8.4. In trapezoid $A B C D$, point $M$ is the midpoint of the lateral side $C D$. Rays $B D$ and $B M$ divide angle $A B C$ into three equal parts. Diagonal $A C$ is the bisector of angle $B A D$. Find the angles of the trapezoid.
Answer: $\angle A=72^{\circ}, \angle B=108^{\circ}, \angle C=54^{\circ}, \angle D=126^{\circ}$.

Solution. Let $\angle A B D=\alpha$, then $\angle A B C=3 \alpha, \angle B A D=180^{\circ}-$ $3 \alpha, \angle B D A=\angle D B C=2 \alpha$ (see Fig. 8.4).

In triangle $B C D$, segment $B M$ is both a bisector and a median, therefore, this triangle is isosceles: $B D$ 
$=B C$. Then $\angle B C D=\angle B D C=\left(180^{\circ}-\angle D B C\right): 2=90^{\circ}-\alpha$.

Moreover, $\angle B C A=\angle D A C=\angle B A C$, therefore, $B A=B C$. Thus, triangle $B A D$ is also isosceles, so $\angle B A D=\angle B D A$, which means $180^{\circ}-3 \alpha=2 \alpha$, from which $\alpha=36^{\circ}$. Therefore, $\angle A=72^{\circ}, \angle B=108^{\circ}, \angle C=54^{\circ}, \angle D=126^{\circ}$.

Grading Criteria.

“+" A complete and well-justified solution is provided

" $\pm$ " A generally correct solution is provided, containing minor gaps or inaccuracies (for example, only angles $A$ and $B$ are correctly and justifiedly found)

"Ғ” Only the isosceles nature of triangles BCD or $A B C$ is proven

“-" Only the answer is provided

“-" An incorrect solution is provided or it is missing