through the extremities of the other three, and be described about the square ABCD. PROP. X. PROB. To describe an isosceles triangle, having each of the angles at the base double of the third angle. Take any straight line AB, and divide (11. 2.) it in the point C, so that the rectangle AB.BC may be equal to the square of AC; and from the centre A, at the distance AB, describe the circle BDE, in which place (1. 4.) the straight line BD equal to AC, which is not greater than the diameter of the circle BDE; join DA, DC, and about the triangle ADC describe (5. 4.) the circle ACD; the triangle ABD is such as is required, that is, each of the angles ABD, ADB is double of the angle BAD. E Because the rectangle AB.BC is equal to the square of AC, and AC equal to BD, the rectangle AB.BC is equal to the square of BD; and because from the point B without the circle ACD two straight lines BCA, BD are drawn to the circumference, one of which cuts, and the other meets the circle, and the rectangle AB.BC contained by the whole of the cutting line, and the part of it without the circle, is equal to the square of BD, which meets it; the straight line BD touches (37. 3.) the circle ACD. And because BD touches the circle, and DC is drawn from the point of contact D, the angle BDC is equal (32. 3.) to the angle DAC in the alternate segment B D of the circle, to each of these add the angle CDA; therefore the whole angle BDA is equal to the two angles CDA, DAC; but the exterior angle BCD is equal (32. 1.) to the angles CDA, DAC; therefore also BDA is equal to BCD; but BDA is equal (5. 1.) to CBD, because the side AD is equal to the side AB; therefore CBD, or DBA is equal to BCD; and consequently the three angles RDA, DBA, BCD, are equal to one another. And because the angle DBC is equal to the angle BCD, the side BD is equal (6. 1.) to the side DC; but BD was made equal to CA; therefore also CA is equal to CD, and the angle CDA equal (5. 1.) to the angle DAC; therefore the angles CDA, DAC together, are double of the angle DAC; but BCD is equal to the angles CDA, DAC (32. 1.); therefore also BCD is double of DAC. But BCD is equal to each of the angles BDA, DBA, and therefore each of the angles BDA, DBA, is double of the angle DAB; wherefore an isosceles triangle ABD is described, having each of the angles at the base double of the third angle. "COR. 1. The angle BAD is the fifth part of two right angles. "For since each of the angles ABD and ADB is equal to twice the angle BAD, they are together equal to four times BAD, and therefore all "the three angles ABD, ADB, BAD, taken together, are equal to five "times the angle BAD. But the three angles ABD, ADB, BAD are equal to two right angles, therefore five times the angle BAD is equal to "two right angles; or BAD is the fifth part of two right angles." "COR. 2. Because BAD is the fifth part of two, or the tenth part of "four right angles, all the angles about the centre A are together equal to "ten times the angle BAD, and may therefore be divided into ten parts "each equal to BAD. And as these ten equal angles at the centre, must "stand on ten equal arcs, therefore the arc BD is one-tenth of the cir"cumference; and the straight line BD, that is, AC, is therefore equal to "the side of an equilateral decagon inscribed in the circle BDE.” 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 the circle ABCDE. Describe (10. 4.) an isosceles triangle FGH, having each of the angles at G, H, double of the angle at F; and in the circle ABCDE inscribe (2. 4.) the triangle ACD equiangular to the triangle FGH, so that the angle CAD 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 (9. 1.) the angles ACD, CDA by the straight lines CE, DB; and join AB, BC, ED, EA. ABCDE is the pentagon required. F B A Because the angles ACD, of CAD, and are bisected by the straight lines CE, DB, the five angles DAC, ACE, ECD, CDB, BDA are equal to one another; but equal angles stand upon equal arcs (26. 3.); therefore the five arcs AB, BC, CD, DE, EA are equal to one another; and equal arcs are subtended by equal (29. 3.) straight lines; therefore the five straight lines AB, BC, CD, DE, EA are equal to one another. Wherefore the pentagon ABCDE is equilateral. It is also equiangular; because the arc AB is equal to the arc DE; if to each be added BCD, the whole ABCD is equal to the whole EDCB; and the angle AED stands on the arc ABCD, and the angle BAE on the arc EDCB: therefore the angle BAE is equal (27. 3.) to the angle AED: for the same reason, 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 shewn that it is equilateral. Wherefore, in the given circle, an equilateral and equiangular pentagon has been inscribed. Otherwise. "Divide the radius of the given circle, so that the rectangle contained "by the whole and one of the parts may be equal to the square of the other 66 "(11. 2.).. Apply in the circle, on each side of a given point, a line equal to the greater of these parts; then (2. Cor. 10. 4.), each of the "arcs cut off will be one-tenth of the circumference, and therefore the arc made up of both will be one-fifth of the circumference; and if the straight line subtending this arc be drawn, it will be the side of an equilateral pentagon inscribed in the circle." 66 66 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, by the last proposition, be in the points A, B, C, D, E, so that the arcs AB, BC, CD, DE, EA are equal (11. 4.); and through the points A, B, C, D, E, draw GH, HK,.KL, LM, MG, touching (17. 3.) the circle; take the centre F, and join FB, FK, FC, FL, 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 (18. 3.) 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; and because FCK is a right angle, the square of FK is equal (47. 1.) 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, of which the square of FC is equal to the square of FB; the remaining square of CK is therefore equal to the remaining square of BK, and the straight line CK equal to BK: and 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 (8. 1.) 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 same reason, the angle CFD is double of the angle CFL, and CLD double of CLF: and because the arc BC is equal to the arc CD, the angle BFC is equal (27. 3.) to the angle CFD and BFC is double of 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; and 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 other sides are equal (26.1.) to the other sides,and the third angle to the third angle; therefore the straight line KC is equal to CL, and the angle FKC to the angle H B G A T M F D K FLC and because KC is equal to CL, KL is double of KC; in the same manner, it may be shewn that HK is double of BK; and because BK is equal to KC, as was demonstrated, and KL is double of KC, and HK double of BK, HK is equal to KL; in like manner, it may be shewn 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; and in like manner it may be shewn, that each of the angles KHG, HGM, GML is equal to the angle HKL or KLM; therefore the 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. PROP. XIII. PROB. To inscribe a circle in a given equilateral and equiangu.ar pentagon. Let ABCDE be the given equilateral and equiangular pentagon; it is required to inscribe a circle in the pentagon ABCDE. Bisect (9.1.) 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; therefore, 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 (4. 1.) to the base FD, and the other angles to the other angles, to which the equal sides are opposite; therefore the angle CBF is equal to the angle CDF: and because the angle CDE is double of CDF, and CDE 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: B from the point F draw (12. 1.) FG, FH, FK, FL, FM perpendiculars to the straight 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 G H K M E D angles of one equal to two angles of the other, and the side FC, which is opposite to one of the equal angles in each, is common to both; therefore, the other sides shall be equal (26. 1.), cach 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 centre F, at the distance of one of these five, will pass through the extremities of the other four, and touch the straight lines AB, BC, CD, DE, EA, because that the angles at the points G, H, K, L, M are right angles, and that a straight line drawn from the extremity of the diameter of a circle at right angles to it, touches (1. Cor. 16. 3.) the circle; therefore each of the straight lines AB, BC, CD, DE, EA touches the circle; wherefore the circle is inscribed in the pentagon ABCDE. 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. B A E F Bisect (9. 1.) the angles BCD, CDE by the straight lines CF, FD, and from the point F, in which they meet, draw the straight lines FB, FA, FE to the points B, A, E. It may be demonstrated, in the same manner as in the preceding proposition, that the angles CBA, BAE, AED are bisected by the straight lines FB, FA, FE: and because that 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 side CF is equal (6. 1.) to the side FD: in like manner it may be demonstrated, 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 centre F, at the distance of one of them, will pass through the extremities of the other four, and be described about the equilateral and equiangular pentagon ABCDE. PROP. XV. PROB. D To inscribe 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 of the circle ABCDEF, and draw the diameter AGD: and from D, as a centre, 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 centre of the circle ABCDEF, GE is equal to GD: and because D is the centre 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 (Cor. 5. 1.); and the three angles of a triangle are equal (32. 1.) to two right angles; therefore the angle EGD is the third part of two right angles in the same manner it may be demonstrated that the angle DGC is also the third part of two right angles; and because the straight line GC makes with EB the adjacent angles EGC, CGB equal (13. 1.) to two right angles; the remaining angle CGB is the third part of two right angles; therefore the angles EGD, DGC, CGB, are equal to one another; and also the angles vertical to them, BGA, AGF, FGE (15. |