## Task B-3.7.

Given is a parallelogram $A B C D$ with an area of $75 \sqrt{3}$, where $|A B|=10$ and $\varangle A B C = 120^{\circ}$. On side $\overline{A D}$, point $E$ is given, and on side $\overline{B C}$, point $F$ is given such that segment $\overline{E F}$ is parallel to side $\overline{A B}$, and the area of parallelogram $A B F E$ is twice the area of parallelogram $C D E F$. Determine the measures of angles $\varangle F A B$ and $\varangle E F D$.
## First Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_3d5a33db4db92485fe9eg-15.jpg?height=471&width=760&top_left_y=912&top_left_x=588)

Let $\alpha = \angle FAB$, $\beta = \angle EFD$. Denote the areas of parallelograms $ABFE$ and $EFCD$ as $P_1$ and $P_2$ respectively. We have $P_1 : P_2 = 2 : 1$, which means $|AB| \cdot |AE| \sin 60^\circ = 2 |EF| \cdot |ED| \sin 60^\circ$. From $|AB| = |EF| = 10$, it follows that $|AE| = 2 |ED|$.

Furthermore, $P_{ABCD} = |AB| \cdot |AD| \sin 60^\circ$, so from $75 \sqrt{3} = |AB| \cdot |AD| \sin 60^\circ$, it follows that $|AD| = 15$.

Now, $|AD| = |AE| + |ED| = 3 |ED|$, so $|ED| = 5$ and $|AE| = 10$.

Thus, parallelogram $ABEF$ is a rhombus, and its diagonals bisect the angles. Therefore, $\alpha = \frac{60^\circ}{2} = 30^\circ$.

2 points

Now consider triangle $DEF$. We know that the measure of the angle at vertex $E$ is $60^\circ$. Drop a perpendicular $v$ from vertex $D$.

![](https://cdn.mathpix.com/cropped/2024_05_30_3d5a33db4db92485fe9eg-15.jpg?height=294&width=871&top_left_y=2080&top_left_x=524)

$\sin 60^\circ = \frac{v}{|ED|} \Rightarrow v = |ED| \sin 60^\circ = 2.5 \sqrt{3}$ and $\cos 60^\circ = \frac{x}{|ED|} \Rightarrow x = |ED| \cos 60^\circ = 2.5$

Now, $10 - x = 7.5$, so $\tan \beta = \frac{v}{10 - x} = \frac{2.5 \sqrt{3}}{7.5} \Rightarrow \beta = 30^\circ$.

2 points