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.