4. The bisector of angle $A$ in parallelogram $A B C D$ intersects side $B C$ at $M$ and the bisector of $\angle A M C$ passes through point $D$. Find angles of the parallelogram if it is known that $\angle M D C=45^{\circ}$.
4. Answer: $\angle A=60^{\circ}$ and $\angle B=120^{\circ}$. Let $\angle D A M=\angle B A M=\alpha$. Then, since $A D \| B C$, we have also $\angle A M B=\alpha$. Then $\angle A M C=180^{\circ}-\alpha$ and hence $\angle C M D=$ $90^{\circ}-\frac{\alpha}{2}$. But then $\angle M D A=90^{\circ}-\frac{\alpha}{2}$. Then $\angle M D C=\angle C D A-\angle M D A=$ $\left(180^{\circ}-2 \alpha\right)-\left(90^{\circ}-\frac{\alpha}{2}\right)=90^{\circ}-\frac{3}{2} \alpha$. Hence $90^{\circ}-\frac{3}{2} \alpha=45^{\circ}$ and $\alpha=30^{\circ}$.