In parallelogram $A B C D$, point $E$ is the midpoint of side $A D$, and point $F$ is the foot of the perpendicular dropped from vertex $B$ to line $C E$.

Find the area of triangle $A B F$, if $A B=a, \angle B A F=\alpha$.

#
Extend $C E$ to intersect line $A B$ at point $K$ (see figure). From the equality of triangles $A K E$ and $D C E$, it follows that $A K=C D=A B$. Therefore, $F A$ is the median of the right triangle $K F B$ drawn to the hypotenuse, so $F A=a$. Thus, $S_{A B F}=1 / 2 A B \cdot A F \sin \angle B A F=1 / 2 a^{2} \sin \alpha$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ccb5665c9ecd710ea159g-37.jpg?height=597&width=775&top_left_y=-1&top_left_x=647)

## Answer

$1 / 2 a^{2} \sin \alpha$.