Dem.—Join OC, OD, OE. Then the triangles ABO, CBO have the side AB equal to BC (hyp.), BO common, and the angle ABO equal to CBO (const.). Hence the angle BAO is equal to BCO [I. iv.]; but the angle BAE is equal to BCD (hyp.); and since BAO is half BAE (const.), BCO is half BCD. Hence CO bisects the angle BCD. In like manner it may be proved that DO bisects CDE, and EO the angle DEA. Again, because the angle EAB is equal to ABC, their halves are equal. Hence OAB is equal to OBA; therefore [I. vi.] OA is equal to OB. In like manner the lines OC, OD, OE are equal to one another and to OA. Therefore the circle described with O as centre, and OA as radius, will pass through the points B, C, D, E, and be described about the pentagon. In the same manner a circle may be described about any regular polygon. Propositions xiii., xiv. are particular cases of the following theorem:— “A regular polygon of any number of sides has one circle inscribed in it, and another described about it, and both circles are concentric.” PROP. XV.—Problem. In a given circle (ABCDEF) to inscribe a regular hexagon. Sol.—Take any point A in the circumference, and join it to O, the centre of the given circle; then with A as centre, and AO as radius, describe the circle OBF, intersecting the given circle in the points B, F. Join OB, OF, and produce AO, BO, FO to meet the given circle again in the points D, E, C. Join AB, BC, CD, DE, EF, FA; ABCDEF is the required hexagon. Dem.—Each of the triangles AOB, AOF is equilateral (see Dem., I. i.). Hence the angles AOB, AOF are each one-third of two right angles; therefore EOF is one-third of two right angles. Again, the angles BOC, COD, DOE are [I. xv.] respectively equal to the angles EOF, FOA, AOB. Therefore the six angles at the centre are equal, because each is one-third of two right angles. Therefore the six chords are equal [III. xxix.]. Hence the hexagon is equilateral. Again, since the arc AF is equal to ED, to each add the arc ABCD; then the whole arc FABCD is equal to ABCDE; therefore the angles DEF, EFA which stand on these arcs are equal [III. xxvii.]. In the same manner it may be shown that the other angles of the hexagon are equal. Hence it is equiangular, and is therefore a regular hexagon inscribed in the circle. Cor. 1.—The side of a regular hexagon inscribed in a circle is equal to the radius. Cor. 2.—If three alternate angles of a hexagon be joined, they form an inscribed equilateral triangle. Exercises. 1. The area of a regular hexagon inscribed in a circle is equal to twice the area of an equilateral triangle inscribed in the circle; and the square of the side of the triangle is three times the square of the side of the hexagon. 2. If the diameter of a circle be produced to C until the produced part is equal to the radius, the two tangents from C and their chord of contact form an equilateral triangle.