lines CE, DB; and join AB, BC, DE, EA. ABCDE is the Book IV. pentagon required. Because each of the angles ACD, CDA is double of CAD, and they are bifected by the ftraight lines CE, DB, the five angles DAC, ACF, ECD, : d G E B d 26.3. H C € 29.3. CDB, BDA are equal to one another But equal angles ftand upon equal arches; therefore the five arches AB, BC, CD, DE, EA are equal to one another: And equal arches are fubtended by equal eftraight lines; therefore the five ftraight lines AB, BC, CD, DE, EA are equal to one another. Wherefore the pentagon ABCDE is equilateral. It is alfo equiangular; because the arch AB is equal to the arch DE: If to each be added BCD, the whole ABCD is equal to the whole EDCB: And the angle AED ftands on the arch ABCD, and the angle BAE on the arch EDCB; therefore the angle BAE is equal to the f 27.3. angle AED: For the fame reafon, 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 infcribed. Which was to be done. T PROP. XII. PROB. O defcribe an equilateral and equiangular penta- See N. gon about a given circle. Let ABCDE be the given cirole; it is required to defcribe an equilateral and equiangular pentagon about the circle ABCDE. Let the angles of a pentagon, infcribed in the circle by the laft propofition, be in the points A, B, C, D, E, fo that the arches AB, BC, CD, DE, EA are equal; and through the a 11.4. points A, B, C, D, E draw GH, HK, KL, LM, MG, touching b the circle; take the centre F, and join FB, FK, FC, FL, b 17. 3. FD: And 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 c 18. 39 angles at C is a right angle: For the fame reason, the angles at the points B, D are right angles: Join BC; and becaufe the angles CBK, BCK are less than two right angles, CK, BK do Book IV. do meet *, if produced: For the fame reafon, LM meets KL, ~ GM, and GH meets GM, HK: And becaute BF is equal to *12.Ax. 1. FC, and FK common to the triangles BFK, CFK, the two BF, A E M F D FK are equal to the two CF, FK; and the bafe BK is equal to dCor.37.3. the bafe KC, because they are drawn from the point K to e 8. 1. touch the circle; therefore the angle BFK is equal e to the angle KFC, and the angle BKF to FKC; wherefore the angle BFC is double of the angle KFC, and BKC double of FKC: For the fame reason, the angle CFD is double of the angle CFL, and CLD double of CLF: And because the arch BC is equal f 27.3. to CD, the angle BFC is equal to CFD; and BFC is double of KFC, and CFD double of CFL; therefore the angle KFC is equal to CFL; and the right angle FCK is equal to FCL: Therefore, in the two triangles H FKC, FLC, there are two angles of one equal to two of the other, each to each, and the fide FC, B adjacent to the equal angles, is common to both; therefore the, other fides fhall be equal to the other fides, and the third angle to the third angle: Therefore KC is equal to CL, and the angle FKC to FLC: And becaufe KC is equal to CL, KL is double of KC: In the fame manner, it may be fhown, that HK is double of BK: And becaufe BK is equal to KC, as was demonstrated, and that KL is double of KC, and HK double of BK, HK fhall be equal to KL: In like manner, it may be fhown, that GH, GM, ML are each of them equal to HK or KL: Therefore the pentagon GHKLM is equilateral. It is alfo equiangular; for, fince the angle FKC is equal to FLC, and that HKL is double of FKC, and KLM double of FLC, as was before demonftrated, the angle HKL is equal to KLM : And in like manner, it may be fhown, that each of the angles KHG, HGM, GML is equal to HKL or KLM: Therefore the pentagon GHKLM is equiangular: And it is equilateral, as was demonftrated; and it is defcribed about the circle ABCDE. Which was to be done. g 26. I. T K C L O infcribe a circle in a given equilateral and equiangular pentagon, Let Let ABCDE be the given equilateral and equiangular penta- Book IV. gon; it is required to infcribe a circle in it. 2 A Bifect the angles BCD, CDE by the ftraight lines CF, DF, a 9. 1. and from the point F, in which they meet, draw the ftraight lines FB, FA, FE: Therefore, fince BC is equal to CD, and CF common to the triangles BCF, DCF, the two fides BC, CF are equal to the two DC, CF; and the angle BCF is equal to DCF; therefore the base BF is equal to FD, and the angle b 4.1. CBF to CDF: And because the angle CDE is double of CDF, and that CDE is equal to CBA, and CDF to CBF; CBA is alfo double of CBF; therefore the angle ABF is equal to CBF; wherefore the angle ABC is bifected by BF: In the fame manner, it may be demonftrated, that the angles BAE, AED are bifected by AF, FE: From F draw B FG, FH, FK, FL, FM perpendiculars to AB, BC, CD, DE, EA: And because the angle HCF is equal to KCF, and the right angle FHC equal to FKC; in the triangles FHC, FKG there are two angles of one equal to two of the H G M C K D C 12. I other, and the fide FC, oppofite to one of the equal angles in each, is common to both; therefore the other fides fhall be equal d, each to each; wherefore the perpendicular FH is equal d 26, 1. to FK: In the fame manner, it may be demonftrated, that FL, FM, FG are each of them equal to FH or FK: Wherefore the circle described from the centre F, at the diftance of one of these five, fhall pass through the extremities of the other four: And because the angles at the points G, H, K, L, M are right angles, and that a straight line from the extremity of a diameter at right angles to it, touches the circle; therefore each of e 16. 3o the ftraight lines AB, BC, CD, DE, EA touches the circle; wherefore it is inscribed in the pentagon ABCDE. Which was to be done. PROP. XIV. PROB. T O defcribe a circle about a given equilateral and Let ABCDE be the given equilateral and equiangular pentagon; it is required to defcribe a circle about it. Bifect the angles BCD, CDE by the ftraight lines CF, FD, a 9. 1. and from the point F, in which they meet, draw the ftraight lines b 6. 1. equal E F BOOK IV lines FB, FA, FE to the points B, A, E. It may be demonftrated, in the fame manner as in the preceding propofition, that the angles CBA, BAE, AED are bifected by the straight lines FB, FA, FE: And because the angle BCD is equal to the angle CDE, and that FCD is the half of the angle BCD, and CDF the half of CDE; the angle FCD is equal to FDC; wherefore the fide CF is B to the fide FD: In like manner, it may be demonftrated, that FB, FA, FE are each of them equal to FC or FD: Therefore the five ftraight lines FA, FB, FC, FD, FE are equal to one another; and the circle defcribed from the centre F, at the distance of one of them, fhall pass through the extremities of the other four, and be defcribed about the equilateral and equiangular pentagon ABCDE. Which was to be done. a 1. 3. b 1. T PROP. XV. PROB. C O infcribe an equilateral and equiangular hexagon in a given circle. Let ABCDEF be the given circle; it is required to inscribe an equilateral and equiangular hexagon in it. Find the centre G, and draw the diameter AGD; and in 4. the circle place DC, DE, each equal to DG, and join EG, CG, and produce them to the points B, F; and join AB, BC, EF, FA: The hexagon ABCDEF is equilateral and equiangu lar. Because DE, EG are each of them equal to DG, the triangle EGD is equilateral; and therefore its three angles EGD, GDE, DEG are equal to one ano B c Cor. 5.1.ther; and the three angles of a tri- thefe f f15. 1. g 26. 3. 29. 3. thefe are equal the vertical angles BGA, AGF, FGE: BOOK IV. Therefore the fix angles at the point G are equal to one another: But equal angles ftand upon equal arches; therefore the fix arches AB, BC, CD, DE, EF, FA are equal to one another: And equal arches are fubtended by equal * straight lines; therefore the fix ftraight lines are equal to one another, and the hexagon ABCDEF is equilateral. It is also equiangular; for, fince the arch AF is equal to ED, to each of thefe add the arch ABCD; therefore the whole arch FABCD hall be equal to the whole EDCBA: And the angle FED ftands upon the arch FABCD, and the angle AFE upon EDCBA; therefore the angle AFE is equal to FED: In the 1 27. 34 same manner, it may be demonftrated, that the other angles of the hexagon ABCDEF are each of them equal to the angle AFE or FED: Therefore the hexagon is equiangular; and it is equilateral, as was shown; and it is infcribed in the given circle ABCDEF. Which was to be done. COR. From this it is manifeft, that the fide of the hexagon is equal to the ftraight line from the centre, that is, to the femidiameter or radius of the circle. And if through the points A, B, C, D, E, F there be drawn ftraight lines touching the circle, an equilateral and equiangular hexagon shall be described about it, which may be demonftrated from what has been faid of the pentagon; and likewise a circle may be infcribed in a given equilateral and equiangular hexagon, and circumfcribed about it, by a method like to that used for the pentagon. PROP. XVI. PROB. O infcribe an equilateral and equiangular quin Tdecagon in a given circle. Let ABCD be the given circle; it is required to infcribe an equilateral and equiangu lar quindecagon in the circle |
