Book IV. PROP. XIII. PROB, To inscribe a circle in a given equilateral and equian gular pentagon. 2. 9. I. ner it 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 A M be demonstrated that the B the equal angles in each, is common to both; therefore the other d. 26.1. fides 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 center F, at the distance of one of these five, shall pass thro’ the extremities of the other four, and touch the straight lines AB, BC, CD, DE, C. 12. 1. 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- m er of a circle at right angles to it, touches the circle. there- c. 16. 3. fie each of the straight lines AB, BC, CD, DE, EA touches the wherefore it is inscribed in the pentagon ABCDE. Which TO PROP. XIV. PROB. angular 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, FD, and a. 9. 1. А. F E D equal to FDC; wherefore the side CF is equal to the side FD. in like manner it may be demonstrated b. 6.1. 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 center F, at the distance of one of them, shall pass thro' the extremities of the other four, and be described about the equilateral and equiangular pentagon ABCDE. Which was to be done. PROP. To infcribe an equilateral and equiangular hexagon in Book IV. PROP. XV. PROB. O inscribe Sec N. a 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 describe 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. Because G is the center of the circle ABCDEF, GE is equal to GD, and because 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 isosceles a. 5. 1. triangle are equal". and the three angles of a triangle are equal to b. 32. 1. two right angles; therefore the angle EGD is the third part of two right angles. in the same manner it may A F B with EB the adjacent angles EGC, CGB G equal to two right angles ; the re- E D H c. 26. 3. stand 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 subtended by equal f straight lines ; therefore the fix straight lines are equal to one another, and the hexagon ABCDEF is equilateral. It is also equiangular; for since 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 infcribed in the given circle ABCDEF. Which was to be done. COR. From this it is manifeft, that the side of the hexagon is equal to the straight line from the center, that is, to the semidiameter of the circle. And if thro' the points A, B, C, D, E, F there be drawn straight lines touching the circle, an equilateral and equiangular hexagon shall be described about it, which may be demonstrated from what has been said of the pentagon; and likewise a circle may be inscribed in a given equilateral and equiangular hexagon, and circumscribed about it, by a method like to that used for the pentagon. PROP. XVI. PROB. "O inscribe an equilateral and equiangular qyindeca- See N. gon in a giren circle. Let ABCD be the given circle; it is required to inscribe an equilateral and equiangular quindecagon in the circle ABCD. Let AC be the side of an equilateral triangle inscribed « in the a. 2 4. circle, and AB the side of an equilateral and equiangular pentagon inscribed in the same; therefore of such equal parts as the whole b. 11. 4. circumference ABCDF contains fifteen, the circumference ABC, being the third part of the whole, contains five; and the circumference AB, А which is the fifth part of the whole, contains three ; therefore BC their difference contains two of the fame B F parts. bisect CBC in E; therefore E BE, EC are, each of them, the fifteenth part of the whole circumfe. С D rence ABCD, therefore if the straight lines BE, EC be drawn, and straight lines equal to them be placed d around in the whole circle, an equi- , d. 1. 4. lateral and equiangular quindecagon shall be inscribed in it. Which was to be done. And c. 30. 3. Book IV. And in the same 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 shall be described about it, and likewise, as in the pentagon, a circle may be inscribed in a given equilateral and equiangu. lar quindecagon, and circumscribed about it. THE |