H A E M D KC L equal to the remaining square of BK: and so B K is equal to C K. And becaufe F B is equal to Fc, and FK is common, the two fides BF, F K are equal to the two fides CF, FK; and the base BK is equal to the base CK: Therefore [by 8. 1.] the angle B F K will be equal to the B angle KFC, and the angle BK F to the angle FK C: Therefore the angle BFC is double to the angle KFC: And the angle B KC double to the angle F K c. By the fame reason the angle CFD is double to the angle CF L, and the angle C L D double to the angle c L F. And because the part B C of the circumference is equal to the part C D of it; the angle B FC [by 27. 3.] will be equal to the angle C F D. But the angle B FC is double to the angle KFC, and the angle D F C double to the angle L FC. Therefore the angle KFC is equal to the angle CFL. Wherefore F K C, FLC are two triangles having two angles of the one equal to two angles of the other, and one fide of the one equal to one fide of the other, viz. È c which is common to them both; therefore [by 26. 1.] the remaining fides of the one will be equal to the remaining fides of the other, and the remaining angle of the one equal to the remaining angle of the other: Wherefore the right line K c will be equal to the right line CL, and the angle F K c equal to the angle F L C. And because K c is equal to CL; K L will be double to K c. By the fame reafon HK will be proved to be double to B K. Again, because BK has been proved to be equal to K c, and K L is double to K c, and H K to B K; HK will be equal to After the manner we demonftrate that GH, GM, M L are each of them equal to HK, or K L: therefore the pentagon GHK L M is equilateral. I fay it is alfo equiangular. For because the angle F K c is equal to the angle FLC, and it has been proved that H K C is double to the angle FKL; and DLC is double to the angle KLF: the angle HKL will be equal to the angle KLM. We demonftrate after the fame manner that the angles KHG, HG M, GML are each equal to the angles H KL, KLM: Therefore the five angles G H K, HK L, KLM, L MG, MGH are equal to one another. Wherefore the N 4 KL. pen pentagon G H K L M is equiangular. But it has also been proved to be equilateral, and is circumfcribed about the circle A B C D E. Which was to be done. PRO P. XIII. PRO BL. To infcribe a circle in a given equilateral and equiangugular pentagon. Let the given equilateral and equiangular pentagon be A B C D E. It is required to inscribe a circle in the penta gon A B C D E. B H M E 1.] each of the angles B C D, CDE by the And from the point F wherein they lines F B, F A, FE. And becaufe B C is equal to CD, and C F is common, the two fides B C, C F are equal to the two fides, DC, CF, and the angle B CF is equal to the angle DCF: Therefore the bafe B F is [by 4. 1.] equal to the base D F, and the triangle Br C is equal to the CK D triangle D C F, and the remaining angles equal to the remaining angles, which are oppofite to the equal fides: Therefore the angle C B F, will be equal to the angle CD F. And because the angle CDE is double to the angle C D F, and the angle C D E equal to the angle ABC, but the angle CDF equal to the angle CBF: The angle C B A will be double to the angle c BF; and accordingly the angle A B F equal to the angle F B C : Therefore the angle A B C is bifected by the right line B F. In like manner we demonftrate that the angles B AE, A ED are each of them bifected by the right lines F A, F E. And fo from the point F draw FG, FH, FK, FL, FM perpendicular to the right lines A B, B C, CD, DE, E A. And because the angle HCF is equal to the angle K CF, but the right angle FHC is equal to the right angle FKC: The two triangles F H C, F K C, will have two angles of the one equal to two angles of the other, and one fide of the one equal to one fide of the other, viz. the common fide Fc, which is oppofite to one of the equal angles : Therefore [by 26. 1.] their remaining fides will be equal, and the perpendicular F H will be equal to the perpendicu lar lar to FK. After the fame manner we demonftrate that FL, FM, FG are each of them equal to FH or FK: Therefore the five right lines FG, FH, FK, FL, FM are equal to one another. Wherefore a circle defcribed with the centre F, and either of the distances F G, F H, F K, FL, F M will also pass thro' the remaining points, and the right lines A B, BC, CD, DE, E A will touch it, because the angles at G, H, K, L, M, are right angles. For if the circle does not touch, but cut them, a right line drawn from one end of its diameter at right angles to that diameter, will fall within the circle. Which [by 16. 3.] has been proved to be abfurd. Therefore a circle defcribed with the centre F, and either of the diftances F G, FH, FK, FL, F M, will not cut the right lines A B, BC, CD, DE, E A: it will therefore touch them. Let it be defcribed, as G H KL M. Therefore a circle is infcribed in a given equilateral and equiangular pentagon. Which was to be demonftrated. To circumfcribe a circle about a given equilateral and equiangular pentagon. Let A B C D E be a given equilateral and equiangular pentagon. It is required to circumfcribe a circle about the pentagon A B C D E. Bifect [by 9. 1.] each of the angles B C D, C D E by the right lines CF, FD; and draw the right lines F B, F A, FE from the point F wherein they meet, to the points B, A, E. Then we demonftrate after the fame manner as in the last propofition, that the angles C BA, BAE, A E D are bisected by the right lines BF, FA, FE. And because the angle B C D is equal to the angle C DE, and the angle F C D is one half of the angle BCD, and C D F one half of the angle CDE; the angle FCD is equal to the angle FDC and fo [by 6. 1.] the B fide c is equal to the fide F D. In like manner we demonstrate that F B, FA, FE are each equal to FC or FD: Therefore the five right lines FA, FB, F A FC, FC, FD, FE are each equal to one another. Wherefore a circle described with the centre F, and either of the distances F A, FB, FC, FD, FE will also pass thro' the remaining points, and be circumfcribed about the equilateral and equiangular pentagon A B C D E. Let the circle be described, and let it be A B C D E. Therefore a circle is circumfcribed about a given equilateral, and equiangular pentagon. Which was to be done. PROP. XV. PROBL. To infcribe an equilateral and equiangular hexagon in a given circle. Let the given circle be A B C D E F. It is required to infcribe an equilateral and equiangular hexagon in the circle A B C D E F. Draw the diameter A D of the circle ABCDEF; and find the centre G of the circle, and about the centre D, with the diftance DG defcribe [by 3. poft.] the circle EG CH and joining E G, C G, produce themto the points B, F Alfo join AB, BC, CD, DE, EF, FA: I fay the hexagon ABCDEF is equilateral and equiangular. F G For because the point G is the centre of the circle ABCDEF, the line GE will be equal to GD. Again, because D is the centre of the circle E G CH; D E will be equal to D G. But GE has been proved to be equal to G D : Therefore GE is equal to D E. Therefore E G D is an B equilatearl triangle; and fo its three angles E G D, GDE, DEG, are equal to one another; becaufe [by 5. 1.] the angles at the bafe of an ifofceles triangle are equal to each other, and [by 32. 1.] the three angles of a triangle are equal to two right angles: Wherefore the angle E G D is one third part of two right H angles. After the fame manner we demonftrate that the angle DGC is one third part of two right angles. And becaufe [by 13. 1.] the right line cG standing upon the right line E B makes the adjacent angles E D EG C, EGC, CGB equal to two right angles, the remaining angle CGB will be one third part of two right angles: Therefore the angles E G D, DG C, C G B are equal to one another. And fo the angles B G A, A GF, FGE at their vertexes [by 15. 1.] are equal to the angles E G D, DGC, CGB: Wherefore the fix angles EG D, DG C, CG B, BGA, AGF, FGE are equal to one another. But [by 26. 3.] equal angles ftand upon equal parts of the circumference: Therefore the fix parts of the circumference a B, BC, CD, DE, EF, F A are equal to one another. And [by 26. 3.] equal parts of the circumference are joined by equal right lines. Therefore thefe fix right lines are equal to one another and fo the hexagon A B C D E F is equilateral: I fay it is equiangular too. For because the part AF of the circumference is equal to the part E D of it, if the common part A B C D of the circumference be added to both, the whole F A B C D is equal to the whole E D C BA; but the angle FED ftands upon the part F A B C D of the circumference, and the angle AFE upon the part E D C BA of it: Therefore the angle AF E [by 27. 3.] is equal to the angle D E F. After the fame manner we demonstrate that every one of the remaining angles of the hexagon ABCDE F, is equal to either of the angles AF E, or F E D. Therefore the hexagon A B C D E F is equiangular: But it has been alfo proved to be equilateral; and is infcribed in the circle A B C D E F. Therefore an equilateral and equiangular hexagon is infcribed in a given circle. Which was to be done. Corollary. From hence it is manifeft that the fide of an hexagon is equal to the femidiameter of the circle. * And if tangents to the circle be drawn thro' the points A, B, C, D, E, F, an equilateral and equiangular hexagon will be circumfcribed about the circle, as is evident from what has been faid of the pentagon: And moreover a circle may be infcribed in a given hexagon, or circumfcribed about one after the like manner as we have infcribed or circumfcribed a pentagon in or about a given circle. Infcribed in a circle. PROP. |