in the same manner, it may be shown that 11. therefore The angles at H, K, F, are right angles; 12. The quadrilateral figure FGHK is rectangular; and it was demonstrated to be equilateral; therefore (I. Def. 30.) 13. FGHK is a square; and it is described about the circle ABCD. therefore PROP. VIII.-PROBLEM. To inscribe a circle in a given square. Let ABCD be the given square; it is required to inscribe a circle in ABCD. Bisect (I. 10.) each of the sides AB, AD, in the points F, E; and through E draw (I. 31.) EH parallel to AB or DC; and through F draw FK parallel to AD or BC. Q.E.F. B H K C Therefore (I. 34.) 1. Each of the figures AK, KB, AH, HD, AG, GC, BG, GD, is a parallelogram, and (I. 34.) their opposite sides are equal; and because AD is equal (I. Def. 30.) to AB, and that AE is the half of AD, and AF the half of AB, (Ax. 7.) 2. AE is equal to AF; wherefore the sides opposite to these are equal, viz. 3. FG is equal to GE; in the same manner it may be demonstrated that 4, GB, GK, are each of them equal to FG or GE; 5. 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, shall pass through the extremities of the other three, and touch the straight lines AB, BC, CD, DA, because the angles at the points E, F, H, K, are (I. 29.) right angles, and that the straight line which is drawn from the extremity of a diameter, at right angles to it, touches the circle (III. 16. Cor.); therefore 6. Each of the straight lines AB, BC, CD, DA, touches the circle EFHK, which therefore is inscribed in the square ABCD. PROP. IX.-PROBLEM. 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. E Q.E.F. D B Because DA is equal (I. Def. 30.) 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 (I. 8.) 1. 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 2. The angles ABC, BCD, CDA, are severally bisected by the straight lines BD, AC; therefore, because the angle DAB is equal (I. Def. 30.) to the angle ABC, and that the angle EAB is the half of DAB, and EBA the half of ABC; (Ax. 7.) 3. The angle EAB is equal to the angle EBA; wherefore (I. 6.) 4. The side EA is equal to the side EB. In the same manner it may be demonstrated that 5. The straight lines EC, ED, are each of them equal to EA or EB; therefore 6. 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, shall pass through the extremities of the other three, and be described about the square ABCD. Q.E.F. PROP. X.-PROBLEM. To describe an isosceles triangle, having each of the angles at the base double of the third angle. Take any straight line AB, and (II. 11.) divide it in the point C, so that the rectangle AB, BC, may be equal to the square of CA; and from the centre 4, at the distance AB, describe the circle BDE, in which place (IV. 1.) the straight line BD equal to AC, which is not greater than the diameter of the circle BDE; join DA; the triangle ABD is such as is required, that is, each of the angles ABD, ADB, is double of the angle BAD. E Join DC, and about the triangle ADC describe (IV. 5.) the circle ACD. And because the rectangle AB, BC, is equal (Constr.) to the square of AC, and that AC is equal to BD, (Ax. 1.) 1. The rectangle AB, BC, is equal to the square of BD; с B D 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 BĎ which meets it; (III. 37.) 2. The straight line BD touches the circle ACD; and because BD touches the circle, and DC is drawn from the point of contact D, (III. 32.) 3. but (I. 5.) 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 (Ax. 2.) 4. The whole angle BDA is equal to the two angles CDA, DAC; but the exterior angle BCD is equal (I. 32.) to the angles CDA, DAC; therefore also (Ax. 1.) 5. BDA is equal to BCD; 6. BDA is equal to the angle CBD, because the side AD is equal to the side AB; therefore (Ax. 1.) 7. CBD, or DBA, is equal to BCD, and consequently 8. The three angles BDA, DBA, BCD, are equal to one another: and because the angle DBC is equal to the angle BCD, (I. 6.) 9. The side BD is equal to the side DC: but BD was made equal to CA; therefore also (Ax. 1.) 10. CA is equal to CD, and the angle CDA is equal (I. 5.) to the angle DAC; therefore 11. The angles CDA, DAC, together are double of the angle DAC: but BCD is equal (I. 32.) to the angles CDA, DAC; therefore also 12. BCD is double of DAC: and BCD was proved to be equal to each of the angles BDA, DBA; therefore 13. 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. Q.E.F. PROP. XI.-PROBLEM. 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 (IV. 10.) an isosceles triangle FGH, having each of the angles at G, H, double of the angle at F; and in the circle ABCDE inscribe (IV. 2.) the triangle ACD, equiangular to the triangle FGH, so that the angle CAD may 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 (I. 9.) the angles ACD, CDA, by the straight lines CE, DB; and join AB, BC, DE, EA. ABCDE is the pentagon required. G H Because each of the angles ACD, CDA, is double of CAD, and that they are bisected by the straight lines CE, DB, 1. The five angles DAC, ACE, ECD, CDB, BDA, are equal to one another; but equal angles stand (III. 26.) upon equal circumferences; therefore 2. The five circumferences AB, BC, CD, DE, EA, are equal to one another; and equal circumferences are subtended (III. 29.) by equal straight lines; therefore 3. The five straight lines AB, BC, CD, DE, EA, are equal to one another. Wherefore 4. 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, (Ax. 2.) 1. 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 (III. 27.) The angle BAE is equal to the angle AED: 2. for the same reason, 3. Each of the angles ABC, BCD, CDE, is equal to the angle BAE, or AED: therefore 4. 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. Q.E.F. PROP. XII.-PROBLEM. 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 (IV. 11.); and through the points A, B, C, D, E, draw (III. 17.) GH,`HK, KL, LM, MG, touching the circle; the figure GHKLM shall be the pentagon required. G 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 |