Dem.—Since ABC is an isosceles triangle, and the angle B is right, each of the other angles is half a right angle; therefore BAO is half a right angle. In like manner ABO is half a right angle; hence the angle BAO equal ABO; therefore [I. vi.] AO is equal to OB. In like manner OB is equal to OC, and OC to OD. Hence the circle described, with O as centre and OA as radius, will pass through the points B, C, D, and be described about the square. PROP. X.—Problem. To construct an isosceles triangle having each base angle double the vertical angle. Sol.—Take any line AB. Divide it in C, so that the rectangle AB.BC shall be equal to AC2 [II. xi.]. With A as centre, and AB as radius, describe the circle BDE, and in it place the chord BD equal to AC [i.]. Join AD. ADB is a triangle fulfilling the required conditions. Dem.—Join CD. About the triangle ACD describe the circle CDE [v.]. Then, because the rectangle AB.BC is equal to AC2 (const.), and that AC is equal to BD (const.); therefore the rectangle AB.BC is equal to BD2. Hence [III. xxxvii.] BD touches the circle ACD. Hence the angle BDC is equal to the angle A in the alternate segment [III. xxxii.]. To each add CDA, and we have the angle BDA equal to the sum of the angles CDA and A; but the exterior angle BCD of the triangle ACD is equal to the sum of the angles CDA and A. Hence the angle BDA is equal to BCD; but since AB is equal to AD, the angle BDA is equal to ABD; therefore the angle CBD is equal to BCD. Hence [I. vi.] BD is equal to CD; but BD is equal to AC (const.); therefore AC is equal to CD, and therefore [I. v.] the angle CDA is equal to A; but BDA has been proved to be equal to the sum of CDA and A. Hence BDA is double of A. Hence each of the base angles of the triangle ABD is double of the vertical angle. Exercises. 1. Prove that ACD is an isosceles triangle whose vertical angle is equal to three times each of the base angles. 2. Prove that BD is the side of a regular decagon inscribed in the circle BDE. 3. If DB, DE, EF be consecutive sides of a regular decagon inscribed in a circle, prove BF − BD = radius of circle. 4. If E be the second point of intersection of the circle ACD with BDE, DE is equal to DB; and if AE, BE, CE, DE be joined, each of the triangles ACE, ADE is congruent with ABD. 5. AC is the side of a regular pentagon inscribed in the circle ACD, and EB the side of a regular pentagon inscribed in the circle BDE. 6. Since ACE is an isosceles triangle, EB2 − EA2 = AB.BC—that is = BD2; therefore EB2 − BD2 = EA2—that is, the square of the side of a pentagon inscribed in a circle exceeds the square of the side of the decagon inscribed in the same circle by the square of the radius. PROP. XI.—Problem. To inscribe a regular pentagon in a given circle (ABCDE).