therefore (E. iv. 6.) AK : KL :: AF : FH: therefore (E. xxiii. 5.) But (E. iv. 6.) therefore (E. ix. 5.) AK: AC :: KF : FH: AK: AC :: KF: FG; And since (E. ii. 6.) CG: GK :: AF: FK, it may, in like manner, be shewn that GI=GF; and (constr.) GI is parallel to FH; therefore (E. xxxiii. 1.) IH is equal and parallel to GF; therefore the figure FHIG is an equilateral parallelogram; and its angles GFH, FGI, have been shewn to be right angles; therefore (E. xxxiv. 1.) all its angles are right angles; therefore (E. xxx. Def. 1.) FHIG is a square. PROP. XXXII. 40. PROBLEM. To inscribe a square in a given trapezium. Let ABCD be the given trapezium: It is required to inscribe in it a square. Since (E. xxxiv. Def. 1.) ABCD is not a parallelogram, one pair, at least, of its opposite sides must meet if they be far enough produced; let, therefore, DA and CB be produced so as to meet in 7: Take any straight line ƒg and upon it describe (E. xlvi. 1.) the square fghi; join f, h; and upon hf, hg, and hi describe (E. xxxiii. 3.) segments of circles, ich, fth, and gbh, capable of containing angles equal, respectively, to the s T, B, and C, and let k, l, and m, be the several centres of the circles; draw km, and divide it (E. x. 6.) in p, so that mp: pk :: CB: BT; also join p, l; through h draw (E. xii. 1.) chq perpendicular to pl produced, and meeting it in q; also let cq, produced, meet the circumference fth in t, the circumference gbh in b, and the circumference ich in c: Again, divide (E. x. 6.) BC in H, so that BH: HC :: bh: hc; make (E. xxiii. 1.) at the point H, in BH, the BHG= bhg, the BHF bhf, and the CHI= chi; lastly, join F, G and F, I: Then is the inscribed figure FGHI a square. = For draw (E. xii. 1.) kr and pq, perpendicular to tc: Then, since (constr. and E. iii. 3.) bh=2qh, and hc=2hs, it is manifest that be=2q8; and, in the same manner, it may be shewn that tb=2rq; : therefore (E. xv. 5.) tb be: rq: qs: But (constr. and E. x. 6.) rq: qs: kp: pm :: TB: BC; therefore (E. xi. 5.) tb bc :: TB: BC. Again (constr. and Supp. xxvi. 1.) the As gbh, GBH are equiangular, as are, also, the as ich, ICH; therefore (E. iv. 6.) hg: hb :: HG : HB: Also (E. iv. 6.) hc hi :: HC: HI; therefore (E. xxii. 5.) hg: hi :: HG : HI. But (constr.) hg=hi; .. HG=HI; and it is manifest, also, from the construction, that the ▲ GHI= <ghi, of the square fghi; therefore the 4 GHI is a right angle. : Again, since (constr.) bh hc :: BH: HC, therefore (comp. and div.) th: bh: TH: BH. Lastly, (constr. and Supp. xxvi. 1.) the two as tfh, TFH, are equiangular, as are, also, the two as bhg, BHG; : hg :: FH : HG. therefore (E. xxii. 5.) fh Wherefore, the two as fhg, FHG, having their sides about the equal 4s fhg, FHG, proportionals, are (E. iv. 6.) equiangular; therefore the FGH is a right angle; and (E. iv. 6.) FG= GH, because (constr.) fh=gh: And, as hath been shewn, the 48 FGH, GHI, are right angles; therefore (E. xxviii. 1.) GF is parallel to HI. It has been shewn, also, that HI-HG; therefore (E. xxxiii. 1. and E. xxxiv. 1.) the figure FGHI is equilateral and rectangular: That is (E. xxx. Def. 1.) it is a square. PROP. XXXIII. 41. PROBLEM. To determine the locus of the summits of all the triangles which can be described on a given base, so that each of them shall have its two sides in a given ratio. Let AB be a finite straight line: It is re F B quired to determine the locus of the summits of all the triangles which can be described upon AB, as a base, having their two remaining sides, in each, in a given ratio to one another. Divide (E. x. 6.) AB in C, so that AC shall be to CB in the given ratio; from the greater segment AC, cut off CD=CB; find (E. xi. 6.) a third proportional to AD and CB, and in AB, produced, make BK equal to it; from the centre K, at the distance KC, describe the circle CPE: The circumference CPE is the locus which was to be determined. For, take any point P, in the circumference CPE, and draw PA, PB, PC, and PK: Then since; (constr.) AD: CB :: CB : BK, therefore (E. xviii. 5.) AD+CB or AC: CB:: CK: BK; therefore (E. xvi. 5.) AC: CK :: CB : BK ; therefore (E. xviii. 5.) AK : CK :: CK : BK; that is, (E. xv. Def. 1.) AK: KP :: KP : KB ; therefore (E. vi. 6.) the two 48 APK, BPK, are equiangular; therefore (E. iv. 6.) PA : PB :: AK : PK or CK: And it has been shewn that AK: CK :: CK : BK :: AC ; CB ; : therefore (E. xi. 5.) PA : PB :: AC : CB. And (constr.) AC is to CB in the given ratio; therefore PA is to PB in the given ratio, wherever, in the circumference CPE, the point P is taken”. PROP. XXXIV. 42. PROBLEM. The base, the perpendicular distance of the vertex from the base, and the ratio of the two sides of a triangle being given, to con struct it. Draw (E. xxxi. 1. and E. xi. 1.) a straight line parallel to the given base, and at a perpendicular distance from it equal to the given perpendicular distance; draw, (Supp. xxxiii. 6.) the locus of the summits of all the triangles, which can be described on the given base, having their sides to one another in the given ratio; and it is manifest that the point, in which this locus meets the line drawn parallel to the base, will be the summit of the triangle which was to be described. If the given ratio be a ratio of equality, the locus to be determined is, manifestly, the straight line drawn at right angles to AB, through the point which divides AB into two equal parts. |