4.4. In parallelogram $A B C D$, the angle at vertex $A$ is $60^{\circ}$, $A B=73$, and $B C=88$. The bisector of angle $A B C$ intersects segment $A D$ at point $E$, and ray $C D$ at point $F$. Find the length of segment $E F$.

Answer, option 1. 9.

Answer, option 2. 13.

Answer, option 3. 12.

Answer, option 4. 15.

![](https://cdn.mathpix.com/cropped/2024_05_06_8c9b9114972c62456a9cg-04.jpg?height=448&width=556&top_left_y=895&top_left_x=750)
Solution 1. Lines $A D$ and $B C$ are parallel, therefore

$$
\angle A B C=180^{\circ}-\angle B A D=120^{\circ}
$$

(angles $A B C$ and $B A D$ are consecutive interior angles). Then $\angle A B E=60^{\circ}$ and $\angle C B E=60^{\circ}$, since $B E$ is the bisector of angle $A B C$. In triangle $A B E$, the angles at vertices $A$ and $B$ are both $60^{\circ}$. Therefore, the angle at vertex $E$ is also $60^{\circ}$, making this triangle equilateral. Thus, $B E=A B=78$. Also, in triangle $B C F$, the angles at vertices $B$ and $C$ are both $60^{\circ}$. Therefore, this triangle is also equilateral, so $B F=B C=87$. Thus, $E F=B F-B E=87-78=9$.