Track / Stage 5 / 90 of 400 #690 of 1964

Problem 690

AIME late

Geometry Difficulty 5.2 Find the answer

In triangle $ABC$ , the bisector $BD$ is drawn, and in triangles $ABD$ and $CBD$ - the bisectors $DE$ and $DF$ respectively. It turned out that $EF \parallel AC$ . Find the angle $DEF$ . (I. Rubanov)

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

Answer: 45 degrees. Solution: Let segments $B D$ and $E F$ intersect at point $G$ . From the condition, we have $\angle E D G = \angle E D A = \angle D E G$ , hence $G E = G D$ . Similarly, $G F = G D$ . Therefore, $G E = G F$ , which means $B G$ is the bisector and median, and thus the altitude in triangle $B E F$ . Therefore, $D G$ is the median and altitude, and thus the bisector in triangle $E D F$ , from which $\angle D E G = \angle E D G = \angle F D G = \angle G F D$ . Since the sum of the four angles in the last equality is 180 degrees, each of them is 45 degrees.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.