Bisect AB, AC in the points D, E, (1. 10.) and from these points draw DF, EF at right angles to AB, AC; (1.11.) DF, EF produced meet one another: for, if they do not meet, they are parallel, wherefore AB, AC, which are at right angles to them, are parallel; which is absurd: let them meet in F, and join FA; also, if the point F be not in BC, join BF, CF. Then, because AD is equal to DB, and DF common, and at right angles to AB, therefore the base AF is equal to the base FB. (1. 4.) and FA, FB, FC are equal to one another: wherefore the circle described from the center F, at the distance of one of them, will pass through the extremities of the other two, and be described about the triangle ABC. Q.E.F. COR. And it is manifest, that when the center of the circle falls within the triangle, each of its angles is less than a right angle, (III. 31.) each of them being in a segment greater than a semicircle; but, when the center is in one of the sides of the triangle, the angle opposite to this side, being in a semicircle, (III. 31.) is a right angle; and, if the center falls without the triangle, the angle opposite to the side beyond which it is, being in a segment less than a semicircle, (III. 31.) is greater than a right angle: therefore, conversely, if the given triangle be acute-angled, the center of the circle fails within it; if it be a rightangled triangle, the center is in the side opposite to the right angle; and if it be an obtuse-angled triangle, the center falls without the triangle, beyond the side opposite to the obtuse angle. PROPOSITION VI. PROBLEM. To inscribe a square in a given circle, Let ABCD be the given circle. It is required to inscribe a square in ABCD. A B C Draw the diameters, AC, BD, at right angles to one another, (III. 1. and I. 11.) and join AB, BC, CD, DA. The figure ABCD shall be the square required. Because BE is equal to ED, for E is the center, and that EA is common, and at right angles to BD; the base BA is equal to the base AD: (1. 4.) and, for the same reason, BC, CD are each of them equal to BA, or AD; therefore the quadrilateral figure ABCD is equilateral. It is also rectangular; for the straight line BD being the diameter of the circle ABCD, BAD is a semicircle; wherefore the angle BAD is a right angle: (III. 31.) for the same reason, each of the angles ABC, BCD, CDA is a right angle: therefore the quadrilateral figure ABCD is rectangular: and it is inscribed in the circle ABCD. Q.E.F. Draw two diameters AC, BD of the circle ABCD, at right angles to one another, and through the points A, B, C, D, draw FG, GH, HK, KF touching the circle. (III. 17.) The figure GHKF shall be the square required. Because FG touches the circle ABCD, and EA is drawn from the center E to the point of contact A, therefore the angles at A are right angles: (III. 18.) for the same reason, the angles at the points B, C, D are right angles; and because the angle AEB is a right angle, as likewise is EBG, therefore GH is parallel to AC: (1. 28.) for the same reason AC is parallel to FK: and in like manner GF, HK may each of them be demonstrated to be parallel to BED: therefore the figures GK, GC, AK, FB, BK are parallelograms; and therefore GF is equal to HK, and GH to FK: (1.34.) and because AC is equal to BD, and that AC is equal to each of the two GH, FK; and BD to each of the two GF, HK: GH, FK are each of them equal to GF, or HK; for GBEA being a parallelogram, and AEB a right angle, and in the same manner it may be shewn that the angles at H, K, F, are right angles : therefore the quadrilateral figure FGHK is rectangular : and it is described about the circle ABCD. Q.E.F. Bisect each of the sides AB, AD in the points F, E, (I. 10.) and through E draw EH parallel to AB or DC, (1. 31.) and through F draw FK parallel to AD or BC: therefore each of the figures AK, KB, AH, HD, AG, GC, BG, GD is a right-angled parallelogram; and their opposite sides are equal: (1. 34.) and because AD is equal to AB, (1. def. 30.) and that AE is the half of AD, and AF the half of AB, wherefore the sides opposite to these are equal, viz. FG to GE: therefore the four straight lines GE, GF, GH, GK are equal to one another; and the circle described from the center G at the distance of one of them, will 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 right angles, (1. 29.) 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 each of the straight lines AB, BC, CD, DA touches the circle, which therefore is inscribed in the square ABCD. Q.E.F. PROPOSITION IX. PROBLEM. It is required to describe a circle about ABCD. D B Join AC, BD, cutting one another in E: and because DA is equal to AB, and AC common to the triangles DAC, BAC, (1. def. 30.) 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; (1. 8.) 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 equal to the angle ABC, (1. def. 30.) and that the angle EAB is the half of DAB, and EBA the half of ABC; therefore the angle EAB is equal to the angle EBA; (ax. 7.) wherefore the side EA is equal to the side EB: (1. 6.) 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 center E, at the distance of one of them, will pass through the extremities of the other three, and be described about the square ABCD. Q.E.F. PROPOSITION 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 divide it in the point C, (II. 11.) so that the rectangle AB, BC may be equal to the square on CA; and from the center 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; (IV. 1.) and join DA. Then 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. (IV. 5.) And because the rectangle AB, BC is equal to the square on AC, and that AC is equal to BD, (constr.) the rectangle AB, BC is equal to the square on BD: (ax. 1.) 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 on BD which meets it; therefore the straight line BD touches the circle ACD: (111. 37.) and because 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: (III. 32.) to each of these add the angle CDA; therefore the whole angle BDA is equal to the two angles CDA, DAC: (ax. 2.) but the exterior angle BCD is equal to the angles CDA, DAC; (1. 32.) therefore also BDA is equal to BCD: (ax. 1.) but BDA is equal to the angle CBD, (1. 5.) because the side AD is equal to the side AB; therefore CBD, or DBA, is equal to BCD; (ax. 1.) and consequently the three angles BDA, DBA, BCD are equal to one another: and because the angle DBC is equal to the angle BCD, therefore also CA is equal to CD, (ax. 1.) and the angle CDA equal to the angle DAC; (1. 5.) therefore the angles CDA, DAC together, are double of the angle DAC: but BCD is equal to the angles CDA, DAC; (1. 32.) therefore also 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 has been described, having each of the angles at the base double of the third angle. Q.E.F. PROPOSITION XI. PROBLEM. To inscribe an equilateral and equiangular pentagon in a 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; (IV. 10.) and in the circle ABCDE inscribe the triangle ACD equiangular to the triangle FGH, (Iv. 2.) 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; |