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placed in the same straight line, be three given straight lines: It is required to find a point within the AABC, from which if straight lines be drawn to A, B and C, the A 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 CEL to BC; in like manner, 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.)

▲ BMC: A AMB::PQ: QR: Whence it follows, also, that the A AMB: ▲ 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

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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.) ÉGF 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 circumfe rence 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.

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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 .. (constr. and E. 34. 1.) the right Again the AFH,

AC at right; at F and Gare AKL (E. 29. 1.)

are equiangular, as are, also, the KFG, KAC; .. (E. 4. 6.) AK: KL:: AF: FH: And (constr.) KL: AC:: KF: AF;

.. (E. 23. 5.)

But (E. 4. 6.)

AK: AC::KF:FH:
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 ; is an equilateral; and its been shewn to be right ;.. (E. 34. 1.) all its are right; .. (E. 30. def. 1.) FHIG is a

square.

...the figure FHIG GFH, FGI, have

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 equal, respectively, to the T, B, and C, and let k, l, and m, be the se

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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.) chq 1 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. 10. 6.) BC in H, so that BH: HC:: bh: hc; make (E. 23. 1.), at the point H, in BH, the

4BHF bhf, and the

BHG = bhg, the

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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 bc2qs; and, in the same manner, it may be shewn that tb = 2rq; .. (E. 15. 5.) tb: bc:: rq: qs:

But (constr. and E. 10. 6.)

rq: qs :: kp: pm: : TB: BC; ..(E. 11. 5.) tb: bc:: TB: BC.

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