# Problem 5. (3 points)

In trapezoid $ABCD$ with base $AD$, the diagonals are the angle bisectors of $\angle B$ and $\angle C=110^{\circ}$. Find the degree measure of $\angle BAC$.

#
# Solution:

$\angle B D A=\angle C B D$ as alternate interior angles, and $\angle C B D=\angle A B D$, since $B D$ is the bisector. Therefore, triangle $B D A$ is isosceles, that is, $A B=A D$. Similarly, $C D=A D$, so our trapezoid is isosceles (equilateral) and $\angle B=\angle C=110^{\circ}$.

$$
\angle B A C=180^{\circ}-\angle B-\angle B C A=180^{\circ}-110^{\circ}-55^{\circ}=15^{\circ} .
$$

Answer: 15