placed in the same straight line, be three given straight lines: It is required to find a point within the A ABC, from which if straight lines be drawn to A, B and C, the 4 shall thereby be divided into three parts that are to one another as PQ, QR, and RS. Through A draw (E. 31. 1.) DAE parallel to BC, and from B and C draw (E. 11. 1.) BD and CE 1 to BC; in like manner, describe upon AB another rectangle ABGF, about the A ABC; divide (E. 10. 6.) DB in H, so that PS : PQ:: DB:BH; divide, also, BG in K, so that PS: QR::BG: BK; through H draw HI parallel to BC, and through K draw KL parallel to BA, and let HI and KL cut one another in M: Then is M the point which was to be found. For draw MA, MB, and MC: And since (E. 41. 1.) each of the rectangles DBCE, ABGF, is double of the A ABC, they are equal to one another; also (E. 41. 1.) AK is double of the A AMB, and HC is double of the A BMC: But (E. 1. 6.) HBCI : DBCE:: HB: DB:: PQ:PS; and ABGF: ABKL:: KB:GB:: PS : QR; .. (E. 22. 5.) HBCI : ABKL:: PQ:QR; ..(E. 41. 1. and E. 15. 5.) A BMC:A AMB:: PQ: QR: Whence it follows, also, that the A AMB: A AMC::QR: RS. Prop. XXX 38. PROBLEM. To divide a given circular arch into two parts, so that the chords of those parts shall be to each other in a given ratio. Let EKF be the given circular arch : It is re G quired to divide it into two parts, the chords of which shall be to one another in a given ratio. Join E, F; and describe (E. 25. 3.) the circle KEGF, of which EKF is a given segment; bisect (E. 30. 3.) EGF in G; divide (E. 10. 6.) EF in H, so that EH shall be to HF in the given ratio ; draw GH, and produce it to meet the circumference in K; lastly join E, K and F, K. Then, since (constr. and E. 27. 3.) the < EKF is bisected by KHG, .. (E. 3. 6.) KE: KF :: EH : HF; that is, (constr.) KE: KH in the given ratio, PROP. XXXI. 39. Problem. To inscribe a square in a given trapezium, which has the two sides about any angle equal to one another, and the two sides about the opposite angle also equal to one another. KA=KC, and also the side LA = LC: It is required to inscribe in AKCL a square. Draw the diameters of the figure, AC and KL; divide (E. 10. 6.) AK in F, so that AF: FK:: AC: KL; draw (E. 31. 1.) FG parallel to AC, and GI and FH parallel to KL, and join H, I: Then is the inscribed figure FHIG a square. For (S. 1. 3. cor.) KL bisects AC at right L; .: (constr. and E. 34. 1.) the L at F and G are right 1 : Again the A AFH, AKL (E. 29. 1.) are equiangular, as are, also, the AKFG, KAC; i. (E. 4. 6.) AK: KL:: AF:FH: And (constr.) KL: AC:: KF: AF; (E. 23. 5.) AK:AC::KF:FH: But (E. 4. 6.) AK: AC:: KF: FG; .: (E. 9. 5.) FG = FH: And since (E. 2. 6.) CG: GK:: AF:FK, it may, in like manner, be shewn that GI=GF; and (constr.) GI is parallel to FH; .. (E. 33. 1.) IH is equal and parallel to GF; .. the figure FHIG is an equilateral ; and its | GFH, FGI, have been shewn to be right L; ;. (E. 34. 1.) all its are right 1;:. (E. 30. 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. 34. def. 1.) ABCD is not a , one pair, at least, of its opposite sides must meet if they be far enough produced; let, .., DA and CB be produced so as to meet in T: Take any straight line fg and upon it describe (E. 46. 1.) the square fghi ; join f, h; and upon hf, hg, and hi describe (E. 33. 3.) segments of circles, ich, fth, and gbh, capable of containing 4 equal, respectively, to the 2 T, B, and C, and let k, l, and m, be the se veral centres of the circles ; draw km, and divide it (E. 10. 6.) in p, so that mp: pk :: CB: BT; also join p, l; through h draw (E. 12. 1.) cho 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 circumfe. rence ich in c: Again,' divide (E. 10. 6.) BC in H, so that BH: HC:: bh: hc ; make (E. 23. 1.), at the point H, in BH, the 2 BHG = L bhg, the 2-BHF = L bhf, and the < CHI = 2 chi; lastly, join F, G and F, I: Then is the inscribed figure FGHI á square. For draw (E. 12. 1.) kr and pq, 1 to tc : Then, since (constr. and E. 3. 3.) bh = 2qh, and hc = 2hs, it is manifest that bc = 298; and, in the same manner, it may be shewn that tb = 2rq; :. (E. 15.5.) tb:bc:: r9:qs : But (constr. and E. 10. 6.) rq:qs :: kp:pm:: TB:BC; .:(E. 11.5.) tb:bc:: TB: BC. |