$5. 1. equals to the angle CBD, because the side AD is equal Book IV. to the side AB; therefore CBD, or DBA, is equal to BCD; and consequently the three angles BDA, DBA, BCD, are equal to one another; and because the angle DBC is equal to the angle BCD, the side BD is equal to the side DC; "6. 1. but BD-was made equal to CA, therefore also CA is equal to CD, and the angle CDA equals to the angle DÁC; therefore the angles CDA, DAC together, are double of the angle DAC: But BCD is equal to the angles CDA, DAC; therefore also BCD is double of DAC, and BCD is equal to each of the angles BDA, DBA; each therefore of the angles BDA, DBA is double of the angle DAB, wherefore an isosceles triangle ABD is described, having each of the angles at the base double of the third angle. Which was to be done. PROP. XI. PROB. To inscribe an equilateral and equiangular pentagon in a given circle. Let ABCDE be the given circle; it is required to inscribe an equilateral and equiangular pentagon in the circle ABCDE. F B a Describe a an isosceles triangle FGH, having each of the 10. 4. angles at G, H, double of the angle at F; and in the circle ABCDE inscribeb the triangle ACD equiangular to the 2. 4. triangle FGH, so that the angle CAD be equal to the angle at F, and each of the angles ACD, CDA equal to the angle at G or H, wherefore each of the angles ACD, CDA is double of the angle CAD. Bisect c the angles ACD, CDA by the straight lines CE, DB;G D E and join AB, BC, DE, EA. ABCDE is the pentagon required. 9. 1. Because each of the angles ACD, CDA is double of CAD, and are bisected by the straight lines CE, DB, the five angles DAC, ACE, ECD, CDB, BDA are equal to one another'; but equal angles stand upon equal circum- * 26. 3. ferences; therefore the five circumferences, AB, BC, CD, e 29.3. BOOK IV. DE, EA are equal to one another: And equal circumferences are subtended by equal straight lines; therefore the five straight lines AB, BC, CD, DE, EA are equal to one another. Wherefore the pentagon ABCDE is equilateral. It is also equiangular; because the circumference AB is equal to the circumference DE: If to each be added BCD, the whole ABCD is equal to the whole EDCB: And the angle AED stands on the circumference ABCD, and the angle BAE on the circumference EDCB; therefore the 27.3. angle BAE is equal to the angle AED: For the same reason, each of the angles ABC, BCD, CDE is equal to the angle BAE, or AED: Therefore the pentagon ABCDE is equiangular; and it has been shown that it is equilateral. Wherefore, in the given circle, an equilateral and equiangular pentagon has been inscribed. Which was to be done. PROP. XII. PROB. To describe an equilateral and equiangular pentagon about a given circle. Let ABCDE be the given circle; it is required to describe an equilateral and equiangular pentagon about the circle ABCDE. b Let the angles of a pentagon, inscribed in the circle, by the last proposition, be in the points A, B, C, D, E, so that a 11. 4. the circumferences AB, BC, CD, DE, EA are equala; and through the points A, B, C, D, E, draw GH, HK, KL, 17.3. LM, MG, touching the circle; take the centre F, and join FB, FK, FC, FL, FD: And because the straight line KL touches the circle ABCDE in the point C, to which FC 18. 3. is drawn from the centre F, FC is perpendicular to KL, therefore each of the angles at C is a right angle: For the same reason, the angles at the points B, D are right angles: And because FCK is a right angle, the square of FK is d 47. 1. equal to the squares of FC, CK: For the same reason, the square of FK is equal to the squares of FB, BK: Therefore the squares of FC, CK are equal to the squares of FB, BK, of which the square of FC is equal to the square of FB; the remaining square of CK is therefore equal to the remaining square of BK, and the straight line CK equal to BK: And because FB is equal to FC, and FK common to e the triangles BFK, CFK, the two BF, FK are equal to the Book IV. two CF, FK; and the base BK is equal to the base KC; therefore the angle BFK is equal to the angle KFC, and 8. 1. the angle BKF to FKC; wherefore the angle BFC is double of the angle KFC, and BKC double of FKC: For the same reason, the angle CFD is double of the angle CFL, and CLD double of CLF: And because the circumference BC is equal to the circumference CD, the angle BFC is equal f (27. S. to the angle CFD; and BFC is double of the angle KFC, and CFD double of CFL; therefore the angle KFC is equal to the angle CFL; and, the right angle FCK is equal to the right angle FCL: Therefore, in the two triangles FKC, FLC, there are two angles of one equal to H B A G E M two angles of the other, each K to each, and the side FC, which is adjacent to the equal angles in each, is common to both; therefore the other sides shall be equals to the 26. 1. other sides, and the third angle to the third angle: Therefore the straight line KC is equal to CL, and the angle FKC to the angle FLC: And because KC is equal to CL, KL is double of KC; In the same manner it may be shown that HK is double of BK: And because BK is equal to KC, as was demonstrated, and that KL is double of KC, and HK double of BK, HK shall be equal to KL: In like manner, it may be shown that GH, GM, ML are each of them equal to HK, or KL: Therefore the pentagon GHKLM is equilateral. It is also equiangular; for, since the angle FKC is equal to the angle FLC, and that the angle HKL is double of the angle FKC, and KLM double of FLC, as was before demonstrated, the angle HKL is equal to KLM: And in like manner it may be shown, that each of the angles KHG, HGM, GML is equal to the angle HKL or KLM: Therefore the five angles GHK, HKL, KLM, LMG, MGH, being equal to one another, the pentagon GHKLM is equiangular: And it is equilateral, as was demonstrated; and it is described about the circle ABCDE. Which was to be done. BOOK IV. a 9. 1. PROP. XIII. PROB. To inscribe a circle in a given equilateral and equiangular pentagon. Let ABCDE be the given equilateral and equiangular pentagon; it is required to inscribe a circle in the pentagon ABCDE. Bisect the angles BCD, CDE by the straight lines CF, DF, and from the point F, in which they meet, draw the straight lines FB, FA, FE: Therefore, since BC is equal to CD, and CF common to the triangles BCF, DCF the two sides BC, CF are equal to the two DC, CF; and the angle BCF is equal to the angle DCF; therefore the base 4.1. BF is equal to the base FD, and the other angles to the other angles, to which the equal sides are opposite; therefore the angle CBF is equal to the angle CDF: And because the angle CDE is double of CDF, and that CDE is equal to CBA, and CDF to CBF; CBA is also double of the angle CBF; therefore the angle ABF is equal to the angle CBF; wherefore the angle ABC is bisected by the straight line BF: In the same t manner it may be demonstrated, that the angles BAE, AED, are bisected by the straight lines AF, FE: From 12. 1. the point F, draw FG, FH, B H C A M E L FK, FL, FM perpendiculars to the straight lines AB, BC, CD, DE, EA: And because the angle HCF is equal to KCF, and the right angle FHC equal to the right angle FKC; in the triangles FHC, FKC there are two angles of one equal to two angles of the other, and the side FC, which is opposite to one of the equal angles in each, is 26. 1. common to both; therefore the other sides shall be equal, each to each; wherefore the perpendicular FH is equal to the perpendicular FK: In the same manner it may be demonstrated; that FL, FM, FG are each of them equal to FH, or FK: Therefore the five straight lines FG, FH, FK, FL, FM are equal to one another: Wherefore the circle described from the centre F, at the distance of one of these five, shall pass through the extremities of the other Book IV. PROP. XIV. PROB. To describe a circle about a given equilateral and Let ABCDE be the given equilateral and equiangular a Bisect the angles BCD, CDE by the straight lines CF, 9. 1. FD, and from the point F, in which they meet, draw the straight lines FB, FA, FE, to the points B, A, E. It may be det monstrated, in the same manner, as in the preceding proposition, that the angles CBA, BAE, AED are bisected by the straight lines FB, FA, FE: And because the angle BCD is equals to the angle CDE, and that E FCD is the half of the angle BCD, and CDF, the half Eof CBE; the angle FCD is equal to FDC; wherefore the side CF is equal to the side FD: In like manner ↳ 6.1. it may be demonstrated that FB, FA, FE, are each of them equal to FC or FD: Therefore the five straight lines FA, FB, FC, FD, FE are equal to one another; and the circle described from the centre F, at the distance of one of them, shall pass through the extremities of the other four, and be described about the equilateral and equiangular pentagon ABCDE. Which was to be done. |