38. PROBLEM.

PROP. XXX.

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

K

E

H

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.) 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 circumference in K; lastly join E, K and F, K.

Then, since (constr. and E. 27. s.) the EKF is bisected by KHG, .. (É. 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 .. (constr. and E. 34. 1.) the right Again the AFH,

AC at right;

at F and G are 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.) AK: AC:: KF: FH: But (E. 4. 6.)

AK:AC:: KF; FG;
FG = FH:

.. (E. 9. 5.) 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 ;.. (E. 34. 1.) all its are right; .. (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 equal, respectively, to the T, B, and C, and let k, l, and m, be the se

Z

veral centres of the circles; draw km, and divide it (E. 10. 6) in p, so that mp: pk::CB: BT; also join p, 1; 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

BHG = bhg, the chi; lastly, join F, G and F, I: Then is the inscribed figure

< BHF = bhf, and the

FGHI a square.

CHI=

For draw (E. 12. 1.) kr and pq, I to tc: Then, since (constr. and E. 3. 3.) bh=2qh, and he 2hs, it is manifest that be 2qs; 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.

Again (constr. and S. 26. 1.) the are equiangular, as are, also, the

gbh, GBH ich, ICH;

.. (E. 4. 6.) hg: hb:: HG: HB: And (constr.) hb: hc:: HB: HC: Also (E. 4. 6.) hc: hi:: HC: HI ; .. (E. 22, 5.) hg: hi:: HG: HI:

But (constr.) hg=hi; .·. HG=HI; and it is manifest, also, from the construction, that the ▲ GHIghi, of the square fghi; .. the <GHI is a right 2.

Again, since (constr.) bh: hc :: BH: HC,

.. (comp. and div.) th: bh :: TH:BH: Lastly, (constr. and S. 26. 1.) the two tfh, TFH, are equiangular, as are, also, the two BHG;

.. (E. 4. 6.) fh: th:: FH: TH:

And th: bh::TH: BH;

Also (E. 4. 6.) bh: hg:: BH: HG; ... (E. 22. 5.) fh: hg:: FH: HG: Wherefore, the two

sides about the equal

bhg,

fhg, FHG, having their

fhg, FHG, proportionals,

are (E. 4. 6.) equiangular; .. the

right angle; and (E. 4. 6.) FG

FGH is a

GH, because

(constr.) fg = gh: And, as hath been shewn, the FGH, GHI, are right ; .. (E. 28. 1.) GF is parallel to HI.

It has been shewn, also, that HIHG; .. (E. 33. 1. and E. 34. 1.) the figure FGHI is equilateral and rectangular : That is (E. 30. def. 1.) it is a square.