Again, join OF, OG. Now the triangles EOF, AOF have the sides AF, FE equal [III. xvii., Ex. 1], and FO common, and the base AO equal to the base EO. Hence the angle AFO is equal to EFO [I. viii.]. Therefore the angle AFO is half the angle AFE. In like manner AGO is half the angle AGB; but AFE has been proved equal to AGB; hence AFO is equal to AGO, and FAO is equal to GAO, each being right, and AO common to the two triangles FAO, GAO; hence [I. xxvi.] the side AF is equal to AG; therefore GF is double AF. In like manner JF is double EF; but AF is equal to EF; hence GF is equal to JF. In like manner the remaining sides are equal; therefore the figure FGHIJ is equilateral, and it has been proved equiangular. Hence it is a regular pentagon. This Proposition is a particular case of the following general theorem, of which the proof is the same as the foregoing:— “If tangents be drawn to a circle, at the angular points of an inscribed polygon of any number of sides, they will form a regular polygon of the same number of sides circumscribed to the circle.” PROP. XIII.—Problem. To inscribe a circle in a regular pentagon (ABCDE). Sol.—Bisect two adjacent angles A, B by the lines AO, BO; then O, the point of intersection of the bisectors, is the centre of the required circle. Dem.—Join CO, and let fall perpendiculars from O on the five sides of the pentagon. Now the triangles ABO, CBO have the side AB equal to BC (hyp.), and BO common, and the angle ABO equal to CBO (const.). Hence the angle BAO is equal to BCO [I. iv.]; but BAO is half BAE (const.). Therefore BCO is half BCD, and therefore CO bisects the angle BCD. In like manner it may be proved that DO bisects the angle D, and EO the angle E. Again, the triangles BOF, BOG have the angle F equal to G, each being right; and OBF equal to OBG, because OB bisects the angle ABC (const.), and OB common; hence [I. xxvi.] OF is equal to OG. In like manner all the perpendiculars from O on the sides of the pentagon are equal; hence the circle whose centre is O, and radius OF, will touch all the sides of the pentagon, and will therefore be inscribed in it. In the same manner a circle may be inscribed in any regular polygon. PROP. XIV.—Problem. To describe a circle about a regular pentagon (ABCDE). Sol.—Bisect two adjacent angles A, B by the lines AO, BO. Then O, the point of intersection of the bisectors, is the centre of the required circle.