* 8.1. PROP. IX. PROB. To describe a circle about a given square. Let ABCD be the given square; it is required to describe a circle about ABCD. B D Join AC, BD, cutting one another in E: and be+ 30 Def. 1. cause DA is equal to AB, and AC common to the triangles DAC, BAC, the two sides DA, AC are equal to the two BA, AC, each to each; and the base DC is equal to the base BC; wherefore the angle DAC is equal to the angle BAC, and the angle DAB is bisected by the straight line AC: in the same manner, it may be demonstrated that the angles ABC, BCD, CDA are severally bisected by the straight lines BD, AC: therefore, because the angle DAB is + 30 Def. equal to the angle ABC, and that the angle EAB is the half of DAB, and EBA the half of ABC; the angle EAB is equal to the angle EBA; wherefore the side EA is equal to the side EB: in the same manner it may be demonstrated that the straight lines EC, ED are each of them equal to EA, or EB: therefore the four straight lines EA, EB, EC, ED, are equal to one another; and the circle described from the centre E, at the distance of one of them, will pass through the extremities of the other three, and be described about the square ABCD. Which was to be done. 1. +7 Ax. 6.1. 11. 2. #1.4. * 5. 4. 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 it in the point C, so that the rectangle AB, BC may be equal to the square of CA; and from the centre A, at the distance AB, describe the circle BDE, in which place* the straight line BD equal to AC, which is not greater than the diameter of the circle BDE; and join DA: the triangle ABD shall be such as is required, that is, each of the angles ABD, ADB shall be double of the angle BAD. Join DC, and about the triangle ADC describe the circle ACD: and because the rectangle AB, BC is + Constr. equal to the square of AC+, and that AC is equal to * * B * * 32. 3. BD, the rectangle AB, BC is equal † to the square of +1 Ax. 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 that 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 the circle ACD: and because ⚫ 37. 3. BD touches the circle, and DC is drawn from the point of contact D, the angle BDC is equal to the angle DAC in the alternate segment 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 +2 Ax. exterior angle BCD is equal to the angles CDA, DAC; 32. 1. therefore also BDA is equal† to BCD: but BDA is †1Ax. equal to the angle CBD, because the side AD is equal * 5.1. to the side AB; therefore CBD, or DBA, is equal to +1 Ax. BCD; and consequently the three angles BDA, DBA, BCD, are equal to one another: and because the angle DBC is equal to the angle BCD, the side BD is equal* to the side DC: but BD was made equal to CA, therefore also CA is equal to CD, and the angle CDA +1 Ax. equal to the angle DAC; therefore the angles CDA, *5.1. DAC together, are double of the angle DAC: but BCD is equal to the angles CDA, DAC; therefore also + 32. 1. BCD is double of DAC: and BCD was proved to be equal to each of the angles BDA, DBA; 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. Which was to be done. * * 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 an isosceles triangle FGH, having each of the angles at G, H, double of the angle at F; and 6.1. * 10. 4. 6 * 2.4. * 9.1. *26.3. * 29.3. † 2 Ax. * 27.3. * 11. 4. FB E in the circle ABCDE inscribe the triangle ACD G H the angles ACD, CDA by the straight lines CE, DB; and join AB, BC, DE, EA. ABCDE shall be the pentagon required. Because each of the angles ACD, CDA is double of CAD, and that they are bisected by the straight lines CE, DB; therefore the five angles DAC, ACE, ECD, CDB, BDA are equal to one another: but equal angles stand upon equal* circumferences; therefore the five circumferences, AB, BC, CD, DE, EA are equal to one another: and equal circumferences are subtended by equal* 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: for, because the circumference AB is equal to the circumference DE, if to each be added BCD, the whole ABCD is equal to the whole EDCB: but the angle AED stands on the circumference ABCD, and the angle BAE on the circumference EDCB; therefore the angle BAE is equal* 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. Which was to be done. 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 circumferences AB, BC, CD, DE, EA are* equal; and through the points A, B, C, D, E draw GH, HK, KL, LM, MG, touching* the circle: the figure GHKLM shall be the pentagon required. * * * 17.3. * 47.1. 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 to KL, therefore 18. 3. 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 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 †1 Ax. squares of FB, BK; of which the square of FC is equal to the square of FB; therefore the remaining square of CK is equal to the remaining square of BK, and †3 Ax. 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, each to each; and the base BK was proved equal to the base KC; therefore the angle BFK is equal to 8. 1. the angle KFC, and the angle BKF+ to FKC: where- +4. 1. fore 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 circumference BC is equal to the circumference CD, the angle BFC is equal 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: and the right angle FCK is equal to the right angle FCL; therefore in the two triangles FKC, FLC, there are two angles of the 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 to the other sides, 26. 1. and the third angle to the third angle: therefore the straight line KC is equal to CL, and the angle FKC to the angle 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 that KL is double of KC, and HK double of BK, therefore HK * H B G A E * 27.3. M D K C +7 Ax. † 6 Ax. † 6 Ax. 9. 1. + Hyp. + Constr. * 4. 1. * 12. 1. : 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 that the angle HKL is double of the angle FKC, and KLM double of FLC, as was before demonstrated, therefore 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. Which was to be done. PROP. XIII. PROB. To inscribe 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. * Bisect 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, each to each; and the angle BCF is equal to the angle DCF; therefore the base BF is equal 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 that CDE is equal to CBA, and CDF to CBF; CBA is also double of the angle CBF; therefore the angle B 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, FE. From the point F, draw* FG, FH, FK, FL, FM perpendiculars to the straight lines AB, BC, CD, DE, EA: and because the angle HCF is equal to H A M E L C KD |