Book IV. a. 9. I. PROP. XIII. PROB. To infcribe 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. Bifect the angles BCD, CDE by the ftraight lines CF, DF, and from the point Fin 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 the angle DCF; therefore the bafe BF is equal b to the base FD, and the other angles to the other angles, to which the equal fides are oppofite; 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 alfo double of the angle CBF; therefore the angle ABF is equal to the angle CBF; wherefore the angle ABC is bifected by the ftraight line BF. in the fame manner it may be demonftrated that the B angles BAE, AED are bifected by the straight lines AF, FE. from the H C. 12. I. point F draw FG, FH, FK, FL, FM perpendiculars to the ftraight C G A M CK D 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 fide FC, which is oppofite to one of the equal angles in each, is common to both; therefore the other d. 26. 1. fides fhall be equal, each to each; wherefore the perpendicular FH is equal to the perpendicular FK. in the fame manner it may be demonftrated 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 center F, at the distance of one of these five, fhall pass thro' the extremities of the other four, and touch the ftraight lines AB, BC, CD, DE, DE, EA, because the angles at the points G, H, K, L, M are right Book IV. angles; and that a straight line drawn from the extremity of the di e er of a circle at right angles to it, touches the circle. there- c. 16. 3. fre each of the straight lines AB, BC, CD, DE, EA touches the ; wherefore it is infcribed in the pentagon ABCDE. Which be done. PROP. XIV. PROB. To defcribe a circle about a given equilateral and equiangular pentagon. 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 straight lines CF, FD, and a. 9. 1. from the point F in which they meet draw the straight lines FB, FA, FE to the points B, A, E. It may be demonftrated, in the fame manner as in the preceeding propofition, that the angles CBA, BAE, AED are bifected by the ftraight 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 B A F E CF is equal to the fide FD. in like manner it may be demonftrated b. 6. 1. 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 described from the center F, at the distance of one of them, shall pafs thro' the extremities of the other four, and be described about the equilateral and equiangular pentagon ABCDE. Which was to be done. PROP. Book IV. See N. a. 5. 1. b. 32. I. T PROP. XV. PROB. 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 center G of the circle ABCDEF, and draw the diameter AGD; and from D as a center, at the distance DG defcribe the circle EGCH, join EG, CG and produce them to the points B, F; and join AB, BC, CD, DE, EF, FA. the hexagon ABCDEF is equilateral and equiangular. A B G Becaufe G is the center of the circle ABCDEF, GE is equal to GD. and becaufe D is the center of the circle EGCH, DE is equal to DG; wherefore GE is equal to ED, and the triangle EGD is equilateral, and therefore its three angles EGD, GDE, DEG are equal to one another, because the angles at the base of an ifofceles triangle are equal. and the three angles of a triangle are equal to two right angles; therefore the angle EGD is the third part of two right angles. in the fame manner it may be demonftrated that the angle DGC is alfo the third part of two right angles. and because the straight line GC makes F with EB the adjacent angles EGC, CGB c. 13. 1. equal to two right angles; the remaining angle CGB is the third part of two right angles; therefore the angles EGD, DGC, CCB are equal to one d. 15. 1. another. and to these are equal the vertical oppofite angles BGA, AGF, FGE. therefore the fix angles EGD, DGC, CGB, BGA, AGF, FGE, are e c d qual to one another. but equal angles e E D H c. 26. 3. ftand upon equal circumferences; therefore the fix circumferences AB, BC, CD, DE, EF, FA are equal to one another. and equal f. 29. 3. circumferences are fubtended by equal f straight lines; therefore the fix ftraight lines are equal to one another, and the hexagon ABCDEF is equilateral. It is alfo equiangular; for fince the circumference AF is equal to ED, to each of these add the circumference ABCD; therefore the whole circumference FABCD shall be equal to the whole EDCBA. and the angle FED stands upon the circumference circumference FABCD, and the angle AFE upon EDCBA; there- Book IV. fore the angle AFE is equal to FED. in the fame manner it may be demonstrated 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 fhewn; and it is inscribed 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 straight line from the center, that is, to the femidiameter of the circle. And if thro' 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 demonstrated from what has been faid of the pentagon; and likewife 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 quindeca- See N. gon in a given circle. Let ABCD be the given circle; it is required to infcribe an equilateral and equiangular quindecagon in the circle ABCD. a Let AC be the fide of an equilateral triangle infcribed in the a. 2 4. circle, and AB the fide of an equilateral and equiangular pentagon inscribed b in the fame; therefore of fuch equal parts as the whole b. 11. 4. circumference ABCDF contains fifteen, the circumference ABC, be ing the third part of the whole, con- E A F C. 30. 3. D lines equal to them be placed around in the whole circle, an equi- d. 1. 4. lateral and equiangular quindecagon fhall be infcribed in it. Which was to be done. And Book IV. And in the fame manner as was done in the pentagon, if thro the points of division made by inscribing the quindecagon, straight lines be drawn touching the circle, an equilateral and equiangular quindecagon fhall be described about it. and likewise, as in the pentagon, a circle may be infcribed in a given equilateral and equiangular quindecagon, and circumfcribed about it. THE |
