In triangle ABC, let AD be the angle bisector of the exterior angle at A, where D is the intersection of the exterior angle bisector with side BC (extended - so C - B - D).
Prove that BD/DC = AB/AC. Hint: Draw a line through C that is parallel to AB to create similar triangles. Let BE be the interior angle bisector of angle B, where E is on AC, and CF be the exterior angle bisector of angle C, where F is on AB (extended - so B - A - F). Use what you proved in Theorem 2.4.15 (similar result for interior angle bisectors) to show that AF/FB * BD/DC * CE/EA = 1. Sketch and figure accurate enough to recognize that the lines AD, BE, CF are concurrent.