Dem.—Because A is the centre of the circle EBF, AB is equal to AE [I. Def. xxxii.]; but AE is equal to D (const.); therefore AB is equal to D. PROP. II.—Problem. In a given circle (ABC) to inscribe a triangle equiangular to a given triangle (DEF). Sol.—Take any point A in the circumference, and at it draw the tangent GH; then make the angle HAC equal to E, and GAB equal to F [I. xxiii.] Join BC. ABC is a triangle fulfilling the required conditions. Dem.—The angle E is equal to HAC (const.), and HAC is equal to the angle ABC in the alternate segment [III. xxxii.]. Hence the angle E is equal to ABC. In like manner the angle F is equal to ACB. Therefore [I. xxxii.] the remaining angle D is equal to BAC. Hence the triangle ABC inscribed in the given circle is equiangular to DEF. PROP. III.—Problem. About a given circle (ABC) to describe a triangle equiangular to a given triangle (DEF). Sol.—Produce any side DE of the given triangle both ways to G and H, and from the centre O of the circle draw any radius OA; make the angle AOB equal to GEF [I. xxiii.], and the angle AOC equal to HDF. At the points A, B, C draw the tangents LM, MN, NL to the given circle. LMN is a triangle fulfilling the required conditions. Dem.—Because AM touches the circle at A, the angle OAM is right. In like manner, the angle MBO is right; but the sum of the four angles of the quadrilateral OAMB is equal to four right angles. Therefore the sum of the two remaining angles AOB, AMB is two right angles; and [I. xiii.] the sum of the two angles GEF, FED is two right angles. Therefore the sum of AOB, AMB is equal to the sum of GEF, FED; but AOB is equal to GEF (const.). Hence AMB is equal to FED. In like manner, ALC is equal to EDF; therefore [I. xxxii.] the remaining angle BNC is equal to DFE. Hence the triangle LMN is equiangular to DEF. PROP. IV.—Problem. To inscribe a circle in a given triangle (ABC). Sol.—Bisect any two angles A, B of the given triangle by the lines AO, BO; then O, their point of intersection, is the centre of the required circle. Dem.—From O let fall the perpendiculars OD, OE, OF on the sides of the triangle. Now, in the triangles OAE, OAF the angle OAE is equal to OAF (const.), and the angle AEO equal to AFO, because each is right, and the side OA common. Hence [I. xxvi.] the side OE is equal to OF. In like manner OD is equal to OF; therefore the three lines OD, OE, OF are all equal. And the circle described with O as centre and OD as radius will pass through the points E, F; and since the angles D, E, F are right, it will [III. xvi.] touch the three sides of the triangle ABC; and therefore the circle DEF is inscribed in the triangle ABC. Exercises. 1. If the points O, C be joined, the angle C is bisected. Hence “the bisectors of the angles of a triangle are concurrent” (compare I. xxvi., Ex. 7).