3. (3 points) The perimeter of parallelogram $A B C D$ is 20. The bisector of angle $B$ intersects lines $A D$ and $C D$ at points $K$ and $L$ respectively. Find $C L$, given that $D K=4$.
Answer: 3 or 7

Solution: $\angle A B K=\angle B L C$, as alternate interior angles. Similarly, $\angle C B L=\angle B K A$. Since points $B, K$, and $L$ lie on the bisector, all four of these angles are equal.

$\angle B K A=\angle D K L$: depending on which of the points $K$ and $L$ lies on the side of the parallelogram and which lies outside, these are either the same angle or two vertical angles. Similarly, $\angle D L K=\angle B L C$, from which the triangle $D L K$ is isosceles, $D K=D L=4$.

The triangles $B C L$ and $B A K$ also turn out to be isosceles: $L=B C$ and $A B=A K$. Therefore, the larger and smaller sides of the parallelogram differ by the length of $D K=D L=4$. Since the perimeter of the parallelogram is 20, these sides are 7 and 3.

$L=B C$, so $L=7$ or $L=3$, depending on whether $B C$ is the larger or smaller side of the parallelogram, and this in turn depends on which of the points $K$ and $L$ lies on the side of the parallelogram and which lies outside.