lines. From g c draw cp perpendicular to ac, and equal to a side of the given square. Take ar to the D\ YF other given line in the given ratio of the parts to be cut off. Join Ep, and produce it; and with the centre a, and radius equal to that C GE other given line, describe a circle cutting ED ins. Join aB; and let it meet p¥, which is parallel to AG,1 F. Draw Fa parallel to cp. ce and BF are the parts requirec to be cut off. For (DF=) CG: FB!:EA: AB, i. e. in the given ratio of thi parts to be cut off; and the difference between the squares 0 4 FA and AG Is aii to the square of «Fr, i. e. to the square o} ! | > i oeeeeeeeeneeetiieatinetttitiemmemtatisen pul IS re tat CD, or the given difference of the squares of the remainders. (42.) From two given lines to cut off' two others so that the 1 re: mainders may have a given ratio, and the sum of the squares ) the parts cut off may be equal to the square of a given line. ——— Let aB be one of the given lines, and in it take ac to the other given line, in the given ratio of the remainders. From c draw cp per- pendicular to aB, and equal to the second given line. Join ap, and draw cz parallel to AD; and with the centre B, and radius equal to the side of the given square, describe a circle, cutting cE in gE. Draw EF parallel to pc. Then Be, Ge will be equal to the parts to be cut off. Jom BE. ‘Che squares of BG, GE are equal to the square oj BE, 2. é. to the given square ; BO Ee ene ee ~~ ——— —_—_ ———. patel Spi a and AG : GF!:AC: CD, 2. €. in the given ratio of the remainders. ¢ (43.) Two points being given in.a given straight line; to defen! mine a third such that the rectangles contained by its distances ect. V.| GEOMETRICAL PROBLEMS. 137 “om each extremity and the given point adjacent to that extremity vay be equal. Let aB be the given straight line, : G F | a Oh fake E A Se ae and p the given points in it. On “oe | Y .c and DB as diameters let circles 4 baa: i; IN " e described, and let er touch them So ey 5 agandr. Bisect EF in G, and let all the perpendicular Gu; His the oint required, | From e draw any lines GNK, GUM cutting the circles. Take \ the centre of the circle AcE, and draw op perpendicular to -K. (Eucl. ii. 6.) the rectangle na, GK together with the square fPN is equal to the square of PG; to each of these add the quare of po; and the rectangle Ne, @k together with the quares of oP, PN (i.e. the square of OC) is aie to the squares fop, PG, i.e. to the square of oG, or to the squares of on, ig. But the square of on is equal to the rectangle cu, HA ogether with the square of oc; whence the rectangle NG, GK is qual to the rectangle cu, HA, togees with the square of ua. In he same manner it may be shown that the rectangle Le, Gm is qual to the rectangle pu, HB together with the square of HG. 3ut since the rectangle ne, Gk is equal to the rectangle Le, em, he rectangle CH, HA is equal to the rectangle Du, uB. Cor. If 1m be a mean proportional between cu and HA; G=GE. (44) Through the point of intersection of two given circles, to traw a line in such a manner that the sum of the respective rect- ingles contained by the parts thereof which are intercepted between he said point and their circumferences, and given lines a and B, aay be equal to a given square. _ Let the two circles c1p, cEK cut each ‘ther in the point c; from c draw the Jameters cD, cr. In cp take the point (such, that cp: cr::a:s. Jom EF; nd on it as a diameter describe a semi- La : : 2 rcle, i which place eG a third propor- 138 GEOMETRICAL PROBLEMS. [Sect. | tional to a and the side of the given square. Draw 1cK paralle to EG; it will be the line required. Join FG, and produce it to u. The angle pic is equal r FGE, 7. €. to FHC, .*. FH 18 parallel to D1; Peer on ar oe | *, the rectangle a, cu, is equal to the rectangle B, c1. No since EG is a third proportional to a and the side of the give) square, the rectangle a, EG will be equal to the given square But the rectangle a, EG, is equal to the ee A, HC, an A, CK, i. e. to ae rectangles B, 1c, and a,cK; .*, the rectangle A, KC, and B, 10, are equal to the given epee. (45.) Through a given point, to draw an indefinite line, such that if lines be drawn from two other given points, and formin| gwen angles with it, the rectangle contained by the segments in tercepted between the given point and the two lines so drawn, sha’ be equal to the square of a given line. Let a be the given point through which the line is to be drawn; B and c the other given points. Join AB, Ac; and on them describe segments of circles ADB, AEC, containing angles equal to the given angles. . B Draw either diameter ar, on which produced take ac suck, that the rectangle ra, ac, may be equal to the given squar Draw GE perpendicular to GF; join EA, and produce it bot! ways ; it is the line required. | Join pr. The angles at a and p being right angles, th triangles AGE, ADF are similar, | . EAS AG!: FA! AD, *, the rectangle EA, AD is equal to the rectangle FA, AG, 4. r to the given square ; and CE, BD form with Ep angles equal t! the given angles. | If ce does not meet the circle, the problem is impossible. — (46.) Through a given point between two straight lines contain ing a gwen angle, to draw a line such that a perpendicular upo oct. V.| GEOMETRICAL PROBLEMS. 139 from the given angle may have a given ratio to a line drawn om one extremity of it, parallel to a line given in position. Let as, ac be the lines forming the 4 tea AR yen angle Bac, and Da point between em, and Ax the line given in position. traw any line FG parallel to an, and x ike An : FGin the given ratio; and with ¢ ae centre A, and radius a4, describe a circle, to which draw 1K atangent. Join Ar; and through p draw uMN parallel to kK, and Lo parallel to re. UN is the line required. For a1 is perpendicular to FK, and .*, AM to LN; and LO is arallel to AB, and FG: LO::; AF: AL?: (AI=) AH; AM, “. AM: LO!:AH: FG, Zz. e. in the given ratio. | (47.) Through a given point between two indefinite straight lines ot parallel to one another, to draw a line which shall be ter- vnated by them, so that the rectangle contained by its segments ‘tall be less than the rectangle contained by the segments of any ther line drawn through the same point. Let aps, ac be the given nes meeting in A. In ac ike any point p, and make foe— AD. Join DE; and rough 1 the given point draw ig parallel to DE. FIG is ae line required. Draw the perpendiculars ro, Go meeting ino. Then since ‘Dis parallel to FG, and the angles AED, ADE are equal, .. \FG and ager are equal. But Aro=aao, each being a right ngle, .. og¢F=oFre,and or=oG; a circle .*. described from ae centre o, and radius 04, will pass through F, and touch aB, .€ in G and F, since the angles at G and F are right angles. set now any other line HKLM be drawn through 1, and ter- ainated by AB, ac. Since all other points in ar but @ are vithout the circle, H is without the circle; .. uM cuts the circle } si “i 140 GEOMETRICAL PROBLEMS. [Sect. y, in K; and for the same reason also in tu. Now the rectang: K I, IL is equal to the rectangle G1, 1r. But the rectangle x, IL is less than the rectangle u1,1M; .*, the rectangle «1, : is less than the rectangle H1, 1m. In the same manner it m be shown that the rectangle @1, 1F is less than the rectang contained by the segments of any other line drawn through, and terminated by aB, Ac. SECTION VI. (1.) To describe an isosceles triangle on a given finite siraig' line. Let aB be the given straight line. Produce it, if necessary ; and make ac and BD, each equal to one EO}, of the equal sides of the triangle. With a and B as centres, and radii Ac, BD, describe circles, cutting yi each other in E; join AE, BE; AEB is the triangle DA B_ required. For Ak = AC p68 D.=- 56. of two squares, whose sides are given. Take a straight line aB terminated at a, and a cut off Ao equal to a side of the greater, and os / | equal to a side of the lesser square. Withoas 4 © = centre, and radius 0 A, describe a circle ocp; and from B dra. Bo at right angles to ap. The square described upon BC the square required. Join oc. (Kucl. i, 48.), the square described upon Be equal to the difference of the squares on oc and oB, i. ¢.on A and OB. | Cor. Hence a mean proportional between the sum and diffe ence of two given lines may be determined. | (2.) To describe a square which shall be equal to the differen | (3.) Yo describe a rectangular parallelogram which shall | ect. VI. | GEOMETRICAL PROBLEMS. 14] mal to a given square, and have its adjacent sides together equal | a given line. Let aB be equal to the given lme. Upon it , D ascribe a semicircle ApB. From a draw ac ‘ama erpendicular to AB, and equal to a side of the «A EB ven square. Through c draw cp parallel to s, and let fall the perpendicular pr. The rectangle contained y AE, EB will be the rectangle required. For the rectangle ar, EB is equal to the square of ED, which equal to the square of Ac, i. e. to the given square; and aB the sum of the adjacent sides an, EB. t (4.) To describe a rectangular parallelogram which shall be ual to a gwen square, and have the difference of its adjacent des equal to a given line. . Let aB be equal to the given line. On it as c ‘ameter describe a circle. From a draw ac at ght angles to aB, and .*, a tangent to the circle 4 A; make ac equal to a side of the given juare. Take o the centre; join co, and pro- aceittop. The rectangle contained by sc, cp is the rect- agle required. For the rectangle Ec, cD, is equal to the square of Ac, 7. e. | the given square ; and the difference of the sides containing 'e rectangle is ED =AB=the given line. | 2 , (5.) To describe a triangle which shall be equal to a given equi- teral and equiangular pentagon, and of the same altitude. Let asnpce be the given pentagon. Join A ‘, AD; and produce cp indefinitely both i | B ays. Through B and £ draw BG, EF re- vectively parallel to ap and ac. Join ar, G AFG is the triangle required. Since AD is parallel to Be, (Eucl. i. 37.) the triangles aBD, GD are equal; and for a similar reason, AEC=AFC; .’. the 142 GEOMETRICAL PROBLEMS. [ Sect. y triangles ABD, AEC are equal to aap, AFC; to these eque, add the triangle Apc; and the pentagon ABDCE is equal; the triangle acr; and they have the same altitude, viz. 1] perpendicular from A upon Dc. lls Let asc be the given isosceles triangle. On ac describe an equilateral triangle a pc, and from p draw DE perpendicular to Ac; it will also bisect Ac and pass through Bs. On DE describe a semicircle; and from B draw BF perpendicular to DE, meeting the circle in F, with the centre £, and radius EF, { describe a circle meeting ED in G; draw @Hu, G1 parallel to D | DC respectively ; the triangle ie is equilateral, and equal ABC. ? | Since GH is vt to AD, and ai to po, the inangle GH, ADC are similar; but apc is equilateral, and .. also @HI) equilateral. | | Also (Eucl. vi, 8. Cor.) ED: EG !! EG: EB, and (Kucl. vi. 2.) ED: EG:: EA! EH, . EG: EB‘: EA! EH, and .*. (Kucl. vi. 15.) the triangles aH, EBA are equal. B GHE=GIE, and BAE=BCBE, .°. also GHI=BAC. i (7.) To describe a parallelogram, the area and perimeter which shall be respectively equal to the area and perimeter of given triangle. Let asc be the given triangle. Produce aB to D, making BD=BC; bisect AD in E; draw BF parallel to ac; and with the centre a, and radius AE, describe a circle cutting BF in G. Join ag; and bisect ac in w. Draw uF paral- lel to AG. AGFH is the parallelogram required. > ect. VI. | GEOMETRICAL PROBLEMS. 143 | For HF=AG=AE, ..HF and AG together are equal to ap, eto ABand Bc together; and@r=an=une, .”, the perimeter f acru is equal to the perimeter of anc; and AGFH is ouble of a triangle on the base au and between the same arallels, and .*. is equal to the triangle a Be. } | | | | | ene eae _(8.) To describe a parallelogram which shall be of given alti- ide, and equiangular and equal to a given parallelogram. } Let ascpv be the given E 1 7 arallelogram, and EF the A B iven altitude. Draw bu and ak ao gat right angles to rx; and i ; the point F, in the line er, r a = ‘ake the angle arr equal to cpa; take re a fourth propor- onal to F1, AD and pc; and from é draw Gx parallel to Ft1, eeting mH produced in H; 1F@H is the parallelogram quired. ? For its altitude is rF; and the angle GFI=CDA, .. FIH= AB; whence the parallelograms are equiangular; and they are yual; since the sides about the equal angles are reciprocally soportional. (Eucl. vi. 14.) ((9.) To describe a square which shall be equal to the sum of ly number of given squares. Let az be a side of one of the given squares. ¥ rom B draw Bc perpendicular to aB, and equal to side of the second square. Join ac; and from c c ‘aw CD perpendicular to it, and equal to a side of e third square. Join ap; and from p draw DE rpendicular to Ap, and equal to a side of the urth. Join an. The square of Az is equal to the squares of B, BC, CD, DE. Since the angles ADE, ACD, ABC are right angles, the square _ AE is equal to the squares of ab, Dx, @. e. to the squares of ‘Cc, CD, DE; and .*, to the squares of AB, BC, CD, DE. And by proceeding in the same manner whatever be the 8 A B 144 GEOMETRICAL PROBLEMS. [ Sect. v| number of given squares, one equal to their sum may by found. Cor. Hence lines may be found, which have the same rati as the square roots of the natural numbers. | (10.) Having given the difference between the diameter an side of a square ; to describe the square. } Let aB be the given difference. Draw Ac,AD, pb E | each making half a right angle with ap; and || complete the square rr. Take AG=the differ- ence between BA and Br. Since the ratio be- ¥ | tween the side of a square and its diameter is given, that of their difference to the diameter is also giver Take .., AH: AB 1: AB: AG; and through u draw HC, Hy perpendicular to ac, AD; cD is the square required. . For pc being a parallelogram is also (Kucl. 1. 46. Cor.) rec) angular ; one the angle pau being half a right angle, is equi to DHA, ... DA=DH; whence ine sides are equal; and th figure is a square. And since BG=BF, HB=HC; and AB the difference between the diameter and side. (11.) To divide a circle into any number of concentric equ| annul. | Let asc be the given circle, and o its B centre. Draw any radius 0 A, and divide it ee into the given number of equal portions in ESSN the points D, B, F,G, &c. On oa describe Bake a semicircle, and draw the perpendiculars DH, El, FK, GL, &c. Join on, O1, &c. » | and with the centre o and radii on, o1, &c. let circles be described; they will divide the circle aBc as_ required. Since the areas of circles are in the duplicate ratio of the radii; the area of the circle whose radius is 04 is to that who radius is on in the duplicate ratio of 0A : OH, 7. e. in the rat of 0A : OD; .*. the area of the first annulus will be to the ar of the circle ane radius isoD::AD: op. And in the san ect. V1. | GEOMETRICAL PROBLEMS. 145 anner the area of the second annulus, will be to the area of e circle whose radius is OD, as DE : OD; and since AD=Dp, e annuli will be equal. The same may be proved of all the st. Cor. The construction will be nearly the same, if it be re- aired to divide the circle into annuli which shall have a given itio; by dividing the radius Ao in that. proportion. | (12.) In any quadrilateral figure circumscribing a circle, the (vosite sides are equal to half the perimeter. a Let Ancp be a quadrilateral figure circum- sibing the circle EFG; its opposite sides are Aika cual to half the perimeter. \ For (Eucl. iii. 36. Cor.) an=au,and pH= “© Hij, AD is equal to az and pe together. In j-><¢<; te same way BC is equal to BE and Gc together, | AD and Bc together are equal to aB and pc together. (13.) If the opposite angles of a quadrilateral figure be equal two right angles, a circle may be described about tt. Let ascp be a quadrilateral figure, whose i Oposite angles are equal to two right angles. / Jom sp; then if a circle be described about ' triangle Bcp it will pass through a. For the zle BCD and the angle in the segment BED, are ether equal to two right angles, and .*, equal to Bop, BAD 3 ence BAD is equal to the angle in the segment BED; and. nust be a point in the circumference; or the circle will be cribed about ABCD. + D 14.) A quadrilateral figure may have a circle described about iM the rectangles contained by the segments of the diagonals be . 4et ABCD be a quadrilateral figure, the rectangles contained L a’ ee 146 GEOMETRICAL PROBLEMS. | Sect. - by the segments of whose diagonals are equal, viz. be the rectangle AE, EC, equal to BE, ED. A} — an Describe a circle about the triangle asc; if it / does not pass through p, let it cut BD in F; then Y (Eucl. iii. 35.) the rectangle BE, EF, is equal to the rectangle AE, EC, ze. to the rectangle BE, ED, by the suppc tion; whence EF is equal to ED, the less to the greater, wh is impossible; .*. the circle must pass through D. | (15.) If from any point within a regular figure circumserid about a circle perpendiculars be drawn to the sides; they. together be equal to that multiple of the semidiameter, which expressed by the number of the sides of the figure. Let asBcp be a regular figure circumscribed about the circle; and from any point o, let per- pendiculars ox, oF, oG, &c, be drawn. Take s the centre of the circle. Join sp, sc, sH. Then the figure will be divided into as many triangles round s and o, as there are sides of the figure ; now the triangle scD : 0cD:: SH: OG; and the same beg perpendicular from s being a radius of the circle. (16.) Ifthe radius of a circle be cut in extreme and mean ra); equiangular decagon inscribed in that circle. | Let ao, the radius of the circle asc, be cut } in extreme and mean ratio in D; AD is equal to (-* ; the side of an equilateral and equiangular deca- *{"" gon inscribed in the circle. In the circle place ¢ ct. VI. | GEOMETRICAL PROBLEMS. 147 cequalto ap; join co. Then (Kucl. iv. 10.) the angles at and c are double the angle at 0; whence aoc is one fifth srt of two right angles, or one tenth part of four right angles, ié. of the angles at 0; and .. ac is the side of a regular (cagon inscribed in the circle. STE Ge ee! Se Oe (a7) Any segment of a circle being described on the base of a langle; to describe on the other sides segments similar to that _ the base. Let asc be a triangle, on the base ac which a segment of a circle ADC is (scribed. Produce AB, CB to E and D. in AD, CE; and through 4, p, B, and (E; B let Adee be described ; the seg- ents ADB, BEC are similar to apc. For the angle apc being in the seg- rents ADB, ADC, those segments are similar. For the same lison the segments ADC, BEC are similar. And since the egles ADC, AEC are equal, .. the segments ADB, BEC are sailar. 18.) If an equilateral triangle be inscribed in a circle; the are described on a side thereof is equal to three times the svare described upon the radius. Let asc be an equilateral triangle inscribed in A aircle, From a draw the diameter ap, and take ( cin the centre; jo BD, BO. Then the angle o\ | PD= BAC=BCA=BDO, ..BD=B8O0; and the Ki Siares of AB, BD are Rl to the square of ap, a : to four times the square of Bo, or Bp; and . the square of aB is equal to three fae the square of BD OBO. 19.) To inscribe a square in a given right-angled isosceles timgle. Let anc bea right-angled isosceles triangle, having the side L2 vi 148 GEOMETRICAL PROBLEMS. [ Sect. | BA=BC. ‘Trisect the hypothenuse ac in the Bp! points p, £; and from pb, & draw DF, EG@ per- JK pendicular to AC; join FG; DFGE is the square SS | required. Since the angle paF is half a right angle, and the angle at) aright angle, .. the angle pra is half aright angle, and equ to DAF; whence DF=pA. In the same manner it may } shown that Ec=rc. But ap=eEc; and .. Fp, DE and y are equal; and (Kucl. i. 33.) FG=pE; .*. the figure is eq lateral. And it is rectangular, (Kucl. i. 46.) since the angl at D and £ are right angles; .*. it is a square. (20.) To inscribe a square in a given quadrant of a circle, Let aos be the given quadrant, whose centre is 0. Bisect the angle aos by the line oc. Draw cE, cp parallel to 0A,OB. DE is a square. For the angle cop=cox, and cpo=cEO, ¢&= since each of them with por make angles equal ¥ to two right angles, and coO_is common, .. cE=cp. And 7 construction CE=oD, and oz=cpD, .°. the figure is equilater. And the angle Dox is a right angle, .*. (Eucl. i. 46. Cor.) alls angles are right angles; and consequently the figure isa square. (21.) Yo inscribe a square in a given semicircle. Let acs be the given semicircle; take o its centre, and from B draw Bp perpendicular and equal to Ba. Join ov, cutting the circum- g__o al ference in £; and from © draw EF perpendi- (/\el \\ cular to AB, and EG parallel to it; draw eu (NJ parallel to nF. Then nu is the square required. “ ™ Join og. Since EG is parallel to an, the angle COH=BC and the angles at H and F are right ete and Go= OE, HOOF. Now EF! FO :: DB: BO, A “.EF=2FO=FH; the figure is .. equilateral; and it is, } construction, rectangular; .*. it is a square. sf ct. V1.| GEOMETRICAL PROBLEMS. 149 Cor. Since FE=2F0, FE’=40F’, and o£’=5oFr’; and if ix be drawn perpendicular to on, of : 0K 3:5: 1. , (22.) To inscribe a square in a given segment of a circle. ‘Let a1B be the segment of a circle, whose aE PMA Ilse AB is bisected in p. From B draw Bc per- a /) ndicular to BA and equal to Bp. Bisect BD () | i and join cE. Draw pe parallel to cE, and A HD Eis ¢tocs. Take pa=pr; draw 1 perpendicular to an, and | parallel to Gr; Join Gi. ¥1 is the square required. Since ep and @F are respectively parallel to cz and cz, GE RD) OB 248 By el \¢rF=2rp=FH. Take o the centre, and draw OL, om per- ndiculars to H1, FG; then since HD=DF, OL=OM, .*. (Kucl. i) 14.),1L=¢8M; but LH=FM, .. IH=GF; whence IG=HF, ad the figure is equilateral; and since the angle at F is a right evle, the figure is rectangular, and .*. is a square. (28.) Having given the distance of the centres of two equal ocles which cut each other; to inscribe a square in the space i'luded between the two circumferences. Let a and B be the centres of two equal circles, g viuich cut each other in c and p. Join AB, and ~,, Lect it in BE; and at the point = make the angle We ¢:F=half a right angle; and from F draw FG@H per- fadicular to as. Make rr=EG; and through 1 -) diw KL perpendicular to AB; join KF, LH. KH is a square. Since EI=EG, BI=AG, and .°, (Kucl. ii. 14.) kL=FH; and try are parallel, ... Kris equal and parallel to Lu, .. KH isa vallelogram. Also since GEF is halfa right angle, and Ear ight angle, .. er@ is half a right angle, and .-. equal to F; whence EG=GF, and rH=1iG. But xF is equal and ‘allel to 1G (Eucl. i. 33); .°. the four sides are equal; and ‘K is aright angle, ., the figure is rectangular (Kucl. i. 46. Cr.), and consequently is a square. - | : phat. 150 GEOMETRICAL PROBLEMS, [ Sect. y (24.) In a given ‘segment of a circle to inscribe a rectanguli parallelogram, whose sides shall have a given ratio. Let asc be the given segment of a circle. 5 Bo | From A draw Ap perpendicular to ac, and > make ap : AC in the ratio of the sides. Com- \qi—¢< Kk) plete the parallelogram ar. Bisect ac in «4, | and join DG; and from F draw FH perpendicular, and F1 me allel to ac. Draw1x parallel to ru; ur is the rectangul: parallelogram required. | Since FH is perpendicular to ac, it is parallel to ap; | and.4 °F Hs 1G. SAD eas | whence FH : HK!:AD: AC, 2. e.in the given ratio. And rua being a right angle, all tk angles of the figure are right angles. (25.) In a given circle to inscribe a rectangular parallelogra: equal to a given rectilineal figure. Let AEB be the given circle; on the diameter D AB describe a rectangular parallelogram aBcp ia equal to the given rectilineal figure; and let the il side pc cut the circumference in x. Join AE, x EB. Draw BF parallel to An, and join AF. FE is the rectangular parallelogram required. \ For the angle EBF is equal to the two angles EBA, ABF, 41. to EBA, BAE, and .*, is a right angle; and the angles pz an BFA are right angles, by construction; .*, also BAF is a righ angle; the figure ArBeE is therefore rectangular; and it | double of aps, and (Eucl. i. 41.) .. equal to ABCD, i. e. to th given rectilineal figure. The given figure must not exceed the square of half a diameter. } | (26.) In a yiven segment of a circle to inscribe an isosceles tr angle, such that its vertex may be in the middle of the chord, an the base and perpendicular together equal to a given line. . > 42,0 lp aa Stak sect. v1. | GEOMETRICAL PROBLEMS. 151 _ Let ane be the given segment. Bisect ac ap, and draw pe at right angles to ac, and qual to the given line. Make pr the half of JB; and join EF, meeting the circle inc. Draw iB ‘parallel to AC; join GD, DB, GDB is the riangle required, _ Since GB is parallel to Ac, it is bisected by OB. Also (Eucl. vi. 2.) Eu is double of Gu, and ., equal to 4B; .. GB and up together are equal to ED, 7.e. to the given ine; and since GH=HB, and the angles at u are right angles, 3D=pDB, .°. the triangle is isosceles. | If er does not meet the segment, the problem is impossible. When the line Fr£ cuts the segment, there are two isosceles tri- ingles p@x, pg that will answer the conditions; when it touches he segment, only one. | | (27.) In a given triangle to inscribe a parallelogram similar to ugiven parallelogram. | _ Let asc be the given triangle. In AB take any point p, and draw pF parallel to 1/4 ie AC; and make the angle FDE equal to one Bet | as ingle of the parallelogram, and take DF : DE - a ea ‘ n the ratio of the sides. Join AF, and pro- luce it to c; draw Gu, ur respectively parallel to FD, DE; and 3K parallel to u1. HX is the parallelogram required. | For ut being parallel to pn, and uG to DF, Hl: DE;:HAs DA:HG ; DP, f p Path cH CG. DE Di, 3. e. in the ratio of the sides. | Also the angle GH1=FrpE= one of the angles of the parallel- ogram, .*, H1K will be equal to the adjacent angle of the par- allelogram, and 1x is similar to the given parallelogram. / ' ; _(28.) In a given triangle to inscribe a triangle similar to a given triangle. i } | | | 152 GEOMETRICAL PROBLEMS. Let asc be the given triangle, in which the triangle is to be inscribed. In aB take any point p, and draw any line pF to the opposite side; and at the points p and F make the angles Fp, DFE equal to two of the angles of the given triangle to which the inscribed one is t be similar, ... the angle at = will be equal to the third angle Join AE, and produce it to G; and from @ draw GH, GI re spectively parallel to ED, EF; join HI. HIG is the triangl required. | Since DE and EFare respectively parallel to ua, @1, the a7 DEF is equal to nai. p Also DE: HG!!AE: AG::EF; GI, | whence (Eucl. vi. 6.) the triangles u@1, DEF are similar, and , HGI similar to the given triangle. scribe a square. ee ae a (29.) In a given equilateral and equiangular pentagon, to aT Let ascpe be the given pentagon. Join EB; and from © draw EF perpendicular and | equal tors. Join ar; and from a, where it © | cuts ED, draw Gu parallel to rr. Draw u1, GK parallel to uB. Join 1K. ux is the square required, Since HG is parallel to nF and utr to BB, F : | ; | i} ‘| HG > BFi,A ARO HI EB, | ] \ but sr=nEB, .. HG=H1I. And since An=AB, -. (Eucl. vi. 2. HE=1B; also Gk and pc being parallel to mB, and DE=BC_ .. EG=BK. The triangles EnG, 1K B, therefore have two side in each and the included angles equal, and .*. HG=1K, and thi — angle E H@=BIK, whence HG and rx are also parallel; there fore also GK is equal to u1; hence the four sides are equal; anc — the angle at n being a right angle, all the angles are right angles and consequently HK is a square. | = KE vI.| GEOMETRICAL PROBLEMS. 153 (80.) In a gwen triangle to inscribe a rhombus, one of whose angles shall be in a given point in the side of the triangle. Let ABC be the given triangle, and p the re given point. Join Bp, and produce it; and nf] Ne with the centre a, and radius ac, Fermi: a - f | PS circle cutting iting. Join AE; and drawpF | /) i parallel to it, FG parallel to ac, and Gu to FD. / FH is the rhombus required. i Since FD is parallel to AE, BF: FD::BA: / AE; and since FG is parallel to ac, 0 | Le | Bre FG*'BA > ACS: BA* AE, . FD = FG; and the sides opposite to these are equal, .. the fier FDHG is a rhombus. | (31.) To inscribe a circle in a given quadrant. Let asc be the given quadrant. Bisect the ® angle acB by the line cp; and at p draw DE touching the quadrant, and meeting ca pro- ducedinr. Make cr=Ar. FromrFdrawrFe ¢ at right angles to ac. Gis the centre of the arcle required. From ¢ draw Gu perpendicular to sc. Join pr. Since the angle pcx is half a right Bae and the angle at p a right angle, peer C= AC—FE;..:, the angle EDF=EFD; whence also GpF=GFp, and cp=GF; and since the angles noe. GCH are equal, and Gc common to “the right-angled triangles G FC, GHC, .. GF=GH; ... the three lines Gp, Gr, GH are equal, and ie circle described from the centre G, and distance of any one of them, will pass through the extremities of the other two, and touch the arc and sides in the points p, r, H, because the angles at those points are right angles. \ _ (82.) To describe a circle, the circumference of which shall pass ‘through a given point, and touch a given straight line in a gwen ‘point. 154 GEOMETRICAL PROBLEMS. Let AB be the given straight line, c the given point in which the circle is to touch it, p the point through which it must pass. Draw co perpendicular to As. Join cp; and at the point Dp make the angle cpbo=pco;; the intersection of the lines co and po is the centre of the circle required. Since the angle pco=cpo, co=po, and .., a circle de-. scribed from the centre o, at the distance op, will pass through | c, and touch the line 4B in c, because oc is perpendicular AB. (33.) Yo describe a circle which shall pass through a given point, have a given radius, and touch a given straight line. 2 Let as be the given straight line, and c the given point through which the circle must pass. In as take any point B; and from it draw BD at right angles to a B, and equal to the given radius; through p draw pe parallel to aB; and with the centre c, and radius equal to the given radius, describe a circle cutting DE in 0. 0 is the centre of the circle required. From o draw oF perpendicular to a8, it is equal to pB, i. & to the given radius ; and the circle described from the centre 0, | and radius or, will touch (Hucl. iii. 16. Cor.) the line a8 in#, and pass through o. & F Br (34.) Zo describe a circle which shall pass through two given points, and touch a given straight line. a Let a, B be the given points, and cp the given © 2 straight line. Join as. And 1. Let cp be parallel to as, Bisect aB in 5, and draw WF perpendicular to AB, and ..tocp. Join FA, and make the angle : FAO=AFO; then will o be the centre of the circle required. _ | Since the angle FAO=AarFo, Ao=or. But AE=EB, and | = Sect. vt. | GEOMETRICAL PROBLEMS. 155 30 is common to the triangles AHO, BEO, and the angles at E ight angles, .. Ao=oB. Whence Ao, oB, OF are all equal ; and the circle described from the centre o, at the distance of any one of them, will pass through the extremities of the other -wo, and touch the line cp, since OF is perpendicular to cp. 2, But if aB is not parallel to cp, let chem be produced to meet in £; and take ‘F a mean proportional between EA and sp. Join FA, FB; and describe a circle ibout the triangle ArB; it will be the circle required. Since EF is a mean proportional between EA and EB, EF ouches the circle (Eucl. iii. 37.) which passes through a and B. (35.) To describe a circle, the circumference of which shall pass through a given point, and touch a circle in a given point ; the two points not being in a tangent to the given circle. _ Let a be the given point in the circumfer- ence of the circle whose centre is 0; B the given point without. Join BA, and produce it to p. Join op; and through a draw 0Az; and draw Be parallel to op, cutting OAE in ‘B. & is the centre of the circle required. Since (Eucl. i. 29.) the angle opa is equal to ABE, and OAD to BAB, .*. the triangles opA, ABE are similar, and op being equal to OA, AE will be equal to eB; a circle .*, described with the centre E, and radius £ a, will pass through s, and touch the circle ApF in the point a, since the line joming the centres passes through a. | $ / - \ _ (86.) To describe a circle the centre of which may be in the per- pendicular of a given right-angled triangle, and the circumference pass through the right angle and touch the hypothenuse. | Let EAD be the given right-angled triangle, having the angle at aaright angle. Make rco=na, Join Ca; and draw co 156 | GEOMETRICAL PROBLEMS. at right angles to ED. The circle described with o as centre, and radius 0 A, will be the circle required. Since EA=EC, the angle EFCA=E AC, and ECO, EAO are equal, being right angles; .°. ocA=oaAc and oA=oc. The circle .-. described from the centre o, and radius 0 a, will pass through the extremity of oc, and touch ED inc, because Co is at right angles to ED. (37.) To describe a circle which shall pass through the extremi- ties of a given line, so that if from any point in its circumference a line be drawn making a given angle with the given line; the rectangle contained by the segment it cuts off and the given line, may be equal to the square of the line drawn from the same poll to the farther extremity of the given line. Let aB be the given line. On it describe a Cc ‘ ae JX segment ofa circle containing an angle equal to fi \ the given angle. Complete the circle; it will be ee NS the one required. x— D From any point c draw cp, making with ap e | the angle apc equal to the given angle; join ca, cB. Since the angle cp A=acpB, and the angle at A is common, the trian- gles ACD, ABC are equiangular, and therefore AB:AC{: AC; AD, whence the rectangle contained by AB, ap, is equal to the Square of AC. (38.) To determine a point in the perpendicular let fall from the | vertical angle of a triangle on the base; about which as a centre a circle may be described touching the longer side, and passing | * through the opposite angular point. Let asc be a triangle, and from s the vertex let pp be drawn perpendicular to ac. In pp take any point x, and from | it draw EF perpendicular to as; and from £ to Bc, draw | - Sect. v1. | GEOMETRICAL PROBLEMS. 157 -EG=EF; from c draw cn parallel to an, and ag from u draw u1 perpendicular to AB; 4 is the 1é ZN | point required. ir a: _ Since EF is parallel to 1, it ie Nis | FE HI iy BE. BH; Aas D and since GE is parallel to uc, | GE:! HC ': BE BH, | VEE HT GR AOS but, by construction, FE=EG, .. HI=HC; and a circle de- scribed from the centre n at the distance u1, will pass through c, and touch 4B in 1, since the angle m1B is a right angle. f (39.) To describe a circle which shall have a given radius, and “its centre in a given straight line, and shall also touch another giwen straight line inclined at a given angle to the former. Let aB be the given line, in which the centre is to be; Be the line which the c Ft _—_f circle is to touch. | Pps | Insc take any point c, and draw cp at » Go right angles to it; and make cp equal to the is ‘given radius. Through p draw po parallel - to cB; 0 is the centre of the circle required. ' Through o draw ok parallel to nc; .*. co is a parallelogram ; ‘whence 0 is equal to Dc, i. e. to the given radius. With the centre o, and radius on, describe a circle; it will touch cx in |B, because co being a parallelogram, and rcp aright angle, | CEO is also a right angle. } (40.) To describe a circle, which shall touch a straight line in a \ gwen point, and also touch a given circle. Let aB be the given line, and c the given point in it, o the ‘centre of the given circle. Draw cp perpendicular to an, and _0£ parallel to cp. Join cx, meeting the circumference in F. Join oF, and produce it to meet cp inp. D is the centre of - the circle required. | | ~ 5 158 GEOMETRICAL PROBLEMS. [ Sect. viv Since the triangles OFF, CFD are similar, i __ ae and OEF=OF, ... FD=DC; consequently a cir- | cle described with the centre p, and radius pF, will pass through c, and touch 4B in c, because the angles at c are right angles; and it will | touch the given circle in F, since the line join- ing the centres passes through F. E (41.) To describe two circles, each having a given radius, which shall touch each other, and the same given straight line on the same side of it. | Let aB be the given straight line. From B ! any point A in it, draw ac at right angles to it, and make ac, aD, equal to the given radu. Produce ca to E, making AE=AD. Draw po parallel to AaB; and with the centre c, and radius GE, describe a circle cutting DO in o. c and o will be the centres of the oid required. . Join co; and draw os perpendicular to as; then DAB: being a right angle, as also ABO, .*. AD is parallel to Bo; and. DO was drawn parallel to aB, .. AO is a porallelorraam and oB=AD. With the centres c and 0, and radii ca, oB describe circles, they will touch a8, since the angles at a and s are right’ angles; they will also touch each other, for co is equal to cz, or to cA and Akg, i. e. to CA and AD, or the sum of the radii. 1 (42.) To describe a circle passing through two given points, and | touching a given circle. | Let a and B be the given points, and cpE the given circle. Describe a circle through A and B, and cutting the given circle in p and E. Join DE, EB, DA, AB. Then the angle EDA= EBA; if ., DE and BA be pro- duced to meet in F, the cee FDA will be similar to the triangle FBE; 28 Sect. VI. | GEOMETRICAL PROBLEMS. 159 | and 4 .-DE RA 2; BF SRB; or the rectangle pF, rx is equal to the rectangle ar, FB. Draw ¥@ a tangent to the given circle; then the square of FG is equal to the rectangle EF, FD, and . en the rectangle BF, FA; whence a circle described through the points a, G, B, will tnireb the given circle, since it touches Fra. ) | (43.) To describe a circle, which shall pass through a given point, and touch a given circle and a given straight line. ' Let asc be the given circle, p the given point, and eF the given straight line. Through o draw AOE perpendi- cularto EF. Join ap; and divide it in '@, so that the rectangle AG, AD, may be equal to the rectangle ac, AE. Through @and p describe a circle touching EF in /F; this will also touch the circle anc. ' Draw the diameter ru; it is (Kucl. ii. 18.) parallel to az. Join ar, meeting the circle in B. Join cs. The triangles ABC, AEF having the angle at a common, and the angles aBc, ) - right angles, are similar; whence AGINAB AR + ARE, to the rectangle AG AD; .°. B is a point in the circle HDF. ‘Take 1 the centre; join os, Bi. Since ac is parallel to F1, ithe angle oAB=BFI; but oAB=OBA, and IFB=IBF, .”. OBA ‘=IBF; and os1 is a straight line, which joins the centres of ‘the two circles, which .°. touch each other. IP the rectangle AB, AF is equal to the rectangle ac, AB, 7. é. (44.) To describe acircle which shall touch a straight line and two circles given in magnitude and position. Let a and B be the centres of the two circles, and cp the line given in position. From B let fall the perpendicular BE, and produce it, making Er = the radius of the circle whose centre is A. Through F draw re parallel tocp. With the } i 160 GEOMETRICAL PROBLEMS. centre B, and radius equal to the difference of the radii of the two circles, describe a circle; through A let a circle be described, touching the line aF and the last described circle (vi. 43.); and let a and u be the points of contact. The centre of this circle will also be the centre of the circle required. Let o be the centre; join 04, 0G, oH; and with the centre o, and radius o1, describe the circle 1kKL. Since LGQ=KH=AIT, ¥, OL=o0K=01; the circle 1KL .*. touches cp in L, and th circle, whose centre is A, in 1; and since 0B is equal to the difference between 04 and HB, 2. e. between 0A and (1A—BRK); or is equal to ok and xB together, .*. it touches the cirele whose centre is B, in K. | (45.) To describe a circle which shall touch two given straight lines, and pass through a given point between them. Let AB, cD be the given lines, and & the given point. Produce the lines to meetin F. Bisect the angle BFp by the _ ayy line r@; and from = draw n@ perpendi- * ~—_ cular to FG, and produce it both ways to B andp. Take GH=GE; and make pI q a mean proportional Beeneen DE and DH; a circle described through the points H, £, 1, will touch cp. For the rectangle D&, nu, is equal to the square of pr. An for a similar reason it will touch aB; since the rectangle BH, BE, is equal to the rectangle ED, DH. | If the lines aB, cD be parallel; through the given point E draw DEHB perpendicular to AB or CD; bisect it in Gg, and make GH=GE. Take DI a mean proportional between px and DH; and a circle described through 1, & and u will be the circle required. | ¥ 1 on Sect. vi. | GEOMETRICAL PROBLEMS. 161 (46.) To describe a circle which shall touch two given straight ines, and also touch a given circle. Let aB, cp be the given straight ines, EFG the given circle, whose entre is 0. Draw H1, KL parallel o the given lines, so that their per- yendicular distances from those lines may be equal to or the radius of the riven circle. By the last problem lescribe a circle touching H1, KL, md passing through o the centre of he given circle. Let p be the cen- ire of this circle; it will also be the centre of the circle -yequired. Join PM, PN, PO. Since these lines are equal, and Ma, RN, )F are also equal by construction, .. Pa, PR, PF are also equal ; md a circle described from the centre p at the distance of any me of them, will pass through the extremities of the other two, ind touch the lines aB, cp, in q@ and R; since the angles at hose points are equal to the angles at m and Nn, and .-. right mgles ; and it will also touch the circle EFG in F, since oP the ine joining the centres passes through F. | (47.) To describe a circle which shall touch a circle and traight line, both given in position, and have its centre also in a nven straight line. | Let the circle whose centre is A, and he straight line Bc be given in position ; ind let cp be the line, in which the cen- re of the required circle is to be. On sc let fall the perpendicular as; and nake BF=AE; through F draw Fe paral- el to Bc, meeting DC in G. _ Join Ga; and draw cu parallel to it, meeting the given circle a # (if the problem be possible). Join an, and let it meet cin o. o is the centre of the circle required. M 162 GEOMETRICAL PROBLEMS. [ Sect. vr; Let fall the perpendicular or. Then (Kucl. vi. 2.) HO: OC‘: AH: GC !: FB: GC by construction, 7: BD : Dc, (Eucl. vi. 2.) 71 10 : oc, by sim. triangles ; | *, HO=103; and a circle described with the centre 0, and radius o1 or o#, will pass through the extremity of the otha and touch the line se in 1, and the circle in w; because the angles at 1 are right angles; and ao the line joining the centres, of the circles passes through H. : (48.) Through two given points within a given circle, to describe a circle, which shall bisect the circumference of the other. Let a and B be the given points within the circle whose centre is 0. Join AO; and pro- duce it indefinitely; and from o draw oc at right angles toit. Join ac; and draw cp at right angles to it, meeting ao produced in pD; aia through A, B, D duserihe a circle; it will Bilder the other in the points E, and F. | Join £0,0F. Then (Kucl. vi. 8.) : AO; 0C 3: 0C: OD, , the rectangle Ao, OD is equal to the square of oc, i. e. to the rectangle EO, oF; whence (Kucl. 11. 35.) EOF is a straight line; and since it passes through the centre of the circle ecr, it will be a diameter of that circle; .. the circumference ECF is equal to the circumference EGF, or the circumference of the given circle is bisected. ; | ; 4 : i (49.) Through two given points without a given circle, to de- scribe a circle, which shall cut off from the given one an are equal to a gwen arc. Let a and B be the given points, and cpr the given circle. Join AB; and bisect it in F; from F draw FG at right angles to as; and from any point G in it, at the distance Ga or GB, describe a circle ABD, cutting the given Sect. vi.| GEOMETRICAL PROBLEMS. 168 ) jircle inc and p. Join pc; and produce it to meet BA in H. from u draw HIE (ii. 20.), so that rm may be equal to the hord of the given arc. Through a, B and & describe a circle ; t will also pass through 1, and cut off the arc required. _ For the rectangle HI, HE is equal to the rectangle Ho, HD, ind .*. also to the rectangle HA, HB, whence I is a point in the uircle ABE. (50.) To describe three circles of equal diameters, which shall ouch each other. | Take any straight line a B, which bisect np; and from the centres Aa and B, with he equal radii ap, BD describe two cir- les. Upon AaB describe an equilateral riangle a BC, cutting the circles in & and +3 and with the centre c, and radius cE cr, describe another circle; these cir- , ‘Jes touch each other as required. | Since AB=AC, and ap is half of AB, ... Az, which is equal oit, is half of ac, and .. AE=xc. In the same manner CF =FB, .. the radii of the circles are all equal. And the circles ouch each other in D, E, F, because the lines joining the cen- Tes pass through those points. (51.) Every thing remaining as in the last proposition; to de- cribe a circle which shall touch the three circles. | Bisect the angles caB, cBA (see last Fig.) by the line ao, sO meeting in o. 0 is the centre of the circle required. Join oc. Since the angle caB=cBA, .*. their halves are qual, or OAB=OBA, .. oA=oB. Also since cA=AB, and .0 is common, and the angle CAO=BAO, .. CO=OB; and he three lines 0A, 0B, OC are equal; parts of which GA, HB, 'I are equal; .*. OG, OH, O1 are equal; and a circle described rom 0 asa centre, with a radius equal to any one of those lines, vill pass through the extremities of the other two, and touch | M 2 | 164 GEOMETRICAL PROBLEMS. | Sect. VI. the circles in the points @, H, 1; because the lines joining | centres pass through those RS | In nearly the same manner a circle may be described which shall touch the three circles on the opposite circumference. (52.) To determine how many equal circles may be placed round another circle of the same diameter, touching each other and the interior circle. | Let A be the centre of the interior circle, and ap its radius. Describe (vi. 50.) the circles DF, EF touching the circle px, and each other. Then the angle at a being one third part of two, or one sixth part of four right angles, subtends an arc ED equal to one sixth of the whole circumference. And the same being true ‘ every other contiguous circle, the number of circles which can be described touching each other and the interior one will be SIX. (53.) To draw two lines parallel to the adjacent sides of a given rectangular parallelogram, which shall cut off’ a portion, whose breadth shall be every where the same, and whose area shall be ” that of the parallelogram in any given ratio. Let ac be the given parallelogram. Pro- duce AaB to D making BD=Bc. On aD describe a semicircle, and produce cB to E; and let the ratio of the part to be cut off, to a the whole, be that of 1: 2. Make Br: BF ::2:n—1; and take BG a mean propor- tional between BE and Br. Biseet ap in 0; and with the centre o, and radius 0G, describe a semicircle, HGI; AH=I1D, will be the breadth of the part to be cut off. Make BL=BI, and Tay HK, LK parallel to the sides of the parallelogram; then ac : HL in the ratio compounded of the ratios of AB: BH and BD: BI, i. e. in the duplicate ratio Sect. vi1.| GEOMETRICAL PROBLEMS. 165 of BE: BG, or the ratio of BE: BF, i. e. in the ratio of as n—l, .. the portion Akc is to Ac in the ratio of 1 : n. (54.) To describe a triangle equal to a given rectilinear figure, raving its vertex in a given point in a side of the figure, and its jase im the base (produced if necessary) of the figure. Let ABcDEF be the given smn ectilineal figure, and pa Reet siven point in cD, which is o be the vertex of the tri- mgle, the base being in AF. foin c A, and draw BG paral- el to it; join CG, PG, PF, >E. Draw cu parallel to pc. Jom px. Draw p1 parallel to px, neeting FE produced ini. Join pr; and draw rk parallel to pr, meeting AFin K. Join pK; HPxK will be equal to aBCDEF. ' Since BG is parallel to ca, the triangles BAG, BCG are equal; he figure therefore is equal to GcpmrF. And since GP is yarallel to cu, the triangles GCP, GHP are equal. Again, since 11s parallel to px, the triangles p1m, pp are equal; ..PDEF s equal to the triangle P1rF, 7. e. to the triangle pKF, since IK is parallel to pr; whence the whole figure ABCDEF is equal to he triangles PHG, PGF, PKF, 7. e. to the triangle PHK. ivy AH G FE K _ (55.) On the base of a given triangle, to describe a quadrilateral igure equal to the triangle, and having two of its sides parallel, ne of them being the base of the triangle; and one of its angles eing an angle at the base, and the other equal to a given angle. _ Let asc be the given triangle, ac its base. ee it the point c make the angle acp equal to vi if NON jhe given angle; and let cp meet Bp drawn / |, |. varallel to ac, in the point p. On Bp de- “Wj 7 cribe a semicircle BED; draw AF parallel to / 'D, and FE perpendicular to Bp; and with he centre p, and radius px, describe the arc 166 GEOMETRICAL PROBLEMS. [ Sect. v1, EG. Draw cu parallel to ar, and ui to ac; Auic will be the figure required. Join uc. Since DG=DBs, BD DG 12D GDF, *, (Eucl. v. 19.) BG > GF 3: DG: DF {: HI: Ac, Now BG: GF !: BH: HA, . BH SHA HT 2 Ae. But the triangles Hc1, Ane are in the proportion of H1 : and the triangles Buc, Auc in the proportion of BH : HA, ", HOI PAHC {! BHO? Alic, or HCI=BHC; .*, ACH, and uci together are equal to ACH and BCH oscars or AHIC=ABC. (56.) A trapezium being given, two of whose sides are poral to describe on one of those sides another trapezium, having it: opposite side also parallel to this, and one of the angles at the basi the same as the former, and the other equal to a given angle. Let ascp be the given trapezium * whose sides AD, BC are parallel. Join ‘Sete BD; and draw CE parallel to it, meet- ing AB produced in &. Then the trian- gles BCD, BED are equal; and .°. the A : triangle AED is equal to aBcp. Hence (vi. 55.) a figure - ADGF may be described equal to ADE, and .. to ADCB. ty 7 (57.) Lf with any point in the circumference of a circle as centre and distance from its centre as radius, a circular arc be describ and any two chords be drawn, one from the centre of the circula arc, and the other through the point where this cuts the are, an parallel to the line joining the centres; the segments of eacl chord intercepted between the circumferences which are concaw to each other, will be equal respectively to those of the om? between the other circumferences. y With any point c in the circumference of the circle ABOa centre, and radius ce equal to the distance from the centre B, i Sect. v1. | GEOMETRICAL PROBLEMS. 167 let a circle prx be described. Join cr, and draw , #2 any chord cra; and through r draw urg@ parallel to ce; then will cF=FuH, and GF=Fa. Produce cE to B, and join HE. And since nGis vl YW, parallel to Bc, the angle FH 1s equal to HEB. Also since the circles are equal, the arc 1B is equal to the arc Fx (ii. 1. Cor. 1.), .”. the angle HEB is equal to FCE, .. FCE=FHE, and HC isa parallelogram ; whence HF=EC=crF. Also since the rectangle CF, FA is equal to the rectangle uF, FG, and HF=CF, .. FA =FG. , Cor. Hence if any number of lines be drawn parallel to Bx, and terminated by the two circumferences, each of them will be equal to BE. (58.) Iftwo diagonals of an equilateral and equiangular penta- gon be drawn to cut one another, the greater segments will be equal to the side of the pentagon; and the diagonals cut one an- other in extreme and mean ratio. Let ABDCE be an equilateral and equiangular A pentagon; draw the diagonals ED, Be cutting ¢ each other in F; EF and FB will be each equal to \ aside of the pentagon; and ED, BC are cut in F, ) in extreme and mean ratio. ' About the pentagon describe a circle. And since AB=CE, the arcs AB, CE are equal; .. AE is parallel to Bc. For the same reason, AB 1s parallel to EF; .. the figure ABFE is a parallelogram ; whence AB=FE, and AE=FB; but AB=AE, “. EF=FB, and each is equal to a side of the pentagon. "Also the angle DCF=CDF=DEC, .°. the triangles pcF, DEC are similar, and BD: CD':) GD : DF; Or ED: EF ?! EF: FD; h . . ° .. ED is cut in extreme and mean ratio. The same may also be proved of Be. (59.) If the sides of a triangle inscribed in the segment of a circle be produced to meet lines drawn from the extremities of the — | 168 GEOMETRICAL PROBLEMS. [ Sect. vi. base, forming with it angles equal to the angle in the segment ; the rectangle contained by these lines will be equal to the square described on the base. “Y . } Let the sides aB, cB of the triangle asc, ED inscribed in the segment aBc, be produced Pp to meet CE, AD, which make with ac, angles B equal to the angle aBc in the segment; the rectangle AD, CE is equal to the square of AC. # Since the angle anc=pac, and the angle at c is common! to the triangles ABc, apc, the triangles are similar. In the: same manner it may be shown that anc, AEC are similar; and *, ADC, AEC are also similar; whence | AD UAC. ACG Of, and the rectangle aD, cx is equal to the square of ac. (60.) If two triangles (one of them right angled) have the same base and altitude, and the hypothenuse intersect a line which is drawn bisecting the right angle; a line passing through this point. of intersection parallel to the base, and terminated by the sides of the other triangle, shall be a side of the square inscribed within it. bs Let apc be a right-angled triangle, and BOG ABC on the same base, have its altitude Be= AD; and let the hypothenuse pc, meet AF « which bisects the angle pac in F; through which draw Gu parallel to ac; 1H will be the side of a square inscribed in the triangle a Bc. From 1 draw 1k perpendicular to ac; then (Eucl. vi. 3.) DA’. ACBe DF EO cop Gia(Gas) 1K, also AC: BE‘!: 1H: (BL=) D@, . e@ @QUO DA: BE?11H: 1K; a But DA=BE, .. IH=IK; andif um be drawn perpendiai to Ac, 1M is a parallelogram, whose sides are equal; and the angles at K and m being right angles (Eucl. i. 46, Cor.) it is a square, p | sect. v1. | GEOMETRICAL PROBLEMS. 169 (61.) If on the side of a rectangular parallelogram as a diameter, » semicircle be described, and from any point in the circumference ines be drawn through its extremities to meet the opposite side roduced; the altitude of the parallelogram will be a mean pro- vortional between the segments cut off. On AB, the side of the rectangular paral- E elogram ABCD, let a semicircle AEB be de- cribed; and from any point 5, draw EA, 3B, and produce them to meet cp produced; ¢ ‘4D will be a mean proportional between ep ind cr. Since De is parallel to BA, the angle DG@A is equal to BAE, ond the angles at p and E are right angles, .. the triangles JAE, DGA are equiangular. In the same manner it may be hown that rcs is equiangular to BA, and .. to DGA; _vhence g i=) a L| GI. DAs (CB=—) Ass OF: :, | (62.) [fon the diameter of a semicircle a rectangular parallelo- ram be described, whose altitude is equal to the chord of half the emicircle, and lines drawn from any point in the circumference ‘0 the extremities of the base intersect the diameter; the squares f the distances of each point of section from the farthest extre- nity of the diameter will be together equal to the square of the Jiameter. which describe the rectangular parallelogram .E, whose side ap is equal to aBa chord of ialf anc; and from any point F in the semi- inele Bey FD, FE cutting the diameter in G nda; the squares of Au and CG are re mct ~ qual to the square of AC. Draw the perpendicular rx; the triangles D@A, DFK being : ~imilar, . DA WAG i EK : KD; and ecu, rKe being similar, i Let asc be a semicircle, on the diameter of : | | 170 GEOMETRICAL PROBLEMS. (CE=)\ DAS GH As OAKS KR; *.. DA’ DAGX CH FK? ){(KEX KD) 1F RE. He Now the square of DE is double of the square of DA, .*. th square of GH is double of the rectangle ac, cu. But th Square of AH is equal to the squares of aG, Gu and twice th rectangle AG, GH, 7. e. to the square of AG and twice the 2 angle AG, GC; ., the squares of Au and Gc are together equa to the squares of AG, Gc, and twice the rectangle AG, GC, 4. € to the square of a.c (Eucl. ii. 4.). Cor. The square of the part of the diameter intercepted b tween the two lines drawn from the point in the semiciralal double of the rectangle contained by the two extreme segments. | (63.) If on the radius drawn from the point of contact of « circle and its circumscribed square, another circle be described. and from any point in the outer circumference a line be drawi through its centre to the inner circumference, and through th same point another line be drawn parallel to the common tangen to the circles, and terminated by the side of the square and it diagonal ; these two lines are equal. Let o be the centre of the circle, circum- scribed by a square, whose diagonal is pr. On Ao describe a circle AoF; and from any point F draw a line rog; and through G@ draw H1 parallel to ap; FG is equal to H1. Join ar; and let H1cut aBin K. Since 1G is parallel to ap, it is perpendicular to AB; .. the angle Gxko is a right angle, and equal to AFo; and the vertical angles at o are equal, and GO=0A; .*. the trian- gles GKO, OFA are equal, and or=ox. But since OB=BE, “. (Bucl. vi. 2.) ok=x1; and .. or=K1, and OG=KH gm FG>=I1H. 4 (64.) Jf two sides of a trapezium inscribed in a circle be pro- duced, and from the same point in one side produced a line be o .) ia Jap produced, draw ru parallel to BE, meeting jcumference in @; the line joining 6, H will jalways pass through the same point. JAI1,pDG. The anglegGpH=GBC in the same seg- Sect. v1. | GEOMETRICAL PROBLEMS. 171 Trawn parallel to the other, intersecting the adjacent side of the ‘trapezium, and a second line to the extremity of that other inter- secting the circumference ; the line joining the two points of inter- \section, will pass through the same point. . Je | Let the two sides ap, Bc, of the trapezium ‘spp inscribed in the circle aBc, be produced, and let them meet in £; and from any point in the side pc in H; and join FB, meeting the cir- Let Gu produced meet the circle ini. Join ment, and .*. is equal to the alternate angle GFH; whence a circle may be described through the points G, H, D, F; and .*. the angle D@H=DFH =pDEB. But the angle pps being always the same, D@1, and . DAT, and also the arc pi will be invariable, and p being a fixed point, 1 must be also; ¢.¢. Gu will always pass through 1. (65.) If the diagonals of a quadrilateral figure inscribed in a circle cut each other at right angles, the rectangles contained by the opposite sides are together double of the quadrilateral figure. Let ancp be a quadrilateral figure inscribed in a circle, whose diagonals ac, BD cut each other at right angles in £; the rectangles contained by * AB, CD, and ap, Bc are together double of the figure. For (Eucl. vi. p.) the rectangles contained by 4B, cD, and AD, BC, are together equal to the rectangle ac, BD, 7. e. to the rectangles contained by Ac, BE and ac,g£D. But the rectangle contained by ac, BE is double of the triangle anc, and the rectangle contained by Ac, ED, is double of apc; hence the rectangles contained by 4B, cp and ap, BC are together double of ABCD. (66.) If a rectangular parallelogram be inscribed in a right- 8 172 GEOMETRICAL PROBLEMS. [ Sect. vi | angled triangle, and they have the right angle common; the rect angle contained by the segments of the hypothenuse is equal to th sum of the rectangles contained by the segments of the sides abou the right angle. Let asc be aright-angled triangle, in which B | the rectangular parallelogram DBEF is in- ‘< scribed, having one of its angles at B; the 7 | rectangle Ar, FC is equal to the rectangles ap, 4 F © © DB and BE, EC together. Draw £G@ perpendicular to rc. The triangles ADF, EFC being similar, ADL AK. EG (An P=) BD. *, the rectangle AD, DB is equal to the rectangle ar, Fa. Ih the same manner, AM ut Dir Ben Ox. C re | .. the rectangle BE, EC is equal to the rectangle ar, Go, whence the rectangles AD, DB and BE, EC are together equa to the rectangles ar, FG and AF, GC, @. €. to the rectangle ar, CyMochatl, 1.) | (67.) If on the diameter of a semicircle, two equal circles be described, and in the curvilinear space included by the three cir- cumferences a circle be inscribed ; its diameter will be to that gy the equal circles in the proportion of two to three. a On AaB the diameter of the semicircle ADB let two equal circles AcE, BCH be de- scribed; and in the curvilinear space let the circle DEG be inscribed; its diameter ; FG ign. Cis hot Let o and 1 be the centres of the circles. Join 01, which will pass through the point of contact n; and ; produce it to kx. From c draw cp perpendicular to aB, which will pass through 0. Then the rectangle K 0, oF is equal to the | square of oc; f and Of :0C 3: 0C: OK, 3 OF} OC 3. OF+O0C : OC+0K::CD > KE+CDi:1:2; | | | { ‘ect. VI. | GEOMETRICAL PROBLEMS. 173 and of: cp:: 1:8, “FG 1 (CD=) Ac ::2:3. | (68.) If through the middle point of any chord of a circle two hords be drawn; the lines joining their extremities will intersect ie first chord at equal distances from the middle point. Let acB be a chord of the circle asp, u isected in c; ; and let DCF, ECG be any nords drawn through c. Join p&G, EF cut- ng AB in 1 and 4; then will cr=cnH. Through u draw x HL parallel to pa, meet- ig DF in K, and GE produced in u. Because H is parallel to a1, the angle HLE=CGI=HFK, and the ertical angles at H are equal, .. the triangles LEH, HKF are Ageneular, 5 BLT: BM: MPR Ses nd the rectangle LH, HK 18 equal to ie rectangle HE, HF, _e. to the rectangle au, 18, or the difference of the squares of ‘candcu. The triangles CID, CHK may in like manner be vroved to be equiangular, as also the triangles CHL, CIG; vence | Kee Ce Se IVs EC, | anes H sh Ons TL? TC, | RK HX Gee Ce LX IG? tc’. But KH xX LH=AC’—HC’, and DIX IG=AC’—IC’, ° AC’—HO? ; HC’:;A0’—IC? : 10’ COMPARE: “2HOst Sect. vit.] GEOMETRICAL PROBLEMS. 179 _ Because cs = BG, and the angle cpp = GBD, .*. BD is per- _pendicular to c@; and .*, FH is parallel to ca. But since ar = FG, .. (Eucl. vi. 2.) ac is bisected by rH; which .°. passes through £. 7 ; : , (7.) If a point be taken without a circle, and from it tangents _ be drawn to the circle, and another point be taken in the circum- ference between the two tangents, and a tangent be drawn to it ; 'the sum of the sides of the triangle thus formed is equal to the sum of the two tangents. a From a given point p let two tangents DA, DB be drawn; and to c any point in the circumfer- ence between them, let a tangent Ecr be drawn. The sum of the sides of the triangle is equal to _ the two tangents DA and Ds. Eiuepince AE = EC, and FC = FB, .*. DE, EF, FD together are equal to Ap and pB together. In the same man- ner, if through any other point in the arc acB a tangent be ‘drawn, it will be equal to the two segments of DA, DB inter- ‘cepted between it, and the points of contact a and B; and the three sides of the triangle so formed will be equal to pa and j DB together. if (8.) Of all triangles on the same base and between the same parallels, the isosceles has the greatest vertical angle. | Let asc be an isosceles triangle on the base Bae oD ‘AC, and between the parallels ac, Bp. It has ja greater vertical angle than any other triangle ADC on the same base, and between the same “4 $ parallels. _ About asc describe a segment of a circle anc; and since B “is the middle point of the arc, and BD is seaestted to AC, BD is jatangent at s. Let the arc cut ap in £; jomec. Then the jangle ABC=AEC, and .°, is greater than ADC. - Cor. Of all triangles on the same base and having the same vertical angle, the isosceles is the greatest. For the i N 2 180 GEOMETRICAL PROBLEMS. [ Sect. Vin. triangle anc has the same vertical angle with anc, and iia — apc on the same base and between the same parallels; but) ADC is greater than AEC, .*, ABC is greater than AEC. it: (9.) If through the vertex of an equilateral triangle a perpemmel cular be drawn to the side, meeting a perpendicular to the Al drawn from its extremity ; the line intercepted between the vertex and the latter perpendicular is equal to the radius of the circum scribing circle. Let BE perpendicular to AaB meet AE, = . . . . B which is perpendicular to the baseac,in B; | BE is equal to the radius of the circle de- scribed about ABC. | BG D. Draw BF, CG perpendicular to the sides; ; and produce cc ton. Then cr is equal to the radius of the circle described about asc; and EBin is a parallelogram. : And since cr is equal to FA, (Eucl. vi. 2.) cr is equal to 1m, i. e. to the opposite side BE; and .*. BE is equal to the radius) of the circumscribing circle. aE (10.) Ifa triangle be inscribed in a semicircle, and a perpen-\ dicular drawn from any point in the diameter, meeting one side,\ the circumference, and the other side produced; the segments cut off will be in continued proportion. : Let asc be a triangle in the semicircle ABC; and from any point p in the diameter, let pF be drawn perpendicular to aD, meeting BC, the circumference, and aB produced, in E, G, F; DE: DG!:DG@: DF. For the angles at £ Bae equal, and the angles at B right angles, .*, the angle ECD is equal to BFD; and the angles at D : are right angles; .. the triangles npc, ApF are similar, and’ therefore a DF: DA!:: DC: DE, but-pA “Det VG PDOs ex @qu0 DF! DG!: DG: DE. kal Sect. V11.] GEOMETRICAL PROBLEMS. 181 , (11.) Ifa triangle be inscribed in a semicircle, and one side be sequal to the semi-diameter ; the other side will be a mean propor- tional between that side and a line equal to that side and the dia- \meter together. | Let asc be a triangle inscribed in the = semicircle, and let Bc be equal to the ‘)semi-diameter ; then will C™ i BC: BA‘: BA: BC+CA. se 2 CHa | Produce ac to p, making cp equal to the semidiameter. Take o the centre. Join BD, BO. Since Bo=BC, the angle BCO is equal-to BoC, 7. e. to OAB and OBA together, or to '2pac. But sca is equal to cpp and cps together, i. e. to 2cpB, since cB=cD; hence the angle BAC=BDC, and BA= \BD; also the triangles BAD, BCD are similar ; |. BC : (BD=) BA!:BA: AD, which is equal to Bc and ca ' (12.) If a circle be inscribed in a right. angled triangle; to ! determine the least angle that can be formed by two lines drawn i from the extremity of the hypothenuse to the circumference of the ‘circle. Let asc be a right-angled triangle, . ,m which a circle DEG is inscribed. /On ac describe a segment of a circle /ADC, which may touch the inscribed £ ‘circle in some point, as p. The lines * ; = )AD, Dc, drawn to this point from A and o, contain an angle less | than the lines drawn to any other point in the circumference of ) For take any other point £, and join aE, EC; produce CE to ne, and join AF. ‘The exterior angle AEC is greater than AFC, }2. e. than apc, which is in the same segment. And the same ymay be proved of lines drawn to every other point in the cir- ‘cumference of the circle prea. . 3 } [ ’ (| ! q (13.) If an equilateral triangle be inscribed in a circle, and i i « 182 GEOMETRICAL PROBLEMS. [ Sect. | through the angular points another be circumscribed ; to determiné the ratio which they bear to each other. | Let apse be an equilateral triangle in- D scribed in the circle, about which another DEF is circumscribed, touching the circle n B | in the points A, B, C. , : | Since pA touches the circle, the angle DAB=ACcB (Kucl. m. 32.); -but aca = ABC; .. DAB=ABC, and they are alternate angles, .. DF 4 parallel to Bc. In the same manner it may be shown that aB is parallel to F5, .. ABCF is a parallelogram; and the triangle ABC is equal toarc. In the same manner ABC may be shown to be equal to each of the triangles ABD, BCE; and .°, it is on fourth of the circumscribing triangle. : (14.) A straight line drawn from the vertex of an equilaterai triangle inscribed in a circle to any point in the opposite circum- ference, is equal to the two lines together, which are drawn from the extremities of the base to the same point. . ¥ Cc rE Let asc be an equilateral triangle inscribed in a circle; from B draw BD to any point p in the circumference. Join AD, CD. BD is equal to AD and cp together. Make pE=DA, and join ag. The angle DAE is equal to the angle Dra; but ADE=ACB in the same seg ment, .. DAE and DEA Ronee are equal to cBA and CAB together; whence DAE=CAB; and taking away the common angle CAE, DAC=EAB; but DCA=EBA, and AC=AB, .. BE=DC; and BD is equal to ap and cp together. (15.) If the base of a triangle be produced both ways, so that’ each part produced may be equal to the adjacent side, and throug: the extremities of the parts produced and the vertex a circle | é described ; the line joining its centre and the vertex of the triangle will bisect the angle at the vertex. F Sect. vit. | GEOMETRICAL PROBLEMS. 183 Let ac aside of the triangle asc be B woduced both ways till ap=as, and 3h=CB; and through D, B, & let a circle ye described, whose centre iso. If os be oined, it will bisect the angle anc. | Jom BD, BE,OA,OD,O&£. Since DA=AB, the angle aBD_ sequal to apB; but the angle opp is equal to ops, and .°. she angle oBA is equal toopa. In the same manner it may de shown that the angle onc is equal to ofc; and since ODA 8 equal to OFC, OBA is equal to oBC; or ABCis bisected by os. Oo | (16.) If an isosceles triangle be inscribed in a circle, and from the vertical angle a line be drawn meeting the circumference and the base ; either equal side is a mean proportional between the seg- iments of the line thus drawn. Let asc be an isosceles triangle inscribed in the circle arc, the side aB being equal to iS “FD Bb ac; and from a draw any line AED, meeting the circumference in §, and cB produced in p; i AB is a mean proportional between pa and ” # AE. |} Join ec. Since an=ac, the angle acB=ABC=AEC in the same segment; and the angle at a is common to the trian- igles Arc, ACD, .*. the triangles are equiangular and similar ; | ADC A CA (17.) If from the extremities of one of the equal sides of an isosceles triangle inscribed in a circle, tangents be drawn to the circle, and produced to meet ; two lines drawn to any point in the circumference from the point of concourse and one point of contact will divide the base (produced if necessary) in geometrical pro- portion. Let cpa be an isosceles triangle in- ‘Scribed in a circle, the side cB being equal to BG; and at B and c let tan- gets BA, CA be drawn, meeting in 4A. From A and B draw AD, BD to any 184. GEOMETRICAL PROBLEMS. [ Sect. v point p in the circumference, cutting the base c@ in and F; CE: CF!: CF: CG. 7 Joincp. The angle asc being equal to BGC, is equal t BCG, and .. C@ is parallel to AB; and the triangles aBc, om are equiangular ; FB Cl; GM, SAyB SB 0s | but (vn. 16.) BF: BC :: BC: BD. | ex equo BF: CG:: AB: BD: EF: FD, since CG is parallel to aB; | but CF : BF?! FD : FG, | oe Gk MOOG ons cer ea ay | whence (Eucl. v. 19.) ce : CF 3: CF: CG. | scribed similar to a segment on the base, and from the extremitie of the base tangents be drawn intersecting their circumferences the points of intersection and the vertex of the triangle will be % the same straight line. | (18.) If on the sides of a triangle, segments of circles be ie On AB, BC, the sides of the triangle | ABC, let the segments apB, BEC be described, similar to arc the segment on ac. At aA andc let tangents ap,cE ? be drawn. Join DB, BE; they are in the same straight line. Since pA touches the circle are, the angle DAC is equal to the angle in the alternate segment age i.e. to the angle in the segment ans. But the angle ap together with the angle in the segment aup, will be equal t two right angles; .. the angles cap, ADB are equal to twi right angles; ... Ac, DB are parallel. In the same manner AC BE may be shown to be parallel; .. Bp and BE being draw from the same point, parallel to the same lines, will also be ‘ the same straight line. A Sect. Vil. ] GEOMETRICAL PROBLEMS. 185 (19.) The centre of the circle, which touches the semacircles de- scribed on the two sides of a right-angled triangle, is im the middle goint of the hypothenuse. ‘On-the sides aps, Bc of the right- ‘angled triangle anc, let semicircles aps, Bec be described. Bisect ac in 0; 0 is the centre of a circle which will touch both the semicircles. _ From o draw Orr, OHD perpen- dicular to the sides. Then on being parallel to Bc, | (Bucl. vi. 2.) AO: OC 1: AH: HB, +, AH=HB, and u is the centre of the semicircle aps. Hence the centre of a circle touching ADB in D is in the line Duo. For the same reason, the centre of a circle touching BEC in E isin the line nro. Also since oD is equal to on and nD toge- ‘ther, i. e. to BF and HB, or EF and Fo together, 7. e. to EO, O if. centre of the circle, which will touch both. ' Cor. The diameter of this circle will be equal to the sides of the triangle together. _ (20.) If on the three sides of a right-angled triangle semicircles | ‘be described, and with the centres of those described on the sides, ‘circles be described touching that described on the base; they will | also touch the other semicircles. On the sides as, Bc of the right- angled triangle anc let semicircles be described; and with the centres p and )#, let circles be described touching that described on the base in F and @; each ‘of these circles will touch the semicircle described on the other side. ') Join opr, OEFG, DE. Since AB, BC are bisected in p and &, DF is parallel to ac, and equal to half /ac,i.e.to Ao oroc. In the same manner OD is parallel to Be, and ok to AB; .. ODBE is a parallelogram, and EB, @. ¢. -EH=op; but or=D8, .. DH=DF, and H is a point in the 186 GEOMETRICAL PROBLEMS. [ Sect. vi: circumference of the circle FHK; and being in the cire ference of BHC, it will be the point of contact, since DE join, the centres. In the same manner it may be shown that ° circle Gi touches the circle ABI in 1. (21.) If from any point in the circumference of a circle per pendiculars be drawn to the sides of the inscribed triangle ; th three points of intersection will be in the same straight line. From p any point in the circumference of the circle ABC, let DE, DF, DG be drawn perpendicular to the sides of the inscribed triangle ACB; join EF, FG; EFG 1s a Straight line. Tein AD, BD, CD. Since the angles DFB, DGB are right angles, a circle may be described about the quad lateral figure DGBF (vi. 18.); and the angle pra is equal t psa. Also since the. angles DFA, DEA are right angles, ¢ circle may be described about the quadrilateral figure DFEA, whence the angles DFE, DAE are together equal to two righ! angles. But apsc being a quadrilateral figure inscribed in ¢ circle, the angle DAc is equal to DBG, 7. e. to DFG; .*. DFE DFG are equal to two right angles; and merc is a straight line.. (22.) The base of a right-angled triangle not being greater thas the perpendicular ; if on any line drawn from the vertex to th base a semicircle be described, and a chord equal to the per. pendicular placed init, and bisected; the point of bisection wih always fall within the triangle. Let asc be a right-angled triangle, of which the side Ac is not greater than Bc. From B let any line Bp be drawn to the base; on which describe a semicircle Bcp, and in it place EF =BC, which bisect in @; the point & is within « the triangle aBe. r Take o the centre of the semicircle; draw ou perpendiculat to BC; jom Ga. Since BC is equal to EF, OH is equal to o@ ect. vu. | GEOMETRICAL PROBLEMS. 187 ha the angles at G and u being right angles, a circle described ‘vith the centre 0, and radius o«, will touch Bc in u, and .. G 3 within the angle ppc. Also since Ac is not greater than BC, yc is less than Bc or EF, .*. EF is nearer to the centre o, than yc is; or G falls above pc and within the angle pcs. | 3, The straight line bisecting any angle of a triangle | Weeribed in a given circle, cuts the circumference in a point, ‘vhich is equidistant from the extremities of the side opposite to he bisected angle, and from the centre of a circle inscribed in the Let asc be a triangle inscribed in the circle ‘cp. Bisect the angles Bac, ABC by the lines “41D, BO, which meet in 0; 0 is the centre of the circle inscribed in the triangle. Join Bp, bc. The lines pB, Dc, DO are equal to each = _ other. D | Because the angles DAB, DAC are equal, BD=pc; and be- cause the angle csp=cAD=DAB, to each of these add the -mgle cxo or its equal 480; and the whole angle op is equal 40 the two ABO, OAB, i. e. to BOD (Hucl. i. 82.); and .. op= DB. | (24.) The perpendicular from the vertex on the base of an equilateral triangle is equal to the side of an equilateral triangle } inscribed i in a circle, whose diameter is the base. '. From c the vertex, let co be drawn perpen- dicular to AB, the base of the equilateral tri- angle ABC; upon AB describe a circle ADB, and let pF be an equilateral triangle inscribed : in it; co will be equal to a side of this eos _ Draw ve bisecting the angle at p, and . bisecting EF at right angles, consequently ‘passing through the centre. Joi eG. The angles aco, AapG being each equal to half the angle of an equilateral triangle, are - equal to each other, and aoc=p£6, each being a right angle, Said AC—=AB=DG, .°. CO=DE. | ‘ z | = h | i 7 188 GEOMETRICAL PROBLEMS. [Sect. vi (25.) If an equilateral triangle be inscribed in a circle, and ti adjacent arcs cut off by two of its sides be bisected; the ly joining the points of bisection will be trisected by the sides. Let asc be an equilateral triangle inscribed 4 E in a circle; bisect the arcs AB, AC in D and E; join DE; it is divided into three equal parts in the points F and a. Since DE and BC cut off equal arcs BD, Cz, | they are parallel, and .*, (Kucl. vi. 2.) ar=ac. Join Bp, Al The angle BFD=AFE, and DBF=AEF in the same segmen and BD=Ak, since they subtend equal arcs; .. pr=ra. I] the same manner it may be shown that ac=ar. Now tk triangle Are being similar to asc is equilateral, .. pF, F¢ GE are all equal, and DE is trisected. | (26.) If any triangle be inscribed in a circle, and from th vertex a line be drawn parallel to a tangent at either extremity the base; this line will be a fourth proportional to the base “ two sides. Let anc be a triangle inscribed in the circle E ABC; and from B let Bp be drawn parallel to B 1 AE a tangent at a; then willac: AB:: BC: BD. Produce cB to meet the tangent ing. Since the angle EAB is equal to the angle in the alter- | nate segment AcB, and the angle AEB is equal to csp, .*. th triangle ABE is similar to cBD, and AE: AB?! CB: CD; but from similar triangles BDC, EAC, AO een ee aD) 1) Be “.e@ @QUO AC: AB‘: CB! DB. a i (27.) If a triangle be inscribed in a circle, and from its verte lines be drawn parallel to tangents at the extremities of its base they will cut off similar triangles. | e fect. V11. | GEOMETRICAL PROBLEMS. 189 | Let asc be a triangle inscribed in acircle, » 38 nd AD, CE tangents at the points A and c. tyom B draw BF, BG respectively parallel to hem ; these lines will cut off the triangles aBr, 3BG, which are similar. | For (Eucl. iii. 82.) the angle acB is equal to DAB, #. e. to he alternate angle ABF; and the angle BAC is equal to BCE, .e. to cBG; whence the triangles aBr, CBG having two angles n each equal, will be equiangular and similar. ' Cor. 1. The rectangle contained by the segments of the base ‘djacent to the angles will be equal to the square of either line lrawn from the vertex. A Cc For if ap and cx be produced, they will meet and form with \c an isosceles triangle, to which Bre is similar, .. BF=BG. | Now AF: BF :: BG: GC +, the rectangle ar, Gc is equal to the rectangle BF, BG, @. @. 0 the square of Br. | Cor. 2. Those segments are also in the duplicate ratio of the -idjacent sides. _ For the triangles apr and cxa are each of them similar to ABC, whence AC: AB?! AB: AF, | “. AC : AF in the duplicate ratio of ac : AB; jor the same reason, | AC : CG in the duplicate ratio of Ac : CB, .. AF: CG in the duplicate ratio of aB : CB. (28.) If one circle be circumscribed and another inscribed in a viven triangle, and a line be drawn from the vertical angle to the centre of the inner, and produced to the circumference of the outer vircle; the whole line thus produced has to the part produced the same ratio that the sum of the sides of the triangle has to the base. } Let asp be a circle circumscribed about the wiangle ABc; 0 the centre of the inscribed circle. Join ao, and produce it to p; then aoD bisects vhe angle BAC. Join BD, DC; and draw BO, CO 0 the centre of the inscribed circle; then | AD DOSSARY Mew CB, . 190 GEOMETRICAL PROBLEMS. | Sect. v1 Draw or, 0G parallel to aB, AC, meeting BD, CD in F ds Gc. The angle ppc=DAC=DAB=DOF, and the angle atD common to the triangles BED, oOFD, and (vu. 20.) Bp=pt *. OF=BE. In the same manner it may be shown that og: ec. Now the trapeziums BAcD, FoGD being similar, an similarly situated, | AD: OD::AB+AC : FO+0G TI:AB+AC : BC. (29.) If in a right-angled triangle, a perpendicular be draw from the right angle to the hypothenuse, and circles inscribe within the triangles on each side of it ; their diameters will be i each other as the subtending sides of the right-angled triangle. Let asc be a right-angled trian- - gle; from the right angle B let fall the perpendicular Bp; and in the triangles ABD, BDC let circles be inscribed ; their diameters are to one another as AB to BC. Bisect the angles BAD, ABD by > D the lines Ao, BO, they will meet in the centreo; in the sam manner lines bisecting DBC, DCB meet in the centre E; dra oF, EG to the points of contact. Now the triangles ABD, BD being similar (Eucl. vi. 8.), .. the triangles aBo, BCE ar similar; whence AB VBOS BOs On; but the triangles opr, EGC are similar, . BO: CE!!! OF: EG?:: 20F : 2EG, . AB; BC :: 20F: 2EG. (30.) To find the locus of the vertex of a triangle, whose bas and ratio of the other two sides are given. Let aB be the given base; divide it in c, Dp so that Ac : CB may be in the given ratio of the sides. Produce as to 0; and take co IN a mean proportional between Ao and BO. A c 5 @ 3 —————— - uF _ << ‘Nect. Vil. | GEOMETRICAL PROBLEMS. 191 With the centre 0, and radius co, describe a circle ; it will be be locus required. | In the arc cp take any point D; join DA, DB, DC, DO. jince OD=OC, | AO: 0D!: 0D: OB, +, the sides about the common angle 0, are proportional, and he triangles ADO, BDO are equiangular ; | fa UDR 6DOL OR. COR OB SAO CO | 2 7; AO—CO : CO—OB?:! AC: CB, .é. in the given ratio. In the same manner, if any other point ye taken in the circumference of the circle, and lines drawn to te they will be in the same given ratio, and .. the circum- erence is the locus required. ' Cor. Since in any triangle, if from the vertex a line be drawn sutting the base in the ratio of the sides, it will bisect the angle, .. the angle Aapc=BDC. (31.) A given straight line being divided into any three parts ; ‘o determine a point such, that lines drawn to the points of section ind to the extremities of the line shall contain three equal angles. Let aB be the given line, and ac, cp, ee oB the given parts. Take co a mean eat i = proportional between ao and po; and with the centre o and radius oc describe A CC DO BE .circle. Produce cB; and make DE a mean proportional be- ween CE and BE; and with the centre £, and radius ED, describe a circle cutting the former-in F; F is the point sequired. Sa rr _ For, as was proved in the last proposition, | AF IPD (AC ..CD, and .. the angle Aarc=crp; and CF: FB‘: CD: DB, *, the angle cFD=DFB; and .-, the three angles Arc, CFD, DFB are equal. (82.) If two equal lines touch two unequal circles, and from tie i | Pal 192 GEOMETRICAL PROBLEMS. [ Sect. vi extremities of them lines containing equal angles be drawn cuttin the circles, and the points of section joined; the triangles : formed will be reciprocally proportional. . Let two equal lines D AB, CD touch two un- i K 7 mms bY equal circles EB F, GDH; BaESS eS and from a and c let x E | lines AIK, AEF, CLM, cou be drawn containing the equal angles KAF, MCH. Joi IE, KF, GL, MH; then will the triangle AKF: CHM } CGL: AIE. . | Since AB is equal to cp, the rectangles Ha, AF, and GC, ; are equal ; A ) (33.) If from an angle of a triangle a line be drawn to cut th opposite side, so that the rectangle contained by the sides includin, the angle, be equal to the rectangle contained by the segments ¢ the side together with the square of the line so drawn; that lin | bisects the angle. OAR? CH GO AE. and for the same reason, AK SCM £2.0 bc eA 1, whence AFX AK ? CHXCM 3: C@XCL?: AEXAIT, . the triangle AKF : CMH !! CGL: AIK, since AK ; CM in the ratio of the perpendicular from K and on AF and cH; and cL: Ar in the ratio of the perpendicula from L and 1. From B one of the angles of the triangle asc, let Bp be drawn, so that the rectangle as, BC may be equal to the rectangle ap, pce together with the square of BD; BD bisects the angle s. For if not, let BE bisect it; the rectangle an, : BC is equal to the rectangle az, EC together with the square a BE. About asc describe a circle, and produce Bp, BE to th circumference in F and G; join rc. The rectangle ap, pe i) equal to the rectangle BD, DF; .*. the rectangle a B, BC is equa to the rectangle BD, pF together with the square of BD. 7. é Sect. Vil. | GEOMETRICAL PROBLEMS. 193 ‘o the rectangle BF, BD. In the same way the rectangle \B, BC is equal to the rectangle BG, BE; whence the rectangle 3G, BE is equal to the recianale BF, BD; a circle may therefore ye described through D, E, G, F; Onenee DEG, DFG are equal o two right angles, 7. e. to DEG, DES and .°. DFG is equal 0 DEB, or to DAB and aB@; and .*. the arc AB equal to the we Bo, which is absurd, eee ae triangle be isosceles. dence ... BG does not bisect the angle; and no other but pean bisect it. (34.) In any triangle, if perpendiculars be drawn from the ingles to the opposite sides ; they will all meet in a point. Let asc be any triangle; and ar, cp yerpendiculars drawn upon the opposite jides, intersecting each other ina. Through i ; draw BGE; it is perpendicular to ac. _, Join FD; ig about the trapezium BFGD _lescribe a erele. The triangles ADG, GFC being equiangular, i AG: GC:: GD: FG, vhence also the triangles AGC, FGD are equiangular ; ; and . he angle ACcD=DFG=ABE; and the angle BAc is common jo the two triangles ABE, ACD; .°. the angle AEB=ADC, 2. @. t is a right angle, and Bz is perpendicular to ac. X | (35.) If from the extremities of the base of any triangle, two verpendiculars be let fall on the line bisecting the vertical angle ; wd through the points where they meet that line, and the point in | he base whereon the perpendicular from the vertical angle falls, a rele be described ; that circle will bisect the base of the triangle. | Let asc be a triangle, whose vertical angle B is bisected by “he line Bp, on which let fall the perpendiculars ar,cE. From let fall sG perpendicular to Ac; a circle described passing hrough E, F, G will also bisect ac. About the triangle apc describe the circle ADB; and from O 194: GEOMETRICAL PROBLEMS. | [ Sect. vin D draw a diameter which will bisect ac in u. Now Wx since the angle A1D is common to the triangles CAN AIF, HID, and the angles a FI, IHD are right angles, s we ~\ *, the triangles AlF, HID, are similar. In the same manner BIG, CEI are similar. Whence eELUX Gea DX UB 2 OL ROT or ek eae and since IDXIB=IAXIC, .. HIXIG=IFXIE, or a cirely passing through &, Fr, @ will pass through nw (vi. 18.), and .* bisect the base ac. HE LD aie, bas and 1G : IB‘: IE: IC, (36.) If from one of the angles of a triangle a straight line b drawn through the centre of its inscribed circle, and a perpendi cular be drawn to this line from one of the other angles ; the poin of intersection of the perpendicular, and the two points of contac of the inscribed circle, which are adjacent to the remaining angie, are in the same straight line. Let asc be a triangle, and o the centre tT ee of its inscribed circle. From s draw pop bik | through the centre; and from c let fall cu F \ perpendicular to it. Let the circle touch the sides of the triangle in F and q; join ; | Wi HG,GF. HGF is a straight line. ia Join oa,oc; and let acircle be described ie about the triangle aBc; join cv. The a triangles OGE, CHE, having the vertical angles at B equal, anc OGE, CHE right angles, are similar, . tw Be OT, sO. re eG, ae and .*. the rectangle cz, He is equal to the rectangle ox, HE q witehice a circle will pass through the points c, 0, G, H, .”. th angle coH=cG@uH. Again (vii. 20.) cp=po, and AG=AF also the angle coo = GArF, .. COE=AGF; whence cGH= AGF; and CG, GA are in the same straight ine) “. FG, GH in ie same straight line. . ‘ Sect. vit. | GEOMETRICAL PROBLEMS. 195 (37.) If from the three angles of any triangle three straight lines be drawn to the points where the inscribed circle touches the sides ; these lines shall intersect each other in the same point. Let asc be a triangle, in which a circle ¥ is inscribed, touching the sides in 5, F, D. Join AF, CE, cutting each other in o. Join BO, and produce it; it will pass through p. For if not let it pass through some other spoint K; draw EG, EH respectively parallel toac, Bc. ‘Then the triangles OfFH, OFC being similar, OE: EH:: 0C;: (CF=) cD; in the same manner, ie EG: EO!: KC: CO, WieG hie. kK Css OD. _ Again, since £4 is parallel to Br, AH > EH >: 4B) (BF=) BE i: AK ! EG, Sarak A Ee BG B, BE is equal to the rectangle ac, BF. | ) (21.) Lf semicircles be described on the segments of the base nade by a perpendicular drawn from the right angle of a triangle; hey will cut off from the sides, segments which will be in the tri- licate ratio of the sides. From the right angle sp let Bp be drawn yerpendicular to ac; and on Ap, vc let yemicircles be described, cutting AB, cB in tand Fr; AE: CF in the triplicate ratio of KB: CB. ) Join DE, DF; they are perpendicular to aB, BC respectively ; *, (Eucl. vi. 8. Cor.) AC: AB ‘i: AB: AD ABAD AD AR: tence AC : AE in the triplicate ratio of Ac : AB. In the same manner it may be shown that AC : CF in the triplicate ratio of ac : cB, <. mm. and ex equo, AE : CF in the triplicate ratio of AB : cB. (22 .) If from any point in the diameter of a semicircle a per- endicular be drawn, and from the extremities of the diameter P 210 GEOMETRICAL PROBLEMS. [ Sect. viny lines be drawn to any point in the circumference, and meeting t H perpendicular ; the rectangle contained by the segments which the cut off from the perpendicular, will be equal to the rectangle cor tained by the segments of the diameter. i From any point p in the diameter ac of the semicircle ABC, let a perpendicular pF be drawn; and to any point B in the circum- ference let the lines AB, cB be drawn, meeting the perpendicular i & and F; the rectangle FD, DE is equal. to the rectangle ap, pc. » ‘ Since the angle ABC in a semicircle is’a right angle, FBE }j also a right angle, and .. equal to rpc; and the verticall} opposite angles at & are equal, ., the angle Bre is equal t ECD, and the triangles FDA, DEC are equiangular, | aid. Bip Aan espe, j .. the rectangle FD, DE is equal to the rectangle ap, pc. (23.) If from the point of bisection and any other point in | given are of a circle, two parallel lines be drawn, the former ter minated by the circumference, the latter by the chord of the are the rectangle contained by these two lines will be equal to tha contained by the lines which join the latter point with eae extremity of the given arc. : From c the middle point of the arc acs, and B | p any other point, let any two parallel lines Le > 4) cz, DF be drawn, of which cx is terminated ch y by the circumference of the circle, and pF by the chord AaB. Join AD, DB; the rectangle cE, DF is equal to the rectangle aD, DB. e Draw DG perpendicular to AB; draw the diameter cH; an jom EH. ‘The angle rp@ being equal to EcH, and the angle at a and £ right angles, the triangles rpG, ECH are equ’ angular, * sw DGDF,); EC-.CH,; | whence the rectangle Ec, DF is equal to the rectangle DG, CF i. e. (Eucl, vi. C.) to the rectangle ap, DB. = | | - { - 3 _ oo 4 , - yi - Sect. v1i1. | GEOMETRICAL PROBLEMS. | 211 | (24.) If two circles cut each other, and from either point of ntersection lines be drawn meeting both circumferences ; the rect- ngles contained by the segments of these lines are to one another n the ratio of the perpendiculars drawn from their intersection vith the inner circumferences upon the line joining the intersec- ions of the circles. , Let the two circles ABC, ABE cut each other oA and B; join AB; and from B draw any two imes BC, BD cutting the circles in b, F, c, D; let all the perpendiculars EG, rH; the rectangles sE, EC and BF, FD are to one another as EG to i". Join ap, Ac. Since the angle AFB=ABB, .*. i\FD=AEC; but ADF=ACE in the same segment, .. the tri- ngles AFD, AEC are similar; and EC: EA !?: FD: FA. But-fCG : BA ‘1 EC X EB: EBA X EB = 8G x diameter of the circle aBE, and FD : FA ?: BF X FD: BF xX FA =FH x diameter of the circle aBE, yhence ECX EB: BFXFD!:EGXD: FHXD!!EG: FH. _(25.) If on opposite sides of any point in the chord of a circle wo lines be taken, one terminating in the chord, the other in the hord produced, whose rectangle is equal to that contained by the egments of the chord; and. the extremities of the lines so taken Je goined to those of any other chord passing through the same oint; the line joining their intersections of the circle will be arallel to the first chord. !-On opposite sides of any point c of the | N hord as of the circle asa let two lines ‘p and cx be taken, such that the rectangle OK €, CE may be equal to the rectangle ac, ALY 'B; and through c let any chord Gor be ‘rawn, and: pr, GE joined, meeting the cir- umference in H andr. Join 1; it will be parallel to an. P2 212 GEOMETRICAL PROBLEMS. [ Sect. vit Since the rectangle DC, CE is equal to the rectangle AC, GB) i. e. to the rentatele GC, CF, &) DOO TS 6o Len, al . the triangles Dor, GCE are equiangular, and the angle FD | 1S a to CGE or FHI in the same segment; .. 111 1s paralle. to AB. | - t 1 4 4 (26.) If from two points without a circle two tangents be dra 2. the sum of the squares of which is equal to the square of the line joining those points; and from one of them a line be drawn cutting the circle, and two lines from the other point to the intersection, with the circumference ; the points in which these two lines cut the circle, are in the same straight line with the former point. From a and B two points without the circle cps let tangents ac, BD be drawn, such that the sum of their squares may be equal to the square of AB. If from a any line AFrE be drawn, and BF, BE joined; the points 4, H, G, will be in a straight line. In as take a point 1, so that the rectangle AB, BI may be equal to the square of Bp. » Join IF IH, AH, HG. Then the rectangles AB, AI and AB, BI are tome a equal to the square of AB, 7. e. to the squares of AC, BD: . the rectangle aB, AI is equal to the square of ac or to i Mi tols AF, AE} : * AR IDAB A 1A, and .*. (Eucl.-vi. 6.) the angle arr is equal to AEB, when also FIB=FHG. Now the rectangle Bi, BA being equal to the square of BD, or to the rectangle BH, BF, if [BE BAY Bay Bin; and .*. the angle AnB is equal to FIB, or FHG; and BHF 18 @ a line, .. AHG is a straight line. j (27.) If from the vertex of a triangle there be drawn a line te any point in the base, from which point lines are drawn paralle: Sect. viit. | GEOMETRICAL PROBLEMS. 218 a the sides; the sum of the rectangles of each side, and its seg- nent adjacent to the vertex will be equal to the square of the line rawn from the vertex together with the rectangle contained by he segments of the base. ' From the vertex a of the triangle anc let a ine AD be drawn to any point p in the base; rom which let pF, DE be drawn parallel respec- lively to aB, Ac; the rectangles BA, AB, and HA, AF will together be equal to the square of ip, and the rectangle Bp, Dc together. _ About asc let a circle be described; and let ap meet the arclein @. Join BG, Gc. From & draw en, making the angle \HE equal to aBG. Produce aB to 1. Because the angles HE, ABG are equal, the points E, B, G, H are in the circum- erence of a circle, .. the rectangle BA, A# is equal to the rect- mgle GA, AH; and the angle END will also be equal to GBI, .€.to ace. And because ac, DE are parallel, the angle EDH s equal to Gac; hence the triangles EDH, GAC are equi- ingular, ands." Al Cie AiG tee D He ind the rectangle aG, pu is equal to the rectangle ac, DE, é. e. 0 the rectangle ac, ar. And because the rectangle BA, AE is ¢qual to the rectangle Ga, Au and the rectangle ca, AF to the vectangle ac, DH, .. the rectangles BA, AE and CA, AF are ogether equal to the rectangles Ga, AH, and GA, DH, i. e. to #A, AD or to the rectangle aD, p@ together with the square of \D; or to the rectangle Bp, pc together with the square of ap. (28.) If on the chord of a quadrantal are a semicircle be lescribed ; the area of the lune so formed will be equal to the area if the triangle formed by the chord and terminating radii of the yuadrant. Let aso be a quadrant, on the chord of which ‘oe et a semicircle ApB be described; the lune apBE US $ equal to the triangle a Bo. ry Since circles are as the squares of their radii, 4 Wie quadrant AEBO; ADC :: AO’; Ac?:: 2:1, 214: GEOMETRICAL PROBLEMS. [ Sect. vil . the quadrant ABO is equal to the semicircle ADB; and Rentini away the part aEBC, the lune apBE is equal iB the teehale ABO. (29.) If from the extremities of the side of a square circles be described with radi equal the former to the side, and the latter te the diagonal of the square; the area of the lune so formed wil, be equal to the area of the square. From p and c the extremities of pc the side of a square, with radii pB and cB, let circles be described, cutting each other again in £; the area of the lune BFE is equal to the square Ac. Join CE, ED. Since Bc is equal to cr, and CD is common, and BD is equal to DE; .*. the angles BCD, ECD are equal; whence BE is a straight line; alk | the angles BDC, EDC are nctels and BDC being half a "righ angle, BDE is aright angle; .. the arc BE is a quadrant; .*/ the lune BFE is equal to the triangle BD& (viii. 28.) 7. e. to the ‘square AC. ; ‘ (30.) If on the sides of a triangle inscribed in a semicircle semicircles be described ; the two lunes formed thereby will toge- ther be equal to the area of the triangle. Let asc be a triangle inscribed in a semi- SS . circle. On As, BC let semicircles app, Ge BEC be described; the lunes ADBF, BGCE are together equal to the triangle a Bc. A Cc Since the areas of circles are as the squares of their diameters, the semi-circle AOR Ota Ba AC. ABS and ABG {BEC ?: Ac’: ac; . ABC! ADB+BEC !: AC’: AB’+BO’ | i.e. in a ratio of equality, .. ABC = ADB + BEC; from these equals take away the segments arB, BGC, and the ae ee as On ; ¢ Sect. vi11. | GEOMETRICAL PROBLEMS. 215 (31.) If on the two longer sides of a rectangular parallelogram as diameters, two semicircles be described towards the same parts; the figure contained by the two remaining sides of the paralleto- gram and the two circumferences shall be equal to the parallelo- gram. Let ascp be a rectangular parallelogram, ‘on the sides AB, Dc of which let semicircles ‘AnB, DFC be described ; the figure DAEBCHFG 4 is equal to ABCD. Since AB=vpc, the semicircles are equal ; ° from each of which take away FGu, and AGFHBE=DGHC; if to these equals be added a pc and Buc, the whole ADGFHCBE will be equal to the whole aBco. (32.) If two points be hans at equal distances from the extre- ,mities of a quadrant, and perpendiculars be drawn from these ipoints to the radius ; the mixtilinear space cut off, shall be equal ‘to the sector which stands on the arc between them. Let two points c, & be taken at equal distances pra from A and z, the extremities of the quadrant AB ; ieee and let fall the perpendiculars cp, HF on Ao. Join *7 "— -co, EO; the figure cpFe is equal to the sector o 5 COE. Since the arc AC=EB, the angle AOC=EOB= Siete and the angles at p and F are right angles, and co=ok8, .. the triangle ‘cop=e5£0F; from each of these take away OFH, .. DFHC= OHE; to each of these add CHE, and CDFE=COE. (33.) If the arc of a semicircle be trisected, and from the points of section lines be drawn to either extremity of the diameter ; the difference of the two segments thus made, will be equal to the sector which stands on either of the arcs. _ Let the arc of the semicircle acs be divided = ¢ D into three equal parts in the points c, and D. L * ; 216 GEOMETRICAL PROBLEMS. [ Sect. vim, the difference of the segments ac and acp is equal to tl Sector COD. | ‘Since the angles CAD, DaB stand on equal circumferenedl : they are equals but the angle DAO=oDA, ... CAE=EDO; 2 id CEA=OED, . . the triangles car, EOD are equiangular; am since OF is drag bisecting the a angle o of the isosceles triangle AoD, it bisects the base, .. AE=ED; and consequently the triangle AEC is equal to Ais triangle Dor; add to each, CED, and CAD=COD, (34.) Ifa straight line be placed in a circle, and on the wail passing through one extremity, as a diameter, another circle b described ; the segments of the two circles cut off by the above straight line will be similar, and in the ratio of four to one. Let ec bea straight line placed in the circle ABC. ‘Take o the centre, and join oc; and upon - ® it describe a semicircle opc. The segments EAC, DGC are similar, and in the ratio of 4: 1. Join op, and produce it both ways to the cir- cumference. Take Fr the centre of the semicircle opc. Join a FD. Then ope being aright angle, ep=D6, and OF=FC, .. (Kucl. vi. 2.) pF is parallel to £0; and CE; EO !: CD: DF, and the segments EAC, DGC are cag whence EAC : DGC :: EC’: cp’ 1: 4: 1. Cor. The segment anc is bisected bg the circumference | DGC. : (35.) [fon any two segments of the diameter of a semicircle semicircles be described ; the area included between the three cir | cumferences wul be boa to the area of a circle, whose diameter y as the mean proportional between the segments. On ap, and pc, segments of ac the dia- -meter of the semicircle anc let semicircles ) Gols, AED, CFD be described; from p draw pB . perpendicular to ac, and .*, a mean propor- tional between ap and cp; the figure | AEDFCB is equal to the circle Besa upon DB. ‘ferences are together equal ie ACB. q ' other in a ands; and from a draw any line ac, ‘cutting the circles in p and c; join DB, BC; the figure p bn cc p is equal to the triangle pBe. Sect. vii. ] GEOMETRICAL PROBLEMS. 217 | Join As, Bc. Since apB is aright angle, the semicircle on AB is equal to those on Ap and ps together ; and the semicircle on BC is equal to those on BD and pc; .*. the semicircles on aB and BC, or the semicircle aBc which is equal to them, will be equal to the semicircles AED, DFc and the circle described jupon DB. From these equals take away the semicircles AED, prc, and the figure AnpDFrcB is equal to the circle described upon DB. (86.) If the diameter of a semicircle be divided into any number of parts, and on them semicircles be described ; their circumfer- ences will together be equal to the circumference of the given semicircle. Let as the diameter of the semicircle ACB be divided into any number of parts in the points D, E; and on AD, DE, EB, let semicircles be described ; their circum- For since the circumferences of circles are as their diameters. ACB: AFD !! AB! AD ACB: DGE:!!AB!: DE ACB !: EHB?:: AB: EB, whence ACB : AFD+DGE+EHB }: AB: AD+DE+EB, in which proportion the third term being equal to the fourth, r ACB=AFD+DGE+EHB. - (87.) If two equal circles cut each other, and from either point ‘of section a line be drawn meeting the two circumferences ; the area cut off by the part of this line between the two circumferences will be equal to the area of the triangle contained by that part and ‘lines drawn to its extremities from the other point of section. Let the two equal circles ApB, AcB cut each Take any points 5, F in the circumferences /AEB, AFB; join AE, EB, AF, FB. Since the ares 8 218 GEOMETRICAL PROBLEMS. [Sect. vit. ADB, AFB are equal, the angles ADB, AFB are equal. But the angles AFB, AEB are equal to two right angles, and .. to ADB, BDC; whence the angle BDC = AEB = AcB, and BD= BG; .. the segment DéB is equal to the segment Bcc; to each of these add péxnc, and the triangle pBc is equal to pbBccD. Cor. If az is a tangent to ADB ata; the area ADdBCCEA will be equal to the triangle ABE. Ee (38.) If two equal circles touch each other externally, and through the point of contact another be described with the same radius; the area contained by the convex circumferences cut off from the touching circles, and the part of the third without them, as equal to the area of the quadrilateral figure formed by lines drawn from the points of intersection to the point of contact, and f to the point where the third circle is cut by a tangent drawn to | eer — the point of contact of the two circles. Let two equal circles touch each other in a; and through the point of contact let an equal circle ABC be described, cutting the former in B and c. Join AB, AC; and to the point a let a tangent AD be drawn; join Bp, pc. The area anne by AEB, AFC and the intercepted arc BDC is equal to the quadrilateral figure aBpc. | Since DA touches the circle Ans, the angle paB is equal to the angle in the alternate segment, and .*, equal to the angle in) the cit BCA, 2. é. equal to the angle Bpa, whence BA=_ BD; .. the segment BEA is equal to the segment BGD; and AEBGDA is equal to the triangle asp. In the same manner it may be shown that Arcupa is equal to the triangle acD; .. AEBGHCFA is equal to the quadrilateral figure aBpe. a (39.) Ifa straight line be divided into any two parts, and upon the whole and the two parts semicircles be described; and from ; the point of section a perpendicular be drawn, on each side of which circles are described touching it and the semicircles; these circles will be equal. | ‘Sect. 1x.] GEOMETRICAL PROBLEMS. 219 Let AB be divided into any two parts inc; and on-aB, AC, CB let Biinicircles be described ; from c ‘draw the perpendicular cp, on each side of which let a circle be de- scribed touching the perpendicular and each of the semicircles. These * circles are equal. Let EFGH touch the perpendicular in u, and the semicircles $n Fand«c. Draw the diameter en parallel to AaB. Join FE, EA; AF will be a straight line (ii. 35). Produce it to meet the perpendicular in p. Join FH, HB; FB will also be a straight ‘line, and perpendicular to ap at the point F. Join £G, GO, ‘ac, GA; Ec and nA will also be straight lines. Produce au ‘tor. Join Br; it will be perpendicular to a1, and pass through 'p, since the perpendiculars to the three sides of the triangle -AHB meet in a point. And since the angles AGC, AIB are “equal, EC is parallel to pB. i . AD: DE}: AB?BC; ,and AC, HE are parallel, | - AD: DE:: AC: EH, SFA BB Cte Ae | He | In the same manner it may be proved that | Hes A Ceo 8 Gas. KE . KL being the diameter of the other circle drawn parallel to a8. | Hence EH=KL, and the circles are equal. SECTION IX. (1:) Given one angle, a side adjacent to it, and the difference of the other two sides ; to construct the triangle. Let cB be equal to the given side; draw the indefinite line ca, making with it an angle equal to the given angle; and Wh ae cut off cp equal to the given itferences | an oN Join pp; and make the angle pBpa equal A D C to BDA; ABC is the triangle required. 220 GEOMETRICAL PROBLEMS. [ Sect. IX. i The angle pBA being equal to aps, the side aD is equal to! AB; and the difference between ca and AB is equal to cp, 4. e to ae given difference. ? - (2.) Given one angle, a side opposite to it, and the difference of the other two sides ; to construct the triangle. In any line ca, (see last Fig.) take cp equal to the given difference ; 3 make the angle cps greater than a right angle by half the given angle ; aye c draw cB equal to the given side, and meeting DB in B; and make the angle pBa equal to Bpa; ABC is the triangle “ants ed. ate the angles ABD, ADB are equal, AB is equal to aD; . the difference between ca and AB is equal to cp, i. e. to the ora difference. Also the angle at a is equal to the difference between the angles cDB, DBA, or CDB, BDA, 2. e. to the give | angle. And cB was made equal to the given side. a (3.) Gaven the base and one of the angles at the base; to con- struct the triangle, when the side opposite the given eo) as equal | to half the sum of the other side and a given line. Let aps be the given base, and asc the © given angle; produce cB to p, making Bp equal to the given line. Join ap; and from B to AD draw BE, equal to half pp. From A draw Ac parallel to BE; ABC is the triangle © required. i For AB and aBcare made equal to the given base and anglay a and since BE is parallel to ac, AC {OD2B Es Bip a la: 5 (4.) Given the base of a right-angled triangle, and the sum of | the hypothenuse and a straight line, to which the perpendicular has a given ratio ; to construct the triangle. Let 1B be equal to the given base. From B draw Bc per= pendicular to it, and such that it may have to the given sum 4 ‘ Sect. 1x. ] GEOMETRICAL PROBLEMS. 221 erect a perpendicular cp equal to the given per- -pendicular ; and take cx equal to the given ditfer- ‘ence between the side and adjacent segment on ‘the given ratio. Join ca, and pro- Dv k duce it; and from B to cp, draw BD Sells equal to the given sum. From a draw fail _AE perpendicular to AB; ABE is the triangle required. c For ak being parallel to Bc, AE!ED!: BC: BD, i. e. in the given ratio; “, AE is equal to the perpendicular; and AB was made equal to the given base. (5.) Given the perpendicular drawn from the vertical angle to the base, and the difference between each side and the adjacent segment of the base made by the perpendicular ; to construct the triangle. From any point c in an indefinite line as, the opposite side of the perpendicular. Also take oS ee CF equal to the other given difference. Join ED, FD; and ‘make the angle FDA=DFA, and EDB=DEB; ADB is the triangle required. Since the angles ADF, AFD are equal, Aap=ArF, .*. the differ- ence between AD, AC is equal to cF, 7%. e. to the given difference. ‘ angle equal to the given angle; and let the base be divided in the given ratio in p. Complete the _ circle; and bisect the arc ACB in C; join cD, and | produce it to H; join AB, EB; AzEB is the triangle required, In the same manner the difference between Bp and BC is equal | to CE, 2. e. to the given difference. (6.) Given the vertical angle, and the base; to construct the triangle, when the line drawn from the vertex cutting the base in | any given ratio, bisects thé vertical angle. Let aB be equal to the given base; and upon it describe a segment of a circle containing an : 222 GEOMETRICAL PROBLEMS. [ Sect. 1x, For AEB is equal to the given angle; and since the are ac =cB, the line Ep, which divides AB in the given ratio in D, makes the angle AED=DEB. (7.) Gaven the vertical angle, and one of the sides containing it ; to construct the triangle, when the line drawn from the veri making a given angle with the base, bisects the triangle. Let aB be equal to the given side; and on it describe a segment of a circle containing an angle equal to the given angle made by the bisecting line with the base; and make the angle anc equal to the given vertical angle. ‘Bisect AB in p; and draw DE parallel to Bc, meeting the circle in ©; join ax, | and produce it to c; aBc is the triangle required. | Jon BE. Since De is parallel to Bc, (Eucl. vi. 2.) ax is equal to Ec, and .. Bx bisects the triangle; and it makes with the base Az an angle equal to the given angle. Also AB a) equal to the given side, and asc to the given angle at the vertex. ) (8.) Given one angle, a side opposite to it, and the sum of the: other two sides ; to construct the triangle. Let a8 be the given side. Upon it describe a segment of a circle containing an angle equal to half the given angle; and from a draw ac equal to the given sum of the two sides; join Bc; and make the angle cBp equal to BcD; ABD is the triangle required. Since the angle DcB is equal to DBC, DB is equal topc; .. AD, DB together are equal to the given sum. ea the angle - ABD is equal to DBC, DCB, 2. e. to twice DcB, and .°. is equal to the given angle. (9.) Given the vertical angle, the line bisecting the base, and the angle which the bisecting line makes with the base; to con- struct the triangle. 7 ~é Sect. 1x.] GEOMETRICAL PROBLEMS. 223 ‘On any line aB describe a segment of E a circle containing an angle equal to the Z Since DG is parallel to Bc, and GF to HI, Bp: Bu (::CG: CH) ::GF: HI?! CF: CI, and since BD, GF and FC are equal, BH, HI and 1c are also equal; whence Au and H1 together are equal to ap; and a1,1H are together equal i to ac; and HAL is equal to the saa angie. - (26.) Given the vertical angle, and the ratio of the sides con- taining it, as also the diameter of the circumscribing circle; to | construct the triangle. | _ On the given diameter describe a circle; and | from it cut off a segment ACB containing an angle equal to the given vertical angle. Divide the | base 8B in D, in the given ratio of the sides; and ‘draw the diameter mF at right angles to AB. Join ED, and let it meet the circumference in €; join Ac, cB; ACB is the triangle required. é, 232 _ GEOMETRICAL PROBLEMS. [Sect. rx, Since AE=EB, the angle acs is bisected by cr, . AC: CB:: AD: DB, zZ. é. in the given ratio. (27.) Given the vertical angle, and the radii of the inscribed and | circumscribing circles ; to construct the triangle. With the given radius describe the circum- scribing circle; and from any point a in it take AB, Ac each equal to the arc subtending at the centre an angle equal to the given angle. Join BC; and parallel to it, at a distance equal to the radius of the inscribed circle draw EF. Join AB; and from a to EF draw Ao=AB, and produce ao to G. Join CG, BG; GcBis the triangle required. For the angle cas (Eucl. iii. 20.) is equal to the angle at the | centre, which stands on ac, and .*, is equal to the given angle, | Also the angle caa=aesB. Join Bo; and since AB=Ao, the | angle AOB=ABO; but AoB is “all to the two oBG, OGB; | and AGB=ABC, ... CBO=0BG, and Bo bisects the angle CBG; | whence 0, the point of intersection of the bisecting lines Ga — and BO, is the centre of the inscribed circle; and its distance from BC was made equal to the given radius. (28.) Given the vertical angle, the radius of the inscribed circle . and the rectangle contained by the straight lines drawn from the» centre of that circle to the angles at the base; to construct the triangle, Let the angle anc be equal to the vertical angle, which bisect by the straight line pp. Let pD (ii. 14.) be the centre of a circle which would touch Ba and Bc; and let & and F be the points of contact; fa DE, DF, and pro- duce BD to G, so that the rectangle DE, DG may be equal to the given rectangle. On DG describe a circle cutting Ba, Bc in A andc; H and inuandi. Join ac (or HI). ABCOor AHI is the triangle required. Join AD, DC; and draw pK perpendicular Sect. 1x. | GEOMETRICAL PROBLEMS. 233 ‘eut off, are equal; .. the angle acp is equal to DCI, 7. €. ACB ‘is bisected by cp, or p is the centre of the circle inscribed in y triangle ABC; .*. K is the point of contact, and DK = DE. But (Eucl. vi. C.) the rectangle ap, Dc is equal to the rectangle DK, DG, 7. e. to the rectangle DE, DG, which is equal to the given rectangle. (29.) Given the base, one of the angles at the base, and the point, in which the diameter of the circumscribing circle drawn \ from the vertex meets the base; to construct the triangle. _ Let ap be equal to the given base, and c the EP ‘given point. Bisect AB in D, and draw DE at Be ‘right angles to as. Upon ac describe a seg- ment of a circle containing an angle equal to Sh twice the difference between the given angle and So B ja right angle; and let it meet DE in o. Join ‘Ao, oc; and produce co to r, making OF=OA; jom AF, FB; AFB is the triangle required. Join os. Because AD=pDB, and DO is common, and the ‘angles at p right angles, «. on=oA=oF; and a circle de- scribed from o as a centre, and radius 0A, will pass through F and 3, and circumscribe the triangle apr. And Foc produced is a diameter. Also the angles aoc, AOF are equal to two ‘right angles, i. e. to Aoc and twice the given angle; .. AOF is equal to twice the given angles; and since it is double of aBF, 'ABF will be equal to the given angle. i (30.) Given the vertical angle, the base, and the difference between two lines drawn from the centre of the inscribed circle to | the angles at the base; to construct the triangle. ‘ _ Take any line aB, unlimited towards B, 4 and cut off ac equal to the given differ- Bice; and at the point c make the angle | Bcp such that its quadruple together with | the given angle at the vertex may be equal r a a a a She & i 2 234 GEOMETRICAL PROBLEMS. [Sect. 1x. to two right angles. From a to cp, draw ap equal to th given base. At the point a make the angle BAE=BAD, and at the point p make the angle cpB=pcs, and BDE=BDA; EDA is the triangle required. "¢ For the angles EAD, EDA being bisected by aB, BD, B is the centre of the inscribed circle. Boor the angle Bop bane equal BDC, BC is equal to BD, and .*. the difference between BA and BD is equal to ac, i. e. to the given difference. Also the angles EAD, EDA are together double of the angles Bap, BDA, i. & of BCD and BCD, and ., are equal to 4B8c0D; whence the angle AED together with 4Bcp will be equal to ig right angles ; and . the angle AED is equal to the given vertical angle. Also AD 1s equal to the given base. : (31.) Given that segment of the line bisecting the vertical ong which is intercepted by perpendiculars let fall upon it from the angles at the base; the ratio of the sides; and the ratio of the radius of the inscribed circle to the segment of the base which is intercepted between the line bisecting the vertical angle and the point of contact of the inscribed circle; to construct the triangle. Let aB be equal to the given segment of vr i the bisecting line; from a and B, on oppo- ie site sides, draw ac, BD at right angles to it. =< On AaB, as hypothenuse, describe a right- Zp iN angled triangle AEB, whose sides are in the A given ratio of the radius of the inscribed ; circle to the segment of the base intercepted between the | bisecting line and the point of contact of the inscribed circle, Divide «8 in F in the ratio of the sides; and through F, draw} CFD parallel to Ar, and meeting the perpendiculars in c and D. Make BG=Bp. Join ca, and ifothice it to meet AB produced , in H; join HD; HCD is the triangle required. | inne co bisecting the angle ucp. Since BG=BD, and BH, is perpendicular to Gp, it bisects the angle GHD; whence o is. the centre of the tSor Dee circle. Toes oO draw oI perpendi- | cular to cp, and .°. parallel to Bx; whence OW? LEY ABRs Sect. 1x.] GEOMETRICAL PROBLEMS. 235 /e.in the given ratio of the radius of the inscribed circle to 4 seoment of the base intercepted between the bisecting line and the point of contact of the inscribed circle. Also from the similar triangles Arc, BFD, . APSR (OOM D.C HED; .. CH : HD in the given ratio of the sides. (82.) Given the line bisecting the vertical angle, and the differ- onces between each side and the adjacent segment of the base made N be the bisecting line; to construct the triangle. _ Let a8 be equal to the given bisecting line. ‘i, Produce it to c, so that Bc may be a fourth | proportional to AB and the given differences. ie On ac, as a diameter, describe a circle ADE; Ni : ‘in which place a line px, passing through B, © eb hii qual to the sum of the given differences ; and 7 fiom A draw AF, AG to the circumference, each equal to DE; ‘and produce them to meet pz produced in w and 1; AHI is the triangle required. - Since px is equal to the sum of the given differences, and the rectangle pB, BE, is equal to the rectangle aB, BC, #. e. to the rectangle contained by the given differences, DB and BE will be the given differences. And since AF=AG, the arcs AF AG are equal and ac being a diameter, rc and cG will also be ‘equal, .*. the angles Fac, CAG are equal, or HAI is bisected by Ac. Also since FA=DkE, and Au and ue will be equal; .. the difference between Au and HB is equal to BE one of the ‘given differences. In the same manner AI=ID; and’ ‘>.’ the ‘difference between at and 1B is equal to Bp, the other given difference. PAN G * we _ (83.) Given one of the angles at the base, the side opposite to tt, ‘and the rectangle contained by the base and that segment of it made by the perpendicular which is adjacent to the gwen angle ; to construct the triangle. Let aB be equal to the given side. Upon it describe a seg- ‘ment of a circle containing an angle equal to the given angle. 236 GEOMETRICAL PROBLEMS. [Sect. 1x Bisect AB in c; and from c draw to the cir- cumference, a line cp such that the difference between the squares of cp and cB may be equal to the given rectangle. Join Ap, pB; ADB is the triangle required. On AB describe a circle ABE. Join BE. z Then the rectangle ap, DE is equal to the rectangle Gp, DF a. €. to the difference of the squares of cp and ce or to the given rectangle. And-Bx is perpendicular to AD; AB is equa to the given side, and apB to the given angle. (34.) Gaven the vertical angle, and the lengths of two lines drawn from the extremities of the base to the points of bisection of the sides ; to construct the triangle. Let a B be equal to one of the given lines; bisect it inc; and on ac describe a segment of a circle containing an angle equal to the given angle. From Ba cut off Bp equal to one third of BA; and-from p to the circle, draw DE equal to one third of the other : given line. Produce Ep, and make pF equal to twice DE Join AE, FB; and produce them to meet in G. Join AF; AFG is the triangle required. a| Join ce. Since Bp is double of pc, and pF double of pg: EC is parallel to r@; and .*. the angle acF is equal to anc, ae, to the given angle. Hence also ar is equal to HG; an _BF being double of xc, is equal to Be; .*. AB and FE, equal to the given lines, are drawn to the points of bisection of the dea (35.) Given the lengths of three lines drawn from the angles; to the points of bisection of the opposite sides; to construct the, triangle. al Describe a triangle anc whose sides are respectively equal) to two thirds of the given lines; and complete the parallelo- gram ABDC. Join Ap; and produce cB to x, making BES] Sect. 1x.] GEOMETRICAL PROBLEMS. 237 po. Join ak, ED; AED is the triangle required. - Produce AaB, DB to F and a@. Since the ‘diagonals of parallelograms bisect each other, Af is equal to HD, and BH tO HC; .. EH is ‘equal to us and su together, 7. e. to Bc and ‘pu or to 3 BC and .°, is equal to one of the given lines. Again, since GB is parallel to ac, (Hucl. vi. 2.) it bisects AX, and BG =iac; but pp=ac, .. De=2ACc, and .*, is equal to another of the given bisecting lines. In the same manner, AF may be » ‘shown to be equal to the other given line, and to bisect DE ‘in F. -(86.) Given the segments of the base made by the perpendicular, a one of the angles at the base triple the other ; to construct the triangle. io Let aB, BC equal to the given segments, ten ‘be placed in the same straight line. Make I BD=BC; bisect AD in E,and Bein F. On ,~—y—¥D Be ‘AD describe a semicircle; and from Fr draw FG at right angles to ap. Join ac, Gp; and let ac meet ‘the perpendicular BH in H. Join uc; auc is the triangle ‘Tequired. Draw, £1 perpendicular to aD; join p1,pH. Then ax being ‘equal to ED and the angles at u right angles, AI=1D, and the jangle 14D=1DA; whence the angle piu is double of DAH. But since EF is equal to FB, and GF parallel to 1m and BH, ... \1c—Gu; and the angles pe1, pax being night angles, D1 is equal to pu, and the angle pu equal to p14, and .., double ‘of pAu. Also since pp=Bc, and the angles at B are right Bt endicular. (37.) The area and hypothenuse of a right-angled triangle being given ; to construct the triangle. 238 GEOMETRICAL PROBLEMS. [ Sect. 1 Let a3 be equal to the given hypothenuse. “Ne Bisect it in c, and on cB describe a rectan- Le \ gular parallelogram cp equal to the given area. On AB describe a semicircle AEB, . cutting the side parallel to as, ing. Join AE, EB; AEB: the triangle required. Jom cE. Since Ac=cB, the triangle ars is double of cE (Eucl. 1. 38.), and ., equal to the rectangle cp, i.e. to th given area. —_—_—_— i (38.) Given one angle, and a line drawn from one of the other bisecting the side opposite to it; to construct the triangle, whe the area is also given. Let aB be the given bisecting line ; and upon it describe a segment of a circle containing an angle equal to the given angle. Upon aB also de- scribe a rectangular parallelogram aBcp equal to the givel area; and let pc meet the circle in ©; join Ba, and produce i making AF = AE; join FB, BE; FEB is the triangle required. For Ba bisects the side pF; the angle BEF is equal to thi given angle; and the triangle BEF is double of Bar (Eucl. i 38.) and .*. equal to ABCD, i. e. to the given area. (39.) In two similar right-angled triangles, the sum of the bas of the one and perpendicular of the other is given; to determim the triangles such that their hypothenuses may contain the righi angle of another triangle similar to them, and the sum of the three areas may be equal to a given area. Let as be equal to the given sum; and p_ ER oa upon it describe a rectangular parallelogram ee . equal to the given area. On ap also describe | a semicircle, cutting cp in £; join AE, EB; 4 Tog the triangles AED, BEC, AEB are the triangles required ; as is evident from the construction. Sect. rx. | GEOMETRICAL PROBLEMS. 239 Lo (40) Given the vertical angle, the area, and the distance between lithe centres of the inscribed circle and the circle which touches the | base, and the two sides produced ; to construct the triangle. ‘Let as be equal to ; a Se i in [the given distance be- ee ab ate ‘8; ween the centres; and ” AE eg erin: jimake the angle BAC } equal to half the given : | angle at the vertex. On ac let fall the perpendicular Bc; and produce ca, till the rectangle Ap, Bc is to the given area, in i the ratio of ac : cs. Complete the parallelogram pEFc; and from the centres A and x, describe two circles touching EF in l@andr. Draw ui, 16 touching the two circles; HIE is the | triangle required. For aB is equal to the given distance between the centres. The angle 1©F is double of aur. ?@. ¢. of Bac. and .*. is equal to the given angle. And from the similar triangles ABC, AED, Me 6 AC CB?2 AD; (DE=) AG:: ADXDC: AGXEF. || But since EF is equal to half the perimeter of the triangle B11, | the rectangle AG, EF is equal to the triangle HHI; whence EHI | is equal to the given area. - (41.) Given the area, the line from the vertex dividing the base | into segments which have a given ratio, and either of the angles | at the base; to construct the triangle. On aB the given line, describe a seg- ment of a circle containing an angle ‘equal to the given angle. Describe a ‘rectangular parallelogram aBep equal l'to twice the given area; and divide Be | in the given ratio at r. Through F draw | eu parallel to an. Join pu, HA; and produce uB so that 1B | may be to pu in the given ratio. Join 143 1A will be the | triangle required. ; } Draw 1k perpendicular to an. The triangles 1K B, BFH are | similar, | : | | | ' 7 240 GEOMETRICAL PROBLEMS. [Sect. 1x; * oT EQ9B By Be BCE Cee Be 4 .. IK=FC; and the triangle 1aB is equal to half of the parallelogram Gc; and the triangle ABH is equal to nae of BG; *, [AH is equal es half the parallelogram ac, and .*. equal to the given area. And aB is equal to the given dividing line, and 1B : BH in the given ratio. (4:2.) Given. the difference between the segments of the bas made by the perpendicular, the sum of the squares of the sides and the area ; to construct the triangle. 7 ‘Take a line AB such that its square may be equal to the difference between half the given sum of the squares, and the square of half the given difference of the segments of the base. Upon aB describe a rectangular parallelogram ABCD equal to the given area; and also on AB describe a semicircle cutting cp in E. Join AE, and produce it both ways; and make ar, a, each equal to half the given difference of the segments, and make EH=EG, Join BF, BH; BFH is the triangle required. Join BG, BE. Since GeE=£H and the angles at x are right angles, ... GB=BH; and the sum of the squares of FB, BH 1s equal to the sum of ihe squares of FB, BG, 2. é. (iv. 30.) is equal to twice the sum of the squares of ra and AB, i. e. by construc- tion, is equal to the given sum. Also the difference between, FE and EH is equal to the difference between rE and EG, i. & to FG, which is equal to the given difference. And the area of Ehe triangle FBH is double of the area of the triangle aBE, and .*", (Kucl. i. 41.) equal to a scp, or the given area. FA (43.) Given the base, one of the angles at the base, and the tif. ference between the side opposite to it and the perpendicular ; to construct the triangle. Take an indefinite line AaB; and from G . any point c in it draw cp perpendicular to it, and equal to the given difference. 7 a From p draw pa making the angle pac wt ‘ S = as : equal to the given angle. Draw pre ao B Sect. 1x. ] GEOMETRICAL PROBLEMS. 24) ‘parallel to as, and equal to the given base. Join AE; to which from p draw pr=pc. Draw £4 parallel to rp, and “meeting Ap produced; EpG will be the triangle required. { _ From a draw Gs perpendicular to as. Since pF is parallel | to GE, and pc to GB, [4 DC: GB: AD: AG 2:.DF.2 GE. _ But vc being equal to pr, GB is equal to @e; whence the difference between GE and Gu is equal to HB, or-D¢, i.e. to the given difference. And the angle @px is equal to Dac, i. e. to the given angle; and DE is equal to the given base. (44) Given the vertical angle, the difference of the base and one side, and the sum of the perpendicular drawn from the angle at the base contiguous to that side upon the opposite side and the 8 segment cut off by it from that opposite side contiguous to the other engle at the base; to construct the triangle. Let as be equal to the given sum, and Bc to the given difference; and let them be placed so as to contain an angle ABC, which with the given vertical angle will be equal to two right angles. Through ce draw DCE parallel to an, and through 8B draw pBr making half a right angle with it. Join ap; and to it from B draw BG equal to Bc; and parallel to these respectively draw aF, /FH; AFH will be’the triangle required. _ Produce ru to 1; and let fall the perpendicular rx. Since BC is parallel to F1, and B@ to AF, BGs FL. BD AP Dee BGs: AR, put BC=BG, ... FI=AF; and HI=BC will be the difference ‘between the base ar and the side ru. Also since the angle ‘ BFK is half a right angle, FK=KB; and AB is the sum of the perpendicular rx and the segment KA. Also BcrH being a Parallelogram, the angle auF, or its vertically opposite 1nB, together with asc will be equal to two right angles; and .-, AHF is equal to the given vertical angle. R ; ’ 242 GEOMETRICAL PROBLEMS. [ Sect. 1x (45.) Given the base, the difference of the sides, and the segment intercepted between the vertex and a perpendicular from one of the angles at the base upon the opposite side ; to construct th triangle. Let aB be equal to the given segment in- tercepted between the perpendicular and the vertical angle ; Bc equal to the given differ- ence of the sides; draw Bp perpendicular to AB, and make BE=BA3;3 join AB, and pro- duce it; and from c to aE draw cr such that its square may be equal to the given squares of the base and segment as; draw FG perpendicular to AB; and make GH=sc. From 8H to BD, draw HD equal to the given base; join ap; AuHD is the triangle required. For ap is made equal to the given segment, and up to the given base. Also since AB=BE, AG=GF, and HB = GGy whence the squares of GA and HB are equal to the squares of GA and Gc, z. e. to the square of cr, or the squares of HD and AB, by construction. But (iv. 29.) the squares of Ap and HB) are equal to the squares of HD and AB; .°. the squares of GA and HB are equal to the squares of ap and HB; whence Ga= AD; and .*. the difference between Au and AD is equal to HG, 1. é. to BC, or the given difference. (46.) Given the vertical angle, the side of the inscribed square, and the rectangle contained by one side and its segment adjacent to the base made by the angular point of the inscribed square. To construct the triangle. Let aB be equal to a side of the inscribed square; and upon it de- scribe a segment of a circle contain- ing an angle equal to the given angle. From a draw ac perpendicular and equal to AB; and through c draw DCE parallel to as. Find o the centre of the circle; and from it to £ Sect. 1x. | GEOMETRICAL PROBLEMS. 243 DE; draw op such that the difference of the squares of op and | the radius of the circle may be equal to the given rectangle. _ Join pA, and produce it tor. Join Fs, and produce it to £; DFE is the triangle required. | From p draw the tangent pc. Join go. Then the squares of DG, Go are together equal to the square of Do, i. e. to the square of Go and the given rectangle; and .*. the square of DG, i. e. the rectangle ap, pF is equal to the given rectangle ; *. DF is a side of the triangle, in which AB is the side of an inscribed square. And FF is equal to the given vertical angle. R.2 APPENDIX. PLANE TRIGONOMETRY. - (1.) Der. 1. Plane Trigonometry is that branch of Mathe- ‘matical science which treats of the measures of angles and the ‘relation between the respective sides and angles of plane Lemma I. (2.) If from the centre of a circle two straight lines be drawn to its circumference, the included angle will be to four right angles, as the intercepted arc is to the whole circumference. Let Ba, Bc be two straight lines drawn from , 2? ¢ the centre B of the circle apr, meeting the cir- cumference in A and c; the angle asc will be ~ B to four right angles as the arc ac is to the whole circumference. - Produce AB till it meets the circumference again in F; and hrough s draw pe perpendicular to AB, meeting the circle in pandx. Then (Eucl. vi. 33.) the angle anc : the right angle ABD::the arc Ac : AD; and quadrupling the consequents, the angle asc : four right angles :: the arc ac ; 44D, or the whole circumference. 246 ELEMENTS OF PLANE TRIGONOMETRY. Lemma II. (3.) If from the common centre of two circles two straight lines be drawn intersecting the circumferences ; the antercepted arc of either circle will be to the intercepted arc of the other circle, as the circumference of the first circle to the circumference of the, second. From s the common centre of two circles are drawn the lines BA, BC, intersecting the circum- ferences in A, C, a, c; the arc ac: ac:: the whole circumference of the first circle : the whole circumference of the second. For (2) the arc ac : the whole circumference (c) of which it is an arc :: the angle anc: four right angles; which is the Cc ves same ratio with that of ac: the whole circumference (c’) of | “ which it is an arc, *. (Kucl*y.'15:)‘Ae ? c $) ae: o’, and alt.ac:ac::c:c’. (4.) Cor. 1. If the circumferences of any two circles be seve- rally divided into the same number of equal parts, whatever number of such parts is contained in any arc (Ac) of one circle, the same number will be contained in that are (ac) of the other circle which subtends the same angle. | . (5.) Cor. 2. Any arc of a circle is the measure of the angle which it subtends at the centre. For (See Fig. 1.) draw any straight line pw to the circumfer- ence; then (Kucl. vi. 83.) the arc ac : the arc aw :: the angle ABC ; the angle a Bz, i. e. in whatever proportion the angle at the centre is increased, the subtending arc is increased in the same proportion *, 1 If the angle be not at the centre, but between it and the circumference + it will be measured by half the sum of the ares intercepted between the sides, and the sides produced. If it be without the circle, it will be measured by half their dif- ference. (ii. 24.) If the angle be formed by a tangent and chord, it will be measured by half the are subtended by the chord. If by a line cutting the circle anda tangent, or by two tangents, it will be measured by half the difference between the convex and concave ares intercepted between the lines. ’ : A ELEMENTS OF PLANE TRIGONOMETRY. 24.7 (6.) Angles at the centres of different circles vary as the arcs which subtend them, directly, and the radii of the circles inversely. | _ For (2) the arc ac: the circumference apr a :: the angle _. the arc Ac ADEA x four right angles; and since the circumference ADEA & AB, and four right angles is an invariable quantity, the angle asc: four right angles. Hence the angle apc = : AC ABC X ——. AB (7.) Till lately, Geometers almost unanimously agreed to _ divide the circumference of the circle into 360 equal parts called - Degrees; each degree into 60 equal parts called Minutes; and _ each minute into 60 equal parts called Seconds, &c. This ‘method appeared to the Greek Geometers to afford some facili- ‘ties for calculations, in consequence of the great number of divisors of 860 and 60; but it is in reality subject to inconve- ‘nience from complicated numbers, and tediousness in opera- tions. Upon the introduction therefore of a new system of _ weights and measures into France, into which they were deci- “mally divided and subdivided, it was thought proper to make a similar division of the quadrant. The quadrant is accordingly divided into 100 minutes, each minute into 100 seconds, &c. > One great advantage of which method is its identity with the common decimal scale of notation; thus 6 deg. 15 min. 53 sec. would be represented by 6.1553 deg. and 34 deg. 5 min. 9 sec. by 34.0509. The length of a degree then on that scale will be less than one on the common scale in the proportion of 90: 100, or 9:10; a minute in the proportion of 90 x 60 : 100 x 100, or _ 27 : 50, and a second in the proportion of 90 x 60 x 60 : 100 x 100 x 100, or 81 : 250. ial q If then n = the number of degrees on the new scale, the cor- ix 7 ; 3 responding number on the former scale will be = — a = : es 9 sy n— 10 Hence .. the following rule, for reducing the new to the old graduation. 248 ELEMENTS OF PLANE TRIGONOMETRY. Express the measure in decimals, and from it subtract _i,th of this number ; mark off the decimals in the remainder, which multiply by 60, and mark off the decimals again ; these again | multiply by 60 and mark off as before, and so on; the whole numbers so obtained will be the degrees, minutes, seconds, &c. on the common scale. Thus to determine how many degrees, minutes, &c. in the. old graduation, correspond to 46 deg. 48 min. 15 sec. in the new. From 46.4315 Subtract 4.64315 41.78835 60 47.380100 60 18.06000 or 41 deg. 47. min. 18.06 sec. (8.) Degrees, Minutes, Seconds, &c. are expressed by °, ’, ”s thus 10°. 15’. 45”. represent an arc of ten degrees, fifteen minutes and forty-five seconds. (9.) Ares and angles are expressed without distinction in | trigonometrical calculations by degrees, minutes, and seconds. (10.) A right angle is measured by an arc of 90°; two right angles by 180°; half a right angle by 45°; and each angle of an equilateral triangle by 60°, (11.) Der. 2. With the centre o and radius y Bp oA describe a circle, and draw the diameters AOC, / BOD at right angles to each other. These will ¢ ae divide the circumference into four equal parts called Quadrants. * (12). Der. 3. The Complement of any arc or angle is the difference between that arc or angle and a quadrant or 90°. Thus taking any point p in the quadrant AB; the complement of the arc Ap is the arc BP; and the complement of the angle ELEMENTS OF PLANE TRIGONOMETRY. 249 aor is the angle Bop. If ap=62°. 16, Bp its complement= 27°. 44’. Also the complement of 35°. 15’. 45” is an are of 54°, 44’.15”. And in general a being any are whatever, its com- plement is 90°—a; or A—90", if a be greater than 90. The two acute angles of a right-angled triangle, being toge- ther equal to a right angle, are complements to each other. (13.) Der. 4. The Supplement of any arc or angle is its defect from a semicircle or 180°. Thus, taking any point Pp, the supplement of the arc ap is PBC, and of the angle aop is the angle poc; the supplement of aBp’ is cp’, and of the angle Aop’ isthe angle cop’. The supplement of 32°. 42’. 18” is an arc of 147°.17’.42”. Ifa be any arc, its supplement is (180—4). Also the supplement of (90—.) is (90+ 4). In every triangle, the sum of the three angles being equal to two right angles, each angle is the supplement to the sum of the other two. (14.) Der. 5. The Chord of an arc is a straight lme joining the extremities of the arc. Thus Ap is the chord of the arc Ap. AB isthe chord of the quadrant AaB. The diameter is the chord of a semicircle ; and the chord of the whole circumference a (0. (15.) If the chord does not pass through the centre, as AP, it is the chord of two unequal arcs, AP and the remaining part of the circumference apcsp. In speaking however of the chord of an arc of a circle, the less arc is generally meant. (16.) Cor. The chord of 60°= radius of the circle. For the triangle Ao Pp is equiangular and .*. equilateral. » The chord of a quadrant = r \/2. (17.) Der. 6. The Sine or Right Sine of an arc is a straight line drawn from one end of the are perpendicular to the dia- _ meter passing through the other end of the arc. _ ‘Thus, if from p, pe be drawn perpendicular to ac, it will be _ the sine of the arc AP less than a quadrant. If pq be perpen- - dicular to Bo, it will be the sine of the arc Bp. Also P’ E, 250 ELEMENTS OF PLANE TRIGONOMETRY. Pp” H’, P’”’ B’” are the sines of arcs ap’, AP”, (et Bae AP” measured from A and terminating in the second, third and fourth quadrants respectively oe: If from a, Aw be drawn perpendicular to op, it will be the sine of ap, and be equal to px, (18.) To trace the changes in magnitude of the Right Sine. The sine begins with the arc; when the arc is nothing, P coinciding with 4, E£ also coincides with it. But as p moves from A towards B, PE continually increases, é. e. as the arc in- creases, the sine increases; and when Pp coincides with B, E coincides with 0, and the sine of 90° becomes radius. When_ the arc is greater than 90°, and terminates in the second qua- drant, as the arc increases, the sine decreases, till it becomes = 0, at the end of the second quadrant, p’ and x’ both coin- ciding with c. Ifthe arc terminates in the third quadrant, the sine continually increases, till at the end of it, it becomes radius; and if in the fourth quadrant, it continually decreases again till it becomes = 0, * Hence it appears that the sine never exceeds radius. '(19.) To trace the changes in the Algebraic Sign of the right sine. If the right sines of arcs in the first quadrant be reckoned positive, they will be positive or negative for arcs which termi- - nate in the other quadrants, according as they are drawn in the same or contrary directions with the former, 2. e. according as they fall on the same or contrary sides of the diameter ac. Hence during the two first quadrants the algebraic sign 1s posi- tive, and for arcs which terminate in the two next it is negative. — (20.) Cor. The sine of an arc is equal to the sine of its sup-. plement. For PE is equal to the sine of csp which is the supplement of ap. Thus 48°. 15’ being the supplement of 131°, 4.5’, the sine of 131°. 45’= the sine of 41°. 15. Also sin. (180—a) = sin. A, and sin. (90+ A) = sin. (90—a),. The exterior angle of a triangle being the supplement of its — interior adjacent angle, their sines are equal. | ELEMENTS OF PLANE TRIGONOMETRY. 251 (21.) Der. 7. The Versed Sine is that part of the diameter passing through the beginning of the arc, which is intercepted ‘between the beginning of the arc and the right sine. _ Thus Ax is the versed sine of the are Ap. AE’, AE ,AE are the versed sines of the arcs Ap’, ap”, ap” respectively. (22.) To trace the changes in magnitude of the versed sine. _ The versed sine begins with the arc, and when the arc is ‘nothing, the versed sine = 0, £ coinciding with a. It increases with the are, till the arc becoming a semicircle, the versed sine ‘becomes the diameter; after which it again decreases, whilst ‘the arcs terminate in the third and fourth quadrants, till at the end of the fourth quadrant it becomes = 0. But as it is always “measured in the same direction from 4, it is positive through all the four quadrants. _ (23.) Cor. Hence (Eucl. vi. 8.) the chord of an arc is a mean : proportional between the diameter and the versed sine of that are. _ (24.) Dur. 8. The Tangent of an arc is a straight line which Biouches the circle at one end of the arc, and is terminated by _ the radius produced through the other end of the arc, _. Thus, if the line Tat’ touch the circle in 4, Z and ap be an arc less than a quadrant, and oP _ joined; since TAO, A oP are less than two right : angles, op and aT will meet above ao; and aT willbe the tangent of ap. But if ap’ be taken an arc greater than a quadrant, and less than a A semicircle, TAO, AOP’ are greater than two right angles, and + pa and p’o will meet on the contrary side of ao, or av’ will be the tangent of app’. In the same manner it may be shown that aT is the tangent of an arc ABcP” terminating in the third quadrant ; and av’ of one terminating in the fourth. (25.) To trace the changes in magnitude of the tangent. The tangent begins with the arc; when the arc is nothing, P coinciding with a, T coincides with it also. Whilst the arc is less than a quadrant, as AP increases, the angle aor, and .°, AT increases, or the tangent increases ; and as the arc approximates 252 ELEMENTS OF PLANE TRIGONOMETRY. to a quadrant, op approximates to the direction oB which is parallel to a, and .*, the tangent approximates to a line greater than any that can be assigned. But if the arc be greater than a quadrant, and less than a semicircle, as the arc increases, BP’ increases, .*, the angle BoP’ or Dov’ increases, and aot’ de- creases, .", PT’ intersects AT in points continually nearer and nearer to A, or the tangent decreases, and at the end of the second quadrant, the tangent = 0. In the same manner it may be shown that the tangent of an arc terminating in the third quadrant continually increases till it approximates to a line greater than any that can be assigned ; and for arcs terminating in the fourth quadrant, it decreases till it becomes = 0. (26.) To trace the changes in the algebraic sign of the tangent. | If the tangent of an arc in the first quadrant be reckoned positive, it will be positive or negative for arcs terminating in the other quadrants, according as it is drawn in the same or contrary directions with the tangents of arcs ending in the first quadrant, 7. e. according as it falls on the same or opposite sides - of the diameter ac. Hence the tangents of arcs ending in the first and third quadrants have a positive sign, and those of arcs ending in the second and fourth, a negative one. (27.) Cor. 1. The tangent of an arc is equal to the tangent of its supplement ; For the angle poc is equal to the vertically opposite angle AOP”; and (24) aT is the tangent of app” which measures _ Aop”. As one of the arcs is less, and the other greater than a quadrant, they will be affected with contrary signs. Hence if a be any angle, tang. (180—a) = — tang. a and tang. (90+ a) =— tang. (90 — a). (28.) Cor. 2. Tang. 45° = radius. For in the triangle OAT, the angle oar is a right angle, and aor= 45°, ». aro=45°= AOT and AaT=A oO, (29.) Der. 9. The Secant of an arc is the line drawn from the centre through the end of the arc, and produced till it meets the tangent ; cal 24. oT is the secant of the arc ap; and the _ AP” terminating in the third and fourth quad- rants will be ot, ov’ respectively. ELEMENTS OF PLANE TRIGONOMETRY. 253 Thus, the construction remaining as in Art. secant of the arc ap’ terminating in the second quadrant is or’. The secants also of arcs ap”, (30.) To trace the changes in magnitude of the secant. ‘When P coincides with a, the secant coincides with the radius _ OA, 7. e. in the beginning of the circle the secant = radius. In ft the first quadrant as AP increases, AT increases (25.) .. OT increases (Eucl. i. 47.) 7. e. as the arc increases, the secant in- _ creases; and when the arc approximates to a quadrant, the secant approximates to a line greater than any that can be assigned. But if the arc be greater than a quadrant and less _ than a semicircle, as the arc increases, T’ approaches nearer to A, and the secant decreases, till at the end of the second quad- rant, it becomes equal to radius. In the same manner, the secant _ of an arc ending in the third quadrant increases from radius till it approximates to a line greater than any that can be assigned ; _ and for arcs terminating in the fourth quadrant, decreases again till it becomes = radius. Hence the secant is never less than the radius. (31.) To trace the changes in the algebraic sign of the secant. If the secants of arcs in the first quadrant are reckoned posi- 4 tive, those of arcs ending in the other quadrants will be positive _or negative according as they are drawn from the centre through ee Se ie ce ee ee _ the end of the arc, or from the end of the arc through the cen- tre. Hence the secants of arcs ending in the first and fourth quadrants have a positive sign; and of those ending in. the second and third, a negative sign. (32.) Cor. 1. The secant of an arc is equal to the secant of its supplement. For oT is the secant of App” which is the supplement of ap. _ As one of the arcs is greater and the other less than a quadrant, the secants will have contrary signs. If a be any angle, sec. (180 — a) = —sec. A, and sec, (90+ a) = — sec, (90 — A). 254. ELEMENTS OF PLANE TRIGONOMETRY. (33.) Cor. 2. Sec. 45° = rvV/2. (34.) Der. 10. 'The Cosine of an arc is that part of the dia- meter passing through the beginning of the arc, which is inter- cepted between the centre and the right sine. Thus 0& is the cosine of the arc aP in the first quadrant; o@ the cosine of the arc BP. Also 0&’, 0B”, 08” are the cosines of arcs AP’, AP”, AP” respectively terminating in the second, third and fourth quadrants. Pr. (35.) To trace the changes in magnitude of the cosine. When P coincides with a, = also coincides with it, and in the beginning of the arc, cosine = radius. In the first quadrant, as AP increases, EO decreases, i. €. as the arc increases, the cosine de- creases; till at the end of the first quadrant, the cosine=0. But if the arc be greater than a quadrant, and less than a semicircle, as A P’ increases, G £’ increases, and .*, as the arc increases the cosine increases, till at the end of the second quadrant it = radius. In the same manner the cosines of arcs terminating in the third quadrant decrease from radius, till at the end of the third quad-— rant the cosine = 0, and in the fourth quadrant increase till at the end of it, cosine = radius. Hence the cosine is never greater than radius. (36.) To trace the changes in the algebraic sign of the cosine. If the cosines of ares ending in the first quadrant be reckoned positive, those of arcs terminating in the other quadrants will be positive or negative according as they are drawn in the same or opposite directions with those of arcs ending in the first quad-_ rant, 7. e. according as they are measured on the same or oppo- site sides of the centre with the cosines of arcs in the first quadrant. Hence the cosine has a positive sign in the first and fourth quadrants, and a negative one in the second and third. (37.) Cor. 1. The cosine of an arc is equal to the cosine of | its supplement. _ than a quadrant. But if the arc asp’ be - greater than a quadrant and less than a semi- _ of the arcs ap”, ap” ending in the third and in B the end of the quadrant a8, Br will be ELEMENTS OF PLANE TRIGONOMETRY. 255 For OX is the cosine of the arc cpp which is the supplement of Ap. But as one of the arcs is less and the other greater than a quadrant, they will be affected with contrary signs. Hence if A be any arc, cos. (180 — a) = — cos. A; and cos. (90 + a) = ~— cos. (90— a). (38.) Cor. 2. The cosine of an arc is equal to the sine of its complement. For o£ the cosine of ap is equal to p@ the sine of pp. Hence cos. A=sin. (90 — a) and cos. (90— a) = sin. a. Ifa = 30°, cos. 30°= sin. 60°, and cos. 60° = sin. 30°. (39.) Cor. 8. Cos. 45° = sin. 45° = Fi (Eucl. i. 47.) (40.) Cor. 4. The versed sine of an arc less than a quadrant is equal to the difference of the radius and cosine; and of an arc greater than a quadrant is equal to their sum. (41.) Der. 11. The Cotangent of an arc is a line touching the circle at the end of the first quadrant and meeting the radius produced through the end of the are. Thus, if 181’ be drawn touching the circle the cotangent of the arc ap which is less circle, B1’ is its cotangent. Also the cotangents fourth quadrants are respectively Bi and Br’. (42.) To trace the changes in magnitude of the cotangent. When P coincides with a, opt falls in the direction OA, and _ being parallel to pr, the cotangent 1s greater than any assignable line. In the first quadrant, as the arc Pp increases, o1 inter- sects the line Br in points continually nearer to B. Hence in arcs less than a quadrant, as the are increases, the cotangent _ decreases, till at the end of the first quadrant it = 0. In arcs greater than a quadrant and less than a semicircle, as the arc _ ABP’ increases, BI’ the cotangent increases; till as the arc ap- _ proximates to a semicircle, the cotangent approximates to a line 256 ELEMENTS OF PLANE TRIGONOMETRY. greater than any that can be assigned. In the same manner the cotangents of arcs ending in the third quadrant continually decrease, till at the end of it, the cotangent = 0; and if the arcs terminate in the fourth quadrant, the cotangent increases again, till it approximates to a line greater than any that can be assigned. | (43.) To trace the changes of the cotangent in the algebraic sign. If the cotangents of arcs ending in the first quadrant be reckoned positive, those of arcs ending in the other quadrants will be positive or negative according as they are drawn in the same or contrary directions with those of arcs ending in the first quadrant, ¢. e. according as they fall on the same side of the diameter Bp with the cotangents of arcs ending in the first quadrant or the contrary. Hence the cotangents of arcs ending in the first and third quadrants have a positive, those of arcs ending in the second and fourth, a negative sign. (44.) Cor. 1. The cotangent of an are is equal to the cotangent of its supplement. For Br is the cotangent of cpp which is the supplement of AP. But these cotangents have contrary signs. If .*, a be any are, cot. (180 — a) = — cot. a, and cot. (90 + a) = — cot. (90 — a). | (45.) Cor. 2. The cotangent of an are is equal to the tan- gent of its complement. For Bi the cotangent of ap is the tangent of Bp which is the complement of ap. If ., a be any arc, cot. A= tang. (90 — a) and cot. (90 — a) = tang. a. (46.) Cor. 3. Cot. 45 = radius. (47.) Der. 12. The Cosecant of an arc is a straight line drawn from the centre through the end of the arc and produced till it meets the cotangent. Thus the arc Ap being less than a quadrant, o1 is its ‘cose- cant. Also or’ is the cosecant of the arc app’ greater than a quadrant and less than a semicircle. And 01, o1’ are the cose- ELEMENTS OF PLANE TRIGONOMETRY. ys b ur cants respectively of the arcs AP”, AP and fourth quadrants. terminating in the third (48.) To trace the changes in magnitude of the cosecant. When P coincides with a, opi falls in the direction of oA which is parallel to s1, and therefore at the beginning of the arc the cosecant is greater than any assignable line. In the first quadrant, as the are A P increases, the line o1 cuts B1 in points continually nearer to B, 7. e. as the arc increases, the cosecant decreases, till at the end of the first quadrant it becomes equal to radius. As the arc increases from a quadrant to a semicircle, I B increases (4:2) and the cosecant increases, till as the arc ap- -proximates to a semicircle, the cosecant approximates .to a line greater than any that can be assigned. If the arc terminates in _ the third quadrant, as the arc increases, the cosecant decreases, till it becomes equal to radius; and during the fourth quadrant it increases again, till it approximates to a line greater than any that can be assigned. Hence the cosecant is never less than radius. (49.) To trace the changes in the algebraic sign of the cosecant. The cosecants of arcs terminating in the first quadrant being . reckoned positive, and drawn from the centre through the ends _of the arcs ; the cosecants of arcs ending in the other quadrants will be positive or negative according as they are drawn from the centre through the ends of the arcs, or from the ends of the arcs through the centre. And hence the cosecants of arcs ending in the two first quadrants have a positive sign, and those of arcs ending in the two last a negative sign. (50.) Cor. 1. The cosecant of an arc is equal to the cosecant of its supplement, and they have the same sign. If .. a be any arc, cosec. (180 — A) = cosec. A, and cosec. (90 + A) = cosec. (90 — a). (51.) Cor. 2. The cosecant of an arc is equal to the secant of its complement; for o1 the cosecant of the arc Aap is the ‘secant of Bp the complement of ap. Hence cosec. a = sec. (90 — a) and cosec. (90 — a) = sec. A. “ae akg 258 ELEMENTS OF PLANE TRIGONOMETRY. (52.) Cosec. 45° = R/2. (33.) (53.) In trigonometrical calculations, when the quantity re- quired is deduced in terms of the sine, the case is ambiguous, since the sine of any arc and its supplement (19) have the same algebraic sign. The ambiguity may however be removed by some other consideration. But if an expression be deduced in terms of a cosine or a tangent, there will be no ambiguity, since a positive cosine (36) or tangent (26) denotes an arc less than 90°, and a negative cosine or tangent indicates an arc greater than 90° and less than 180°. In calculations arcs and angles are generally less than 180°. Prop. lI. (54.) The sines, cosines, tangents, &c. of the same angles at the centres of different circles are proportional to the radii of those circles. Let as, ab be two arcs described with the BF centre c and radii cA, ca, subtending the same NDF angle at c. Draw Bs the sine and ar the tan- gent of AB; Js the sine and at the tangent of ab. Then the triangles css, cbs being similar, BS; 68:3. CB 00, i.e. the sines are proportional to the radii. Also cs : Cs::cB: cd, or the cosines are proportional to the radii. And since C8 : cs::cB:cb::cA: ca; “. (Huck viloias Sagi: ce: cd, or the versed sines are proportional to the radii. Again from the similar triangles arc, atc, AT >; al;:AC : ac, or the tangents are proportional to the radii. And cr: cf::ac: ac, or the secants are proportional to the radii. ey ELEMENTS OF PLANE TRIGONOMETRY. 259 (55.) Cor. 1. Hence et a and .*, if s and s represent ‘ . oe Ss Ss R _ the sines of the same angle to radii R and r, aes and s = 2 5 . mes. itr ly s = rs, ard Sle te And the same is true of “_ any other corresponding lines. Hence if any expressions be ‘calculated for rad. = 1; the corresponding expressions in a circle whose radius is r will be determined by taking each line an R™ part of the former. (56.) Cor. 2. Ifa given radius be divided into any number _ of equal parts, and the sines &c. of every angle be given in such — parts, the sines &c. of any given angle may be found, corres- ponding to another given radius. Prop. II. (57.) To find the relation between the sines, cosines, tangents _&c. of the same angle. ey Let ap be any arc, PE and o& its sine and ‘cosine, draw the tangent av, secant or and ‘cotangent Br. Then Px being parallel to a7, the triangles opz, OAT are equiangular ; or cos. : sine::rad. : tang. sine sine* , “tang. =rad. x ——-= ——., ifr=l. COs. COS. ee.” * In the second quadrant the sine is positive and the cosine negative. Hence : 7 . Hang. (180—a) ee ey eA . — cos. A COS. A Sshown in Art. 26. When the sine and cosine have the same algebraic sign, the tangent will have a positive sign, and when different, a negative one. Also since cos. 90° = 0 and = — tang. a, which confirms what was A 1 ; ‘ sin. 90° = Rr, tang. 90° = 9 — >» Which confirms what was shown in Art. 25. $2 . cet 260 ELEMENTS OF PLANE TRIGONOMETRY. Also from the same triangles, Of : OP!:0A : OT, 2 *k ae “ if R=1. or COS. ° R:.R: Sec. .°. 8€C. = COS. COS. Also from the similar triangles OPE, BOI, PE POs*0B 705, 0F SIL. BR ..COser. 2 ms cose, a Gf w—1): And from the same triangles, PE : EO‘: BO: BI, or sin. : cos.::R : cotang.; cos. /¢cos.J[ . cotang.=R xX —— = ——, ifr=l. sine sine Also from the similar triangles OAT, OBI, TA :AO?:0B: BI, or tang. : R::R : cotang.; R? ] .. cotang. = Pine eseranee Rls (58.) Cor. 1. If either the sine or cosine of an arc A be known, all the rest may be found. . For (Eucl. i. 47.) po? =pr’ + £0’, i. €. R’=sin.” A + COS.” A, “sin. A=,4/(R’—cos.’ a), and cos. A = 4/(R’ — sin.” a), which values may be substituted in the preceding expressions. | (59.) Cor. 2. Since oT?=0A’+ AT’, Sec.” A=R’ + tang.’ A, and .°. sec. A = 4/(R’ + tang.’ A), whence cos. A’ = 2 sa rn et tah ,and sin, A= hte Sc \/(R* + tang.’ A) 4/(R’ + tang.’ A) (60.) Cor. 8. Also tang. A = 4/ (sec.’ A — R’), and cosec.? A=R’+cotang.’ A. (61.) Cor. 4. If the radius be represented by unity, the expressions become, sin. A = 4/ (1 — cos.’ A), cos. A = J/ (1 — sin a), sec. A = 4/ (1 + tang.’ A), tang. A = °. * Hence the secant will have the same algebraic sign as the cosine. + Hence the cosecant will have the same algebraic sign as the right sine. + Hence the cotangent will have a positive sign, when the cosine and sine have ' the same algebraic signs, and a negative one, when different ; or as also appears from the next equation, it will have the same sign as the tangent, ELEMENTS OF PLANE TRIGONOMETRY, 261 ¢ ] / (I - tang.’ A) , tang. A x cotang. A=]. / (sec. A — 1), cos. A = Eas Deg . tang.’ A y/ (1 + tang.” a) Prop. III. (62.) The sine of any arc is equal to half the chord of double the are. - Let the arc ps be double of pa. Join oa, PB intersecting each other in zr. Since ps is double of pa, PA=AB, and .°, the angles por, BOE are equal, but ore is equal to ops, and OE is common to the triangles OPE, OBE, .°. (Eucl. i. 26.) pz is equal to EB, and .. is half of pB; and the angles orp, OF B are equal, i. e. each is a right angle; hence Pre is the sine of the arc ap, and is half the chord of double the arc. (63.) Cor. 1. Hence the sin. 30° = } radius. For (16) the chord of 60° = radius. Also cos. 30° = // (R’—sin.’ 30) = J/ (v=) 4 Die? 2R tang. 80° = —<-, and sec. 30° = “=. ss Si" eo" J/3 ait ak /3 _ Hence sec. 30° is double the tang. 30°. (64) Cor. 2. Sin. 60° = << (38), and cos. 60" = 5. : - . i ll Versed sine 60° = a tang. 60° = rv/38, and sec. 60° = 2p. Prop. IV. (65.) The diameter is to the versed sine of any arc as the square of radius is to the square of the sine of half that arc. : : ts :. ee aa 262 ELEMENTS OF PLANE TRIGONOMETRY. Let ABD be a semicircle, ap its diameter, 5S ACB any arc, whose chord is 4B and versed sine ¢ AE. Join DB, and from the centre o draworc & Tan perpendicular to the chord aB, and meeting the : circumference in C; then AB is bisected in F, and the are AB in c; and aris the sine of as. Also the triangles paB, OAF are similar, whence DA ; AB!:0A : AF. But (Eucl. vi. 8. and v. Def. 10.) DA AHS DAGY A De DARA Kae Ager Ade (66.) Cor. 1. Hence the rectangle contained by the radius and the versed sine of any arc is equal to twice the square of the sine of half that arc. (67.) Cor. 2. And the square of the sine of any arc varies as the versed sine of double that arc. Prop. V. (68.) Radius is to the cosine of any arc as twice the sine of that arc is to the sine of double that are. The same construction remaining, oF is the 15 cosine of Ac; and the triangles 0a F, BA, hay- Va ing the angle at 4 common, and right angles at f iN F and £, are equiangular ; whence 0A : OF}: (BA=) 2AF: BE. (69.) Cor. 1. Hence the rectangle contained by the sine and cosine of any are « the sine of double that arc; for BE « 2AFX OFQKAFX OF, 7. é. sin. 2A oO Sin. AX COS. A. Also R x sin. A=2 sin. 3 A x cos. 2a. (70.) Cor. 2. If from 0, oc be drawn perpendicular to pB, it bisects it, .. DB=2BG=20Fr=2 cos. 1aB; or the chord of an arc is equal to twice the cosine of half the supplemental arc. ELEMENTS OF PLANE TRIGONOMETRY. 263 Prop. VI. + (71.) In any right-angled triangle, radius is to the sine of either of the acute angles, as the hypothenuse is to the side opposite to that angle. _ Let asc be a right-angled triangle, having ae ‘the angle at aa right angle. With the cen- Beh tre c, and radius cp = the tabular radius, de- at scribe an arc DE, meeting cB produceding; € A FD it will be a measure of the angle c. Let fall _ the perpendicular EF, which will be equal to the tabular sine of the angle c. The triangles cer, cBa being similar, CE: EF::CB: BA, or rad. ; sin. C3: GB: BA. _In the same manner, if BA be produced till it be equal to the _ tabular radius, and a circular arc be described with B as a cen- _ tre, it may be shown that rad. : sin. B :: BC: GA. Pe jecon) |.) oRad. 3 cos. ce OB CAs anata. - COS<8 °. OB. HA. fe (/o.) Cor. 2.. If rad. = 1, BA) = cB x sin. C, or cB x COS. B, : and CA = BC xX Sin. B, Or BC X COS. C. Prop. VII. _ (74) In any right-angled triangle, radius is to the tangent of either of the acute angles as the side adjacent to that angle is to the opposite side. Let cas be a right-angled triangle, having the angle at A aright angle. With the centre G _c, and radius cp = the tabular radius, describe B -acircular arc DE meeting cB produced in £; the pe is the measure of the angle c; and drawing p@ perpendicular to pc meeting 264 ELEMENTS OF PLANE TRIGONOMETRY. CE in G, DG will be the tangent of px or of the angle acs. Now the triangles pec, asc being similar, DC: DG i: AC: AB, or rad. ; tang. © ::.AC :_AB, In the same manner it may be shown that rad. : tang. B:: AB ; AC. (75.) Cor. Rad. Y séc. 6:1: AC 2 Be, and rad. : sec. B:!: AB : BC. Propaey Ula: (76.) Having given two parts of a right-angled triangle ; to find the other parts. This resolves itself into four cases. B 1. Having given the hypothenuse cB, and the angle c; to find the rest. Since (71) tabular radius: sin. c:: BO: BA, 6 A and the angle c being given, its sine may be found from the tables, the three first terms in the foregoing proportion are known, .. BA = HORSES may be deter-_ mined. Also AC = 4/ (BC’ — BA’) = 4/{(Bc + BA).(Bo — BA) may be computed, or three terms in the proportion (72) rad. : _ BC X COS. C cos. C :: BC: CA being known, ca = a also determined. And the angle Bn = 90° — c. Ex. Suppose the angle c= 30° and pc = 10 chains. Then 1 sin. 380° = 3r, and cos. 30° = a Rye. 1B ee a = 5, and ca = 4/(15 x 5) = 54/3, orca = 10 x V3 ie 5/3, wa and B = 90° — 30° = 60°. By the same method the triangle may be solved, if the hypo- ELEMENTS OF PLANE TRIGONOMETRY. 265 thenuse Bc and the angle B be given; the point B being now _ considered as the centre. 2. Having given one side Ac and the angle c; to find the rest. Since (74) tabular radius : tang. C1: AC: AB, and the angle c being given, its tangent may be “apa from the tables; the three first terms in the preceding proportion are known, and AC x tang. C | BA = aaa ts ate may be determined. Also (75) tabular radius : sec. ¢ :: AC : CB, inwhich propor- — AC xX Sec. C tion three terms being known, the fourth, csp = = may be determined. Or if in the tables the secants are not computed, tab. cos. c AC XR Cos. C Also the angle B = 90°—c. Ex. If ac = 6 chains, and the angle c = 60°; 7 RIS AC?TCB,..«. CB = » which may be determined. tang. 60° = ,/3.R, .. ap = OXY EE _ 6/5, Also sec. 60° = 2k, .°. CB = eae 12, 6xR or cos. OO" = 1 2,..°, CB = er ao and the angle Bn = 90°—60° = 30°. In the same manner the triangle might be solved, if AB and the angle B were given. If Ac and the angle B were given ;- since the angle c = 90°— B, C is also known, and .*. this is reducible to the second case. 3. Having given the hypothenuse Bc, and one side ac; to find the rest. Here (72) Bc : cA :: rad. : cos. c, in which proportion the RXCA three first terms being known, cos. c = may be com- puted, and .°. the angle c determined from the tables. 266 ELEMENTS OF PLANE TRIGONOMETRY. Also the angle Bn = 90°—c and is .*, known. a And AB = 4/ (BC’—CA’) = 4/ { (BC+CA).(BC—CA) } which © may be found ; _ BCXsIm. C may be Or since rad. : sin. C :: BC: BA, ... BA computed. In the same manner the triangle may be solved, if Bc and aB be given. 4, Having given the two sides Ba and Ac containing the right angle; to find the rest. Here (74) ac ; AB :: rad. : tang. c, in which proportion the three first terms being known, the tang. c = a“ may be computed, and the angle c determined from the tables. Also the angle 8 = 90°—c. And tabular radius : sec. c :: AG : CB, whence CB = aah Js coe may be determined ; CXR Os. C or CB = ~ may be computed. (77.) By Case 2. the height of any object may be determined, the base of which is accessible. From the base a, measure along the horizontal plane any length ac. At c let the angle be observed which the object subtends. Then its height Ba = AC X tang. C ee Prop. IX. (78.) The sides of any triangle are to one another as the sines of the angles opposite to them. Let asc be any triangle, Bp a perpendicular let fall on ac, and 1. Let it fall within the triangle, then B (71) AB: BD:: rad. : sin. A, and BD : BC :: sin. C: rad. LO: -. €® equo per. AB ; BC }: sin. C: sin. A As c Ai Os _* ELEMENTS OF PLANE TRIGONOMETRY. 267 2. But if Bp falls without the triangle, B Ai eee, EAC SHITE A and BD : BC :: sin. BCD; rad. life PAB > BOS sine BCD 2 sina ° . e e__—_—_ | °: sin. BCA ; sin. A, since the A C D angle BCA is the supplement of Bcp (20). aN Prop. X. | (79.) Having given, in any triangle, a side and an angle oppo- site to it, and also one other angle or one other side, the remaining sides and angles may be found. 1. Having given the angles at a and c, and oe the side pc; to find the rest. Since the angles at a and ¢ are given, their sines may be found from the tables, and (78) sin. A; Sin. C }: BC: BA, In which propor- BC X sin. C - tion the three first terms being known, BA=—. . — may be determined. Also the angle B = 180°—(a+c) and may .. be found. Whence also sin. A : sin. B 3: BC: CA, BC X sin. B sin. A In the same manner, if the angles a and B and the side Bc; or the angles B and c and the side ac; or the angles a and c and the side ac were given; the other sides and angles might be found. Ex. 1. Hence may be determined the height of an object, the base of which is inaccessible. Let as be the height; along the horizontal base measure any distance cp, and at c let the vate angle BcD be observed, and at p the angles .. / As BT and .. cA = may be computed. B oe BDC, BDA; then in the triangle BDC, the angle cBD=180°—(snep+s8D¢c) and ., is known; and sin. CBD : Sin. BCD :: CD: DB, 268 ELEMENTS OF PLANE TRIGONOMETRY. sin. BCD : °.. DB = cD x ~———., which may .*. be computed. Hence sin. CBD in the right-angled triangle BAD, rad. : sin. BDA !: BD: BA, BD xX sin. BDA and... BAS rad. Ex. 2. To determine the height ofa cloud a A or any object in the air; let two observers c and D in the same vertical plane take the angles of elevation ACB, ADB (=~ a’ and 6°) and measure cp (=c yards) the distance between them; the height and the distances from them may be found ; for the angle apB being = 6°, acs =a*, ADc = 180° —b, and cap = 8°—a@=a. Hence in the triangle CAD, sin. CAD; sin. C 1: CD: DA, c. Sin. @ sin. d ” which may .*. be determined. Cc D ‘Be which is .*. known. or sin. d: sin.ati:c: DA= And sin. CAD: sin. CDA 1: CD: CA, c. sin. 6 “sin, d known. Hence in the right-angled triangle apB, rad. : sin. ADB :: AD: AB, _ ADXsiN. ADB _ ¢. sin. ax sin. b i.é sin, d: sin. (180—0) ::¢:ca= » which is also ane A B R id Seen Asin.. doo. which may be determined. Ex. 3.. To determine the breadth of a river A and the distance of an object B by its side from another object A on the opposite bank, let a line Be be measured (= a yards) along £—-)— its bank, and by means of a Theodolite let the angles cBA, BCA be measured (=0° and c’); then the angle BAC = 180°—(B+c) = d’. And sin. BAC : sin. C :: BC: AB; @ sin. C sin. d ” Whence in the right-angled triangle aps, rad, ;-sin. B?! AB: AD, which may be determined. or sin.d@: sin.€::@:AB= Se ELEMENTS OF PLANE TRIGONOMETRY. 269 AB.Sin. Ba. sin.cx sin. 6 .. 42S SE R R. sin. d which may be computed. Kx. 4. From B the top of a tower, the angle of depression of a vessel at anchor (1 Bs = a@°) was observed, and at c the bottom of the tower, the angle of depression (ecs = 6°) was again ob- served. The height of the tower (Bc =c) being known; to determine the horizontal distance as, and the height ca of the bottom of the tower above the level of the sea. The angle Bsc = BSA—CSA=HBS— 4 ECS=a—b=d and snc =90—HBS=90 —a, .. in the triangle sBc, sin. BSC ; sin. SBC!:BC : CS bees or sin. d : cos. @ 3: ¢ : C8, an ; C. COS. & ae and? And in the triangle cs, rad. : sin. cSA 1: CS: CA, CS.Sin.CSA_ ¢.cos. ax sin. b R Peat imbesined and rad. : cos. CSA 3: CS: SA, which may be found. - CA => 3 CSXCOS.CSA * €.CcOS. ax COS. b “SAS R RX sin. d 2. Having given the angle at a and the B sides Ac, cB; to find the rest. The angle at a being given, its sine may be 4 found from the tables; and (78) cB: cA:: A Sin. A ; SIN. B; n oR ar ; hence sin. B = Gp X SD. A, and may .*. be determined. But as the sine of an angle is equal to the sine of its supple- ment, the angle B may be greater or less than a right angle, unless BC be greater than ac and consequently the angle a greater than the angle B*. * This ambiguity occurs in oblique-angled triangles whenever the side opposite to the given angle is less than the side adjacent to it. For if cB be less than ac, from ¢ draw cp perpendicular to aB, produced if 270 ELEMENTS OF PLANE TRIGONOMETRY. The angle Bp being found, the angle c=180—(A+B8), and may .*. be determined. Also sin. A : sin. © :: CB: BA, S whence BA = CBX — sin, A may be computed. In the same manner may be solved any case, wherein are given two sides and an angle opposite to one of them. (80). If all the angles of any triangle be given, the ratio of the sides to each other may be found. For the angles a, B and c being given, their sines may be found from the tables, and (78) sin. C ; sin. A:: AB: BO, SIN.28 SIC 2 Ad SUT LAU WATTLE oo ois Cie cas in which proportions the two first terms being given, the ratio of the third to the fourth in each is given. (81.) In any rectilineal triangle, it is sufficient if three out of the six parts which belong to it are known, provided that one of these parts be a side; for if only three angles be given, any triangle equiangular to the given one will satisfy the conditions. Lemna ITI. (82.) Ifthe semi-sum of two quantities be added to their semi- difference, the aggregate will be the greater of the two; if the semi-difference be subtracted from the semi-sum, the remainder will be the less of the two. Let ap and Bc’ representing’ the two°.. quantities be placed in the same straight 4 EH D B C necessary. Make pp’=pp, and join cs. (Eucl.i. 4.) cp’ =cB and the angle cp’p is equal to cBp, which is the supple- ment of aBc. Hence there are two triangles cpa, oB’A, which have one angle and the two sides answering the con- ditions. ELEMENTS OF PLANE TRIGONOMETRY. 271 line, which bisect in p, and cut off Aar=pec and “. DE=pr. Then aB—xsc = AB—AE = 2pDB, whence pp =the semi- difference of the lines; and since ap =the semi-sum, ap+ DB=AB the greater, and AD — DB=pc — pB=BC the less. ___(83.) Cor. If the semi-sum be subtracted from the greater _ quantity, the remainder will be the semi-difference. For AB — AD = DB, the semi-difference. Prop. XI. (84.) In any triangle, if a perpendicular from the vertex be drawn upon the base; the base will be to the sum of the sides as the difference of the sides to the difference or sum of the segments of the base made by the perpendicular, according as the perpen- dicular falls within or without the triangle. Let asc be the proposed triangle, _c the vertex, and ax the base; draw the perpendicular cp cutting the base or base produced in p. With the centre c, and radius cB the less of the two sides, describe a circle cutting the side ac in F, and the base or base pro- duced in G; produce ac ton. Then AH is equal to the sum of the sides AC, cB; and AF to their difference. And because BD = pD@, (Kucl. ITT. 3.) AG is the difference of the segments AD, DB in one case, and their sum in the other. And (Kucl. iii. 36.) the rectangle ap AG is equal to the rectangle aun, AF, sh AB (ABE ii AW igs AGS (85.) Cor. Hence if the three sides of any triangle are given, the three angles may be found. For the three sides being given, the three first terms in the preceding proportion are given, and consequently the fourth, 7. e. the difference of the segments of the base made by the perpendicular when it falls within the triangle, may be found; and the base or sum of the segments 8 272 ELEMENTS OF PLANE TRIGONOMETRY. is also given, .*. the segments themselves aD, DB may be found (82). Hence in each of the right-angled triangles acp, BCD, the two sides being known, the angles at A and B may be deter- mined (76), and .*, the remaining angle at c (13). tall But if the perpendicular fall without the triangle, the fourth term in the proportion or sum of the segments, will be found; and the base aB, which is their difference, is given, .*. the segments may be determined, and the angles as before. Prop. XII. (86.) In any plane triangle, the cosine of an angle is to radius as the difference between the sum of the squares of the sides which contain that angle and the square of the third side, is to twice the rectangle contained by the two first sides. Let asc be any triangle; from B let fall B the perpendicular Bp on ac; then (Kucl. 11. 13). BO’=AB’+AC’—2ACX AD, es 1 ra or 2ACX AD=AB’+AC’—BC’. Now cos. A: rad.::DA:AB!!2ACX AD: 2AC X AB "SAB?+AC?—BC’ : 2ACX AB. Nearly in the same manner it may be shown, if the perpendi- cular falls without the triangle. (87.) Cor. 1. Let a, 6, ¢ be the three sides opposite to the ] -. th Nabe CF oe La Be angles A, B, C;. then cos. A = ape? pate OS oy Woe Onis cos. B = —~——_, and cos. 6 = ————_—— 2ac 2ab Hence the sides of a triangle being given, the angles may be found *, * The preceding expressions not being easy for calculation, values may be de- b2 + c2 — a?\2 2be ) _(a+b+c).(b+e-—a).(a+b—c).(a+tec—b) ti 4b? y2 : [and duced fo the sines, sin.? a = 1 — cos.2a = 1 — ( yy ELEMENTS OF PLANE TRIGONOMETRY. 973. ‘ Prop. XIII. (88.) In any triangle, the sum of any two sides is to their dif- __ ference as the tangent of the semi-sum of the angles at the base is _ to the tangent of their semi-difference. Let asc be any triangle, whose base _isac. With the centre s, and radius _ BC the less of the two sides, describe a r circle meeting the side AB in F and AaB produced in rz andacinp. Join sp, _ CE, CF; and draw re parallel to ac. _ Then since CBE, CFE are on the same base crn, (EKuel. iii. 3 20.) CFE = $CBE= }(caB + BCA); and since Bp = BO, the angle BcD = BD¢, and .*, DBA is equal to the difference of the _ angles Bca, BAC. Hence (Eucl. iii. 20.) FCD and .°, its equal | rg = 3(BcA— Bac). If now with the centre r, and radius FC, a circle be described, ie 7 A D Aree - - « CE : CG:; tang. CFE : tang. cra. But re being parallel to ac one of the sides of the triangle AEC, (Kucl. vi. 2.) : CE: CGi:AE:AF:!AB+BC: AB—BCO, “AB + BC: AB—BC:; tang. CFE: tang.cra :: tang. }(BCcA+BAC) ; tang. 3(BCA—BAC). bude _ (89.) Cor. 1. Hence if the sides an, Bo of any triangle, _ and the angle asc included between those sides, be given, the _ other sides and angles may be found. ; For, from the included angle a Bc, its supplement rpc may _ be determined ; half of which is the semi-sum of the angles at _ the base ; and their semi-difference may be found by the Pro- position. Having ., the semi-sum and semi-difference of the angles at the base, the angles themselves Bac, BCA may be €, he ” * af and pie Alle Aes sip x J i(a +b-+c).(a+b—c).(ape—b).(b+e—a)} | ete (raha (re see oh taro 4b i ty 274 ELEMENTS OF PLANE TRIGONOMETRY. found (82). The three angles and two sides ., being obtained, the third side may be found (78). Or the side may be calculated by means of a subsidiary angle, thus (87), B = a’ + c — 2a¢ «cos. B = (a—c)’ + 2ac. (l1— cos. B) = (a—c)’ + 4ac.sin.’ 3B (70) 4ac.sin.* 5B) a ery teense @ = (a—c)’ x 41 + rar Ee Now since tangents admit of every degree of magnitude, find an . 2 fac. sin. kp angle # whose tangent is ———_——___, te - b=(a—ce). 1+tang.’ xv) = (a—c). 2 and .*. b= (a—c) .4/(1+tang.’ x) = (a €) Set & conse or to radius R = Cat aah which (since x is known from the COs. &@ ; 2 c. sin. . equation tang. 7= a ee) may easily be computed. Ex. To determine the distance between two visible inacces- sible objects. Let a and B be the two objects. Mea- B sure a given line cp (= a) along a hori- “ zontal base, from the extremities of which a and B are visible. At c observe the angles ACB, BCD, and at p the angles Apo, ApB. Then in the triangle acp, the angle cap = 180° — (acp + apc) iy by and: sil; DAG: Sit ADC 2teD cA OF SIN.e eile ar OA, . cA = eS, which may be computed. And in the triangle pcp, the angle cs p=180"° — (cDB+ BCD) = d, whence Sin, CBD ; sn? CDB 7°2.CD > UH, ‘> ELEMENTS OF PLANE TRIGONOMETRY. 275 ' orsin. @ :sin. e€ <: @ 0Be . Gesins o Ce ’. CB= ———-, which may be computed. sin. d And in the triangle Acs, AC+CB!AC@CB :: tang. § (BAC+ABC) ; tang. 4 (BAC # ABC) -: tang. 5 (180—acs) : tang. }(BaAc~ asc), AC ~ CB .“. tang. }(BACHABC) geen Bue cot. § ACB, _ which may be calculated. Hence from the tables the difference of the angles may be found, and the sum being known, the angles themselves may be determined (82). And sin. aBco sin. sin. ACB sin. ABC _ Or AB may be ascertained as before by means of the subsi- diary angle 2; and from the three sides being known the two "remaining angles may be determined. BACB?3AC: AB, *. AB= x ac may be calculated. (90.) Cor. 2. Hence also the sum of the sines of two angles is to the difference of the sines, as the tangent: of half the sum _ of those angles is to the tangent of half their difference. : For ac: BC!: sin. B: sin. a, “ AC + BC: ACwBC :: sin. B + sin. A: sin. Be sin. A and .*, sin. A + sin. B: sin, Asin. B °*: tang. (A + B) [: tang. 2(a ~ B). Prop. XIV. ae q (91.) The sum of the tangents of any two angles is to their dif- : ference as the sine of their sum is to the sine of their difference. _ Let pas, Bac be the angles, and through any point B in as, draw pc per- -pendicular to AB. Make the angle BAC = BAC; then will ac=ac. From c and ¢ draw Ck, cF perpendicular to ap. Then Cre : ¢F :: sin. CAD: sin. cAD : Sin. (CAB+BAD) : sin. (BAD# BAC). T 2 276 ELEMENTS OF PLANE TRIGONOMETRY. And BD: BC :: tang. BAD ; tang. BAG, CD: cp:: tang. BAD+ tang. BAC : tang. BAD ew tang. BAC, And cr, cE being perpendicular to 4, are parallel, whence CE: ¢F::CD: CD or sin. (DAB+BAC) : sin. (DABe BAC) ¢: tang. DAB+tang. BAC : tang. DABe~ tang. BAC. Prop. XV. (92.) In any plane triangle, the rectangle contained by any two sides is to the rectangle contained by the excesses of half the peri- meter above each of them respectively as the square of radwus is 7 to the square of the sine of half the angle contained by those sides. Let asc be any triangle, Bc its base, AB the greater of the sides and ac the less; p= 4 the perimeter of the triangle, then AB X AC: (P — AB) .(P — AC) os eins Peels In az take AD = AC, jo pc; and draw AE perpendicular to pc, and EG parallel to Bc, cutting AB in G. With the centre G, and radius GE, describe a circle cutting AB in L, and AB and EG produced ink and Hu. Join 1B; and produce Ak, HB till they meet in m. Since Aap = Ac, and the angles at © are right angles, the squares of An, ED are equal to the squares of An, EC; and . ED = EC, or DC = 2p. Hence by similar fee Doll DBC, BC = 2GEn = EH, and BC is also parallel to EH, . HBM is parallel to CED, and (Kucl. i. 29.) the angle BME = pEA,or it is a right angle; and ne being a diameter, m is a point in the circle. Now (71) AB: BM :: rad. ; (sin. DAE =) sin. 3 BAC, and AD: DE :: rad. ;: sin. } BAC, .. comp. AB X AD: BM X DE :: R’: sin,’ 3 BAC. - + —- = ore ELEMENTS OF PLANE TRIGONOMETRY. 277 But from the similar triangles pen, pBc, pB = 2G, to each of these equals add ap + ac = 2ap, and BA + AC = 2aqa, to each of which equals add Bc = 2GE = 2ek, “. AB + AC + BOC = QAK,.... AK = P, and BK = AK — AB = P — AB, and BL = DK = AK — AD = Pp — ac. And the rectangle BM-x DE = BM x EC = BM x BH = (Eucl. ni. 35.) BK x BL = (Pp — aB). (p — AC), hence AB x AC: (P — AB). (P — Ac) :: R?: sin’ 4 BAC. (98.) Cor. 1. Hence if the three sides of any triangle be given, the angles may be found. For if the sides opposite to the angles a, B, © be a, 0, ¢ respectively, then sin.? }a = r* x See » may be determined, or the sine of half the _ vertical angle, and consequently the vertical angle itself may be found. Hence the other angle (78). _ kx. Three points a, B, c being _ given in a plane, to determine the _ position of a fourth p, where the angles ADB, ADC, subtended by a and B, a _ and ¢ are a’ and 6° respectively. On aB describe a segment of a cir- cle containing an angle a’, and on ac a segment containing an angle 6°; the point of intersection pD is _ the point required. Let an = c, ac = d, Bc = @, sin. } BAC : = 15 AG dae USA deel » which may be computed, p being = /cd -3-(¢ + 4+ e). Draw the diameter ar, and join rs. In the HAgGeR YN ORR sil, BFA sin. @ Draw the diameter az, and jom cx. In the triangle ace, the angle car = apc — apg = 4 — 90° = g, REX AC) © RKO cOS.CAE COos.g" Da _triangle BAF, AF = may be found. and AE = 278 ELEMENTS OF PLANE TRIGONOMETRY. Join px, pF, they are in the same straight line. (ii. 5.) Hence in the triangle FAB, AE and AF having been computed, and the angle FAKE = BAC + CAE — FAB = fh, and tang. 3 (EFAw# AK # AF AE+AF may be determined. And in the right-angled triangle DAF, AD sian ie ioe ee may be found; which line and the angle : BAD determine the position of p. AEF) = x cotang. }rA; whence the angle AFE (94.) Cor. 2. Also AB: AM :: rad. : cos. 9 BAC. and AD: AE :: rad. ; cos. } BAC, - AB X AD:AM X AE 2: R’: COS. 5 BAC. and since AD = Ac,andAM xX AE =AK X AL=P X (P—BC) AB X Ac: P.(P — BC) !: BR’: COs, 4 BAC, ae a: Gc pebed eet (95.) Cor. 3. AlsoAEn: ED :: R : tang. $ BAC, and AM: MB :: R ; tang. BAC, AE X AM: EDXMB ;:: R’: tang.’5 BAC, or P.(P — BC): (P — AB). (P — AC) :: R’: tang.’ g BAG, “ ik = (p — 6) . (pC) .tang. da =r. Vii Bor eoesag a Prop. XVI. (96.) Having given the sines and cosines of two arcs ; to find the sine of an arc which is equal to their sum. ; Let as, Ac be the two arcs; take AD, AE ¥ their doubles; draw the diameter ar. Join AD, DF, AE, EF, ED. Then the chord ap = 2. sin. AB (62) AE = 2. sin. AC, B < DE = 2. sin. BC, and pF = 2 cos. Ap = 2 cos. AB (70). EF = 2 cos. AC, ELEMENTS OF PLANE TRIGONOMETRY. . 279 Now AFX ED = AD+¥FE+AEX FD (Eucl. vi. p.) or 2 rad. x 2 sin. BC = 2 sin. AB xX 2 cos. AC +2 sin. AC x 2 cos. AB, .. rad. x sin. BC = sin. AB X Cos. AC+Sin. AC X COS. AB. (97.) Cor. 1. Ifrad. =1, AB =a, ac=8, sin. (4 + b) = sin. a x cos. 6+ sin. b x cos. a. (98.) Cor. 2. Ifa=4d, sin. 2a=2 sin. a x cos. a. Hence also sin. 3a = 8 sin. a—4 sin.* a sin. 4¢ = (4 sin. a—8 sin.’ a). cos. a sin. 5a = 5 sin. a— 20 sin.’ 4 +16 sin.’ a. PHODAAGY LL. (99.) Given the sines and cosines of two ares; to find the sine of an are which is equal to their difference. Let as, Ac be the two ares; take ap, AE the ae - doubles ofthem; draw the diameter ar; join AE, AD, FE, FD, DE ae Then the chord ap = 2. ae AB (62) Segemes , D AE =2.smM. AC CB DE = 2.sin. BC, and FD = 2.cos. aB (70) EF =2.cos. AC; and AF X ED =AD X EF — AE X FD (Eucl. vi. p.) or 2 rad. x 2 sin. BC = 2 sin. AB x 2 cos. Ac — 2 sin. AC x . [2 cos. AB, .. rad. X sin. BC = Sin. AB X COS. AC — SiN. AC X COS. AB. (100.) Cor. If rad. = 1, and an =a, ac=4, sin. (ad—6) = sin. a@ x cos. b—sin. b x cos. a. Prop. XVIII. (101.) Given the sines and cosines of two arcs; to find the cosine of an arc which is equal to their sum. 280 ELEMENTS OF PLANE TRIGONOMETRY. Let as, Ac be the two arcs. Take AD, AE the doubles of them. Draw the diameter ar, and make FG=AE. Join FG, GD, DA, AG, Fp. Then Gp is the supplement of DE, and .. its chord pG=2 cos. } DE = 2cos. BC (70). DF = 2 COS. AB. and AG=2 cos. } FG =2 cos. 4AE = 2 COS. AC. AD = 2 sin. AB; and FG = 2 sin. FG=2 sin. AC. And AF X D@ = FD X AG— AD X FG (Eucl. vi. D.) or 2 rad. x 2 cos. BC = 2 cos. AB X 2 COs. AC — 2 SIN. AB X 2 ~ [sin. AC. . rad. X COS. BC = COS. AB X COS. AC—SIN. AB X SIN. AC. [LOcMeoO Reet t fader al eS (fk =a i, cos. (a+) = cos. ax cos. 6— sin. ax sin. b. (103.) Cor. 2. If a= 6, cos. 2a = cos.’ a—sin.’ a = 2 cos.’ a—l, or cos. 2a = 1—2 sin.’ a Also cos. 8a = 4 cos. a—8 cos. a cos. 4a = 8 cos.* a—8 cos.? a+1. cos. 5a = 16 cos.’ a— 20 cos. a+5 cos. a cos. 6a = 82 cos.° a—48 cos.* a+18 cos.’ a—1. Prop. XIX. (104.) Gaven the sines and cosines of two arcs; to find the cosine of an are which is equal to their difference. Let as, ac be the two arcs, take AD, AE F the doubles of them; draw the diameter ac, eras and make GF (on the opposite side) = an. [? D Join AD, AF, DF, DG, FG. The arc DGF is pee the supplement of Dr, which .. is equal to SE AD—AE = 2a B—2AC = QBC, and .. DF = 2 COs. § DE = 2 COS. BC. ELEMENTS OF PLANE TRIGONOMETRY. 281 DG = 2 cos. AB (70) AD = 2.sin. AB (62) AF = 20S. AC FQ = 2. sin. AC. Now AGX DF = DGX AF+ADx GF (Kucl. vi. D.) m™ or 2 rad. x 2 cos. BC = 2 cos. AB x 2 cos. Ac+ 2 sin. AB x 2 ! [sin. AC. *, rad. xX cos. BC = COS. ABX COS. AC+S8iIN. ABX SiN. AC. (105.) Cor: Ifrad. =], an=a,ac=8, cos. (a—b) = cos. ax cos. 6+sin. ax sin. 8. (106.) It may perhaps be objected that the preceding propo- sitions are proved only when the arcs (a) and (4) or even (a+) are less than quadrants. Assuming them to be proved within such a limit that (a) does not exceed a value a, and (0) a value _ 3, it may be proved by means of what has been shown before, that the values of the sines and cosines of the sums are equally _ true, when (a) does not exceed 90° + a and (6) does not m® exceed 33. For let a= 90°+m, .. m = a—90°, which is less than 90°. Now sin. (90°+m-+ 6) = sin. §90°—(m+0)} = cos. (m-+ 0) 7 = cos. mx cos. b—sin. mx sin. b = sin. (90°—m) x cos. b—cos. (90°—m) x sin. b = sin. (180°—a) x cos. b—cos. (180°—a) x sin. 6 = sin. ax cos. b+ cos. ax sin. 6. Also cos. (90+m-+ 6) = — cos. {90— (m+6)} = — sin. (m+) 7 = — sin. mx cos. b—cos. m x sin. 6 = — cos. (90+m) x cos. d—sin. (9O—m) x sin. bd. = — cos. (180—a) x cos. b—sin. (180—a) x sin. b = COS. @ x cos. b— sin. a x sin. 0. Hence .*. these expressions which were demonstrated for (a) less than a and (4) less than 3, are also true when (a) does not exceed 90+a and (4) does not exceed 3. In the very same manner from the preceding, it might be proved that they are true, when (a) does not exceed 90+a and (6) 90+ 3, and so on, @ é. they are true whatever be the values of (a) and (é). The same is equally applicable to the values of sin. (c—4) and cos. (a—d). 282 ELEMENTS OF PLANE TRIGONOMETRY. (107.) Since sin. (80°+a@)=sin. 30° x cos. a+ cos. 80° x sin. a, and sin. (30°—a) =sin. 80° x cos. a—cos. 80° x sin. a, “. Sin. (30° + a) + sin. (80°— a) =2 sin. 80° x cos. a = cos. a (63). and sin. (30°+4@) = cos. a—sin. (30°—a). If .. the sines and cosines of all arcs less than 30° be known, the sines of all arcs above 30° and less than 60° might be found by subtraction. (108.) Since sin. (60°+ a) = sin. 60° x cos. a+ cos. 60° x sin. a, and sin. (60°—a) = sin. 60° x cos. a—cos. 60° x sin. a, . Sin. (60°+ a) — sin. (60°— a) = 2 cos. 60°x sin. a = sin. a, and sin. (60°+ a) = sin. a+ sin. (60°—a). If .*. the sines of arcs less than 60° be known, the sines of arcs greater than 60° may be found, by addition. (109.) Let a = ma and d= a. Then sin, (n+1).a = sin. na x cos. A+cos. NA X SiN. A, and sin. (n—1).A =sin. mA x cos. A—cos. nA X Sin. A, “. sin. (2+1).A+sin. (n—1). A=2 sin. na x cos. A¥, and sin. (n+1).a=2 sin. na x cos. a—sin. (n—1). A. Also sin. (n+1).A=sin. (n—1).a+2 sin. Ax cos. nA. and cos. (v+1).A4=2 cos. vA x cos. A—cos, (n—1). a or = cos. (n—1).A—2 sin. nA x cos, A. (110.) Ifin the equations (97) and (100) a+b=a and a—b=B, then a=} (a+B) and =} (a—s) and sin. (a@+6)=sin. ax cos. 6+cos. ax sin. } sin. (¢—6) =sin. a x cos. b—cos. ax sin. 3, “. sin. (@+6)+sin. (a—b) = 2 sin. ax cos. b and sin. (42+ 6)—sin. (a—b) =2 cos. ax sin. 4, , sm. A+sin. B=2 sin. } (A+B) x cos. } (A—B) and sin. A—sin. B=2.cos. } (A+B) x sin. 4 (A—B). Sa RR a aa SS se * Hence if a series of ares be taken in arithmetic progression, radius is to twice — the cosine of the common difference as the sine of any one are taken as a mean is to the sum of the sines of any two equidistant extremes. Also radius is to twice the cosine of the common difference as the cosine of any one are taken as a mean is to the sum of the cosines of any two equidistant extremes, ee ELEMENTS OF PLANE TRIGONOMETRY. 283 ; : cos. $(A—B : Hence also sin. A+sin. B= < ) x sin, (A+B), cos. 3 (A+B) ; I ° ° sin. ro} A—B ° and sin. A—sin. B ="——— ( ) x sin. (A+B). Hence also sin.? A—sin.’? B=sin. (A+B And in a similar manner, cos.’ B—CoSs.” A=sin. (A+B) X sin. (A—B). (111) To find the sin. (A+B+C). Sin. (A+B+C) = sin. (A+B) x cos. c + cos. (A+B) X sin. c = sin. A X COS. B X COS. C+ SIN. B X COS. A X COS. C + sin. C X COS. A X COS. B— Sin. A X SiN. BX SiN. C. x sin. (A—B.) Prop. XX. (112.) Having given the tangents of two arcs; to determine the tangents of two arcs which may be equal to their sum and dif- ference. Let aB, BC be the given arcs, aB being the greater, then ac is the sum of the arcs (in Fig. 1.) and their difference (in Fig. 2.); atv the tangent of their sum (1) or difference (2), Bp, Bo the tangents of the respective arcs AB, BC. Draw op perpendicular to sa cutting SB or sB produced in @, .*. the triangle oB@ is similar to s#p and .. to spp; hence oB : B@ :! SB: BP, .. OBXBP=sSBXBw#. Also by the similar triangles TAS, ODS, #0 ese 0 ie ak haat bE OD: OP: SD; 8# by the sim. triangles swp, opp, Ge Oia Ans) BG :: SA 1 SB— Bz (Fig. 1.) 1. SA° > SB’—SBXB2 22 SA? > SA*—OBXBP 284 ELEMENTS OF PLANE TRIGONOMETRY. Again (in Fig. 2.) aT : OP ES Ava IISA 2 SB+BH °° SA? :SB’+S8B X B2 ** g A’: SA*+OBX BP. OP SA°=OBX BP Hence aT = SA’? x OBC BP SA?=-0BX BP (118.) Cor. 1. Ifrad.=1, an =a,sc=83, tang.a+tang.d , 1 = tang. a x tang. b © (114.) Cor. 2. Ifa= 4, tang. 2a = _&. tang. a =sA’ x taney. (ab) 1— tang’. a __ 3.tang. a—tang’.a Ua 1—3.tane’? a ” 3 tang. 4a = 4..tang. a—4 tang.’ a 1—6 tang.’ a+ tang.‘ a’ 5 tang. a—10 tang.’ a+ tang.’ a Pehl Ad ada manera Warren s ae 3 5 Parte 6 tang. a—20 tang.’ a+6 tang.’ a P 1—15 tang.’ a+ 15 tang.* a—tang.’ a’ (115.) If (in Art. 113.) a = 45°, 1+tang. d 0 = tang. (45°46) = Testanbass 1 — tang. b and tang. (45 ae b) — I ir tang. 5’ sin. (a+b) __ sin. aXcos. b+ cos. aX sin. b * b b= SS Tene (Gr?) cos.(a+b) cos. aXcos. b—sin. aXsin. b sin. a , sin. b cos. a cos. b tang. a+ tang. b ie sin. @ X sin. 6 1— tang. aX tang. b ~ cos. aX cos. b ELEMENTS OF PLANE TRIGONOMETRY. 285 1 + tang.d 1 — tang. b 1 — tang.d 1 + tang. 6 = 2 tang. 20, .”. tang. (45°+ 5) — tang. (45°— d)= a 2 tang. b "JT — tang. 6 and tang. (45° + 6) = tang. (45° = 6) + 2tang. 2d. If ... the tangents of arcs less than 45° be known, the tangents of arcs greater than 45° may be determined by addition. (116.) Tang. a + cotang. a = 2cosec. 2a. sin. @ cos.a cos.a@ sin. a For tang. a+cotang. a= Mills et COS< 4 Dy : -= ———=2 cosec. 2 a. sin.a@ X COS. a sin. 2a In the same way it may be shown that cotang. a — tang. a = 2cotang. 2a. and sec. a + tang. a = tang. (45° + $a) sin. 2a sin. 2a that tang. a = Tree 7 cotang.a = ie 1 — tang.’ a sec.” a COB. = Testes and sec. 24 = TREAD ae Prop. XXI. (117.) The base and altitude of a triangle being given; to find ats area. (Eucl. i. 41.) The area of the triangle aBc = 3Ac x BD. (See next Fig.). (118.) Cor. 1. If a = the altitude pp, and 4 = the base ab 28 28 Ac, the area s = > ane Coe TF and i= > any two of the terms .*. being known, the third may be found. Prop. XXII. (119.) Two sides of a triangle, and the included angle being given; to find its area. 286 ELEMENTS OF PLANE TRIGONOMETRY. Let AB, Ac and the angle Bac be given. seen Let fall the perpendicular Bp, , AB BD head. a Ria ~~ eAB. xX Silo AD . gS R (if R= 1), e IRAPER. =e A COX BD = Aree UA Groans (120.) Cor. 1. Hence the areas of triangles which have one angle in each equal, are as the products of the sides containing those angles. Which is also true of parallelograms. 258 258 (121.) Cor. 2. ac = —————, aB = ———____, AB X SILA ACXsIn. A : 28 BNC Si. A ee AB X AC Prop. XXIII. (122.) Given two angles and a side of a triangle; to find its area. .- Two angles being given, the third is also known. Let Ac be the given side. ‘ , AC X sin. C Then sin..B 2 sin’ Cf. ACG ha SIn. B : AC X sin. A X SIN. C and BD = AB Xx sin. A = ; sin. B 2 whence the area = }ac? x 7 S © sin. B (123.) Cor. Ifthe angles anc, acs be equal, the area will be =%Ac’ x sin. a. Ifthe angles a and c are equal, the area sin.’ A | aa i ae will be = 4Ac’ x Aw But since the angle Bn = 180° — 2a, sin. B = sin. 2A = 2sin. A X COS. A, sin.? A whence the area = 1 ac? x ————_—_——_ sin. A X COS. A ; sin. A RAPE tae — 5 AC Te aR eR AG” n ae Ae - * “COs. A - eas 8 ELEMENTS OF PLANE TRIGONOMETRY. 287 Prop. XXIV. (124.) Given the three angles, and the altitude of a triangle ; to find its area. Since (122) BD x sin. B = AC x sin. A x sin. c, and ac = 28 ; ; : .. 28 xX sin. A X Sin. C = BD’ ~x SIN. B. BD’ ; . : sin. B ors =ispd? x —-.—. 2 sin. A xX sin. C Prop. XXV. (125.) Given the three sides of a triangle ; to Hing its area. The area of the triangle anc = area ADC + area BDC = AE X DE + ME X DE, (BM being parallel to pc) = am xX DE. But by similar triangles ADE, AMB, AE .ED :. AM : MB, “ AE X AM: ED X AM ?: ED X AM: ED X MB, 2. e. the area of the triangle is a mean proportional between AE X AM and ED xX MB. Now (92.) ED x MB = (Pp — AB). (P — Ac), and AE X AM = AL X AK = P x (P— B6), *, the (area)’ = (Pp — AB). (P — AC).P.(P — BC) ors = 4«/{p.(p — AB). (P — ac). (Pp — Bo). (126.) IfaB = pc = ac,s = das’. /3, the area of an equilateral triangle. If pc = ca,s = 4aB.4/{(2Bc + BA). (2BC — Ba), the area of an isosceles triangle. (127.) To construct the trigonometrical canon. It has been proved (103) that 2cos.?.a =-1 + cos. 24; and therefore if the cosine of any arc be known, the cosine of half that arc may be determined. Now the sine of 80° has been 288 ELEMENTS OF PLANE TRIGONOMETRY. found to be = 4, (R = 1), and the cosine =v ; if then in the formula cos. A = 4/{}. (1 + cos. 2A)t, A = 15°, cos. 15° may be determined. In the same manner from cos. 15°, the cos. 7 30’ may be deduced; and so on, till after 11 divisions, cos. 52” 44” 3 45" is found; from which the sine of this are may be determined. But from the nature of the circle, when the arc is very small, the ratio of the arc to the sine approaches nearly to a ratio of equality, .. 52” 44°” 3” 45%: 1’ :: the sine of the former arc ; the sine of 1’; which .*, may be determined. The sine and cosine of 1’ being ascertained, the sines of 2’, 3’, 4’, &c. may be determined (109) by making n = 1, 2, 3, &c. and the cosines from (58). In this manner the sines and cosines of arcs as far as 30” may be computed. When the arc exceeds 30°, the sines may be computed by Art. (107), and the cosines as before, till the arc is 45°. And since the sine of an arc is equal to the cosine of its complement, the sines and cosines of arcs as far as 90° are determined. Also since the sines and cosines of arcs are equal to the sines and cosines of their supplements ; the sines and cosines of all arcs up to 180° are known. in. A OS. A’ puted. When however they exceed 45°, they are more readily computed from (115) by addition. And the tangent of an arc being equal to the tangent of its supplement, the tangent of all arcs may be determined. Since tang. A = ~ the tangents of all arcs may be com- Hence also the cotangents (45). 1 ’ : Also the sec. A = >; and ... the cosines being known, the secants may be determined. And the secants being known, the cosecants are also determined. , Also the versed sine (= 1 = cosine) may be determined. THE END. GILBERT & RIVINGTON, Printers, St. John’s Square, London. 1 A lp 4 Hb \ A ed AD ‘ v Te a } PUTA coeay Y ciate 3 0112 017248540 “MLL