8.3. Given a parallelogram $A B C D, \angle D=100^{\circ}, B C=12$. On side $A D$ there is a point $L$ such that $\angle A B L=50^{\circ}, L D=4$. Find the length of $C D$.
Answer: 8.

Solution. By the property of a parallelogram, $\angle A B C=\angle D=100^{\circ}, A D=B C=12$ and $C D=A B$. Therefore, $\angle C B L=\angle A B C-\angle A B L=100^{\circ}-50^{\circ}=50^{\circ}$ and $A L=A D-L D=12-4=8$. Since $\angle A L B=\angle C B L$ (as alternate interior angles when $A D$ and $B C$ are parallel and $B L$ is the transversal) and $\angle C B L=\angle A B L=50^{\circ}$, we get that $\angle A L B=\angle A B L$, so triangle $A B L$ is isosceles and $A B=A L=8$. Therefore, $C D=A B=8$.