3. (3 points) The perimeter of parallelogram $A B C D$ is 32. The bisector of angle $C$ intersects lines $A D$ and $A B$ at points $E$ and $F$ respectively. Find the value of $B F$, given that $A E=2$.
Answer: 7 or 9

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

$\angle B F C=\angle A F E$: depending on which of the points $E$ and $F$ lies on the side of the parallelogram and which lies outside, these are either the same angle or two vertical angles. Similarly, $\angle C E D=\angle A E F$, from which it follows that triangle $A F E$ is isosceles, $A E=A F=2$.

The triangles $F B C$ and $C E D$ also turn out to be isosceles: $B F=B C$ and $C D=E D$. Therefore, the larger and smaller sides of the parallelogram differ by the length of $A E=A F=2$. Since the perimeter of the parallelogram is 32, these sides are 7 and 9.

$B F=B C$, so $B F=7$ or $B F=9$, depending on whether $B C$ is the larger or smaller side of the parallelogram, which in turn depends on which of the points $E$ and $F$ lies on the side of the parallelogram and which lies outside.