so that the rectangle, contained under the whole line produced, and the part of it produced, may be equal to the square of L. Bisect (E. 10. 1.) AB in K, and upon AB, as a diameter, describe the circle ABC; from B draw (E. 11. 1.) BD 1 to AB, and make BD=L; join K, D, and let KD cut the circumference in C; from C draw CE to KC, and let CE meet AB produced in E. Then since (constr. and E. 16. 3. cor.) CE touches the circle, and EBA cuts it, .. (E. 36. 3.) AEX EBEC'; but, since the KBD, KCE are right, and the at K is common to the two KBD, KCE, and that (E. 15. def. 1.) the side KD the side KC, .. (E. 26.1.) EC= BD; but (constr.) BD=L; and it has been shewn that AEX EB = EC'; ... AEX EB= the square of L. 94. COR. By the help of this proposition and (E. 14. 2.), a given straight line may be produced, so that the rectangle contained by the whole line thus produced, and the part of it produced, shall be equal to a given rectilineal figure. PROP. LXXIV. 95. THEOREM. If, from the bisection of any given arch of a circle, a straight line be drawn cutting the chord of that arch, or the chord produced, and the circumference also of the circle, the rectangle contained by the two parts of the straight line so drawn, the one lying between the point of bisection and the circumference, the other between the point of bisection and the chord, shall be equal to the square of the chord, of half the arch. Let AB be the chord and let C be the bisec D E B tion, of the arch ACB of the circle ADBC; and, first, let any straight line CD be drawn cutting the chord in E, and then meeting the circumference of the circle in D; also let there be drawn CB, the chord of CB, the half of ACB: Then DCX CE CB2. For join C, A and B, D, and about the ADBE describe (S. 5. 1. cor.) the circle DEB: And because (hyp.) AC=CB, ... (E. 27. 3.) the ABC= 2 CAB; and (E. 21. 3.) the CAB=4CDB; ... the ABC ▲ BDE; ... (S. 59. 3.) the straight line CB touches the circle DEB in B; .. (E. 36. 3.) DCXCE=CB'. PROP. LXXV. 96. PROBLEM. From the bisection of a given arch of a circle, to draw a straight line, such that the part of it intercepted between the chord, or the chord produced, of the given arch and the circumference, shall be equal to a given straight line. Let ACB be a given arch of the circle ADBC; let AB be its chord, and C its bisection, and let LM be a given straight line: It is required to draw from C a straight line such that the part of it between AB, and the circumference ADB shall be equal to LM. Join C, B; and produce (S. 73. 3.) LM to N, so that LNX NMCB'; from C as a centre, at a distance equal to LN, describe a circle cutting ÁDB in D; join C, D, and let CD cut AB in E: Then is ED=LM. For (S. 74. 3.) DC × CE=CB2; and (constr.) LN × NM=CB2; ., DC × CE = LN × NM; but (constr.) DC = LN; .. CE = NM; and .. ED=LM. PROP. LXXVI. 97. PROBLEM. Through any given angle of a given equilateral four-sided figure, to draw a straight line terminated by the sides produced, containing the angle opposite to the given angle, which shall be equal to a given straight line. EF a given straight line: Through any of the angular points of ABCD, as C, it is required to draw a straight line, terminated by AB and AD produced, which shall be equal to EF. Join A, C; upon EF describe (E. 33. 3.) a segment of a circle EKF, capable of containing an equal to the BAD of the rhombus, and complete the circle; bisect (E. 30. 3.) EGF in G; from G draw (S. 75. 3.) GHK so that HK=AC; join E, K; in AB produced, take (E. 3. 1.) AL = KE; join L, C, and produce LC to meet AD produced in M: Then LCM-EF. For join K, F, and G, E, and G, F: And because (E. 32. def. 1.) BA, AC are equal to DA, AC, each to each, and that the base BC, of the ▲ ABC, is equal to the base DC, of the ▲ ADC, .. (E. 8. 1.) the BAC=1 DAC, and the < BAC is,..., the half of the BAD: Again, because (constr.) EG=FG, .:. (E. 27. 3.) the EKG = < FKG, and the EKG is, ..., the half of the EKF, which (constr.) is equal to the BAD; .. the < EKH = < LAC; and (constr.) the two sides EK, KH of the ▲ EKH, are equal to the two sides LA, AC, of the A LAC; .. (E. 4. 1.) LC=EH, and the (E. 13. 1.) the 4 KHF ACL= ‹ KHE; .. been shewn, the 4 CAM = 4 HKF, and the side KH (constr.) of the A KHF is equal to the side AC of the ▲ ACM;... (E. 26. 1.) CM HF; and it has been proved that LC=EH; .. LC + CMEH+HF; that is, LM = EF. |