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 it. Book IV. Describe an isosceles triangle FGH, having each of the a 10.4. angles at G, H double of the angle at F; and in the circle ABCDE inscribeb the triangle ACD equiangular to the b 2.4. triangle FGH, so that the angle CAD may 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 the angles ACD, CDA by the straight lines CE, DB; and join AB, BC, DE, EA. ABCDE is the pentagon required. Because the angles ACD, CDA are each of them 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, therefore the five arches AB, BC, CD, DE, EA are equal to one anotherd; therefore the five straight lines AB, d 26. 3. BC, CD, DE, EA are equal to one another. Wherefore e 29.3. 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. Now the angle AED stands on the circumference ABCD, and the angle BAE on the circumference EDCB; therefore the angle BAE is equal to f 27. 3. the angle AED. For the same reason each of the angles Book IV. 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. a 11. 2. b 1. 4. c 2 Cor. 10. 2. Otherwise. be "Divide the radius of the given circle so that the rectangle contained by the whole and one of the parts may equal to the square of the othera. Apply in the circle, on each side of a given point, a lineb equal to the greater of these parts; then each of the arches cut off will be one-tenth of the circumference, and therefore the arch made up of both will be one-fifth of the circumference; and if the straight line subtending this arch be drawn, it will be the side of an equilateral pentagon inscribed in the circle." a 11. 4. b 17. 3. c 18. 3. 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. Let the angles of a pentagon, inscribed in the circle, be in the points A, B, C, D, E, so that the arches AB, BC, CD, DE, EA are equala; and through the points A, B, C, D, E draw GH, HK, KL, LM, MG, touching the circle; take the centre F, and join FB, FK, FC, FL, FD. Because the straight line KL touches the circle ABCDE in the point C, to which FC 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. Because FCK is a right angle, the square of FK is 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. But the square of FC is equal to the square of FB; therefore the remaining square of CK is equal to the remaining square of BK, and therefore the straight line CK is equal to BK. Because FB is equal to FC, and FK common to the triangles BFK, CFK, the two BF, FK are equal to the two CF, FK; and the base BK is equal to the base KC; therefore the angle BFK is equal to the angle KFC, and the angle BKF to e 8. 1. 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. Because the arch BC is equal to the arch CD, the angle BFC is equal to the angle CFD. But BFC is double of f 27. 3. the angle KFC, and CFD double of CFL; therefore the angle KFC is equal to the angle CFL. Now 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 two angles of the other, each to each, and the side FC, which is adjacent to the equal angles in each, is common to both; therefore the straight line KC is equals to CL, and g 26. 1. the angle FKC to the angle FLC. 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. Because BK is equal to KC, as was demonstrated, and KL double of KC, and HK double of BK, HK is 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 the angle HKL double of the angle FKC, and KLM double of FLC, as was before demonstrated, the angle HKL is equal to KLM. 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 Book IV. 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. PROP. XIII. PROB. a 9. 1. b 4. 1. c 12. 1. 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. 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 BF is equal to the base FD, and the angle CBF is equal to the angle CDF. Because the angle CDE is double of CDF, and because 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 manner it may be demonstrated that the angles BAE, AED are bisected by the straight lines AF, EF. From the point F draw FG, FH, FK, FL, FM perpendicular to the straight lines AB, BC, CD, DE, EA. Because the B H A M angle HCF is equal to KCF, and the right angle FHC equal to the right an gle FKC, in the triangles FHC, FKC there are two angles of one equal to two angles of the other, and the side FC, which Book IV. is opposite to one of the equal angles in each, is common to both; therefore the perpendicular FH is equal to the per- d 26. 1 pendicular 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, will pass through the extremities of the other four; and it will touch the straight lines AB, BC, CD, DE, EA, because the angles at the points G, H, K, L, M are right angles; therefore each of the straight lines AB, BC, CD, DE, EA touches the e Cor. 16.3. circle; wherefore the circle is inscribed in the pentagon ABCDE. Which was to be done. PROP. XIV. PROB. TO describe a circle about a given equilateral and equiangular pentagon. Let ABCDE be the given equilateral and equiangular pentagon; it is required to describe a circle about it. Bisect the angles BCD, CDE by the straight lines CF, a 9. 1, FD, and from the point F, in which they meet, draw the straight lines FB, FA, FE to B E the points B, A, E. It may be |