3. In triangle $ABC$, the bisector $BD$ was drawn, and in triangles $ABD$ and $CBD$ - the bisectors $DE$ and $DF$ respectively. It turned out that $EF \| AC$. Find the angle $DEF$.


Answer: $45^{\circ}$.

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 also the altitude in triangle $B E F$. From this, $D G$ is the median and altitude, and thus also 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 mentioned in the last equality is $180^{\circ}$ degrees, each of them is $45^{\circ}$ degrees.

Comment. A complete and justified solution - 7 points. A generally correct reasoning with minor gaps or inaccuracies - up to 5 points.