8.6. Points $M$ and $N$ are the midpoints of sides $B C$ and $A D$ of quadrilateral $A B C D$. It is known that $\angle B=$ $150^{\circ}, \angle C=90^{\circ}$ and $A B=C D$. Find the angle between the lines $M N$ and $B C$.
Answer: $60^{\circ}$.

Solution. First method. Construct parallelogram $A B M K$ and rectangle $C D L M$ (see Fig. 8.6a). Since $A K\|B C\| L D$ and $A K=B M=$ $M C=L D$, then $A K D L$ is also a parallelogram. Therefore, the midpoint $N$ of its diagonal $A D$ is also the midpoint of diagonal $K L$.

![](https://cdn.mathpix.com/cropped/2024_05_06_63799b39c033d6fb87d1g-3.jpg?height=303&width=765&top_left_y=951&top_left_x=1134)

In triangle $K M L: K M=A B=C D=M L$ and $\angle K M L=\angle K M C-\angle L M C=150^{\circ}-90^{\circ}=60^{\circ}$. Therefore, this triangle is equilateral. Hence, its median $M N$ is also its bisector, so $\angle L M N=30^{\circ}$, which means $\angle B M N=60^{\circ}$.

Second method. Draw perpendiculars $A Q$ and $N P$ to line $B C$ (see Fig. 8.6b). Let $A Q=a, C M=B M=b$. Since $\angle A B Q$ $=30^{\circ}$, then $A B=C D=2 a, B Q=\frac{a \sqrt{3}}{2}$.

By Thales' theorem, $Q P=P C$, so $N P$ is the midline of trapezoid $A D C Q$. Then $N P=\frac{3}{2} a$.

$P M=C P-C M=\frac{1}{2} C Q-b=\frac{(a \sqrt{3}+2 b)}{2}-b=\frac{a \sqrt{3}}{2}$. From triangle $NPM$ by the Pythagorean theorem, we get that $M N=a \sqrt{3}$, then $\angle P N M=30^{\circ}$, and $\angle P M N=60^{\circ}$.

Grading criteria.

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

" $\pm$ " A generally correct solution is provided, containing minor gaps or inaccuracies, for example, in the first method, correct additional constructions are made and the correct answer is obtained using the fact that $AKDL$ is a parallelogram, but this is not separately proven

“Ғ” The idea of additional construction is present, but there is no further progress

“-" Only the answer is provided

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