Plane Geometry - Bruce E. Shapiro

216 SECTION 40. CIRCLES AND TRIANGLESChoose point E ∈ −−→ MN such that ME = AD = BC.Then by construction □AMED and □BMEC are Saccheri quadrilaterals.Let l be the line through the midpoints of AM and DE.Let m be the line through the midpoints of BM and CE.Then by theorem 31.17 l ⊥ DE and m ⊥ CE.By the alternate interior angles theorem (theorem 29.4), l ‖ m.Hence l and m do not intersect. By theorem 40.2 (the circumscribed triangletheorem), triangle △CDE does not have a circumscribed circle.Definition 40.7 A circle is said to be inscribed in a triangle if each of theedges of the triangle is tangent to the circle. The center of the inscribedcircle is called the incenter of the triangle.Theorem 40.8 (Inscribed Circle Theorem a ) Every triangle has aunique inscribed circle. The bisectors fo the interior angles of any triangleare concurrent and the point of concurrency is the incenter of the triangle.a Euclid Book 4 Proposition 4.Proof. (Proof of existence.) By the crossbar theorem, the angle bisector of∠BAC intersects side BC at some point D (figure fig 40.3).By the crossbar theorem, the angle bisector of ∠ABD must intersect segmentAD at some point E.Since A ∗ E ∗ D, E is in the interior of angle ∠ACD.Let F , G, and H be the feet of the perpendiculars dropped from E to eachof the three sides of the triangle (bottom of figure 40.3).Since the interior angle of any triangle is less than 180, we have∠EAB = 1 180∠CAB