and that AE is the half of AD, and AF the half of AB, AE is equal to AF ; wherefore the sides opposite to these are equal, viz. FG to GE; in the same manner, it may be demonstrated, that GH, GK, are each of them equal to FG or GE ; there A E D fore the four straight lines GE, GF, GH, GK, are equal to one another; and the circle described from the centre G, at the distance of one of them, will pass through the extremities of the other three ; and F K will also touch the straight lines AB, BC, CD, DA, because the angles at the points E, F, H, K, are right (29. 1.) angles, and because the straight line which is drawn from the extremity of a diameter, at right B H С angles to it, touches the circle (16.3.) ; therefore each of the straight lines AB, BC, CD, DA touches the circle, which is therefore inscribed in the square ABCD. Which was to be done. 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 it. Join AC, BD, cutting one another in E; and because 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, and the base DC is equal to the base BC ; wherefore the angle DAC is equal (8. 1.) to the angle BAC, and the an А, it E C ABC ; the angle EAB is equal to the angle EBA: and the side EA (6. 1.) 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, must pass through the extremities of the other three, and be described about the square ABCD. Which was to be done. PROP. X. PROB. 1 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. 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 E E 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 B 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 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 CDB, because the side AD is equal to the side AB ; therefore CBD, or DBA is equal to 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 (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. Which was to be done. ** 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 v angle BAD, they are together equal to four times BAD, and there“ fore 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 *s 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 arches, therefore the arch BD " is one-tenth of the circumference; and the straight line BD, that is " AC, is therefore equal to the side of an equilateral decagon inscrib. 66 ed in the cirtle 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 dou F ble of the angle CAD. Bisect E (9. 1.) the angles ACD, CDA by the straight lines CE, DB ; В. and join AB, BC, DE, EA. ABCDE is the pentagon required. Because the angles ACD, Н D CDA are each of them double 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 (26. 3.) arches ; therefore the five arches AB, BC, CD, DE, EA are equal to one another : and equal arches 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 ; berause the arch AB is equal to the : DE ; if to each be added BCD, the whole ABCD is equal to this le EDCB :, I lie angle AED stands on the arch ABCD, ar le BAE arch 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 shown that it is equilateral. Wherefore, in the given circle, an equilateral and equiangular pentagon has been inscribed. Which was to be done. Otherwise. * Divide the radius of the given circle, so that the rectangle con“ tained by the whole and one of the parts may be equal to the square “ of the other (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.) 66 each of the arches cut off will be one-tenth of the circumference, and " therefore the arch made up of both will be one-fifth of the circumfer ence; and if the straight line subtending this arch be drawn, it will ** be the side of an equilateral pentagon inscribed in the circle." 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 arches, 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 becalise 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 arch BC is equal to the arch CD, the angle BFC is equal (27. 3.) to the angle CFD ; and BFG is double of the angle KFC, and CFD double of CFL ; therefore the angle KFC is equal to the angle CFL ; now the G right angle FCK is equal to the right angle FCL; and therefore in the E two triangles FKC, FLC, there are two angles of one equal to two an M gles of the other, each to each, and H the side FC, which is adjacent to F the equal angles in each, is common to both; therefore the other sides B ED 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 FLC: and because KC is equal to CL, KL is double of KC : in the same manner. it may be shown 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 shown 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 shown, 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 (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 |