2. 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)
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.