## 

Side $A B$ of parallelogram $A B C D$ is equal to $2, \angle A=45^{\circ}$. Points $E$ and $F$ are located on diagonal $B D$, such that

$\angle A E B=\angle C F D=90^{\circ}, B F=\frac{3}{2} B E$.

Find the area of the parallelogram.
Let $E F=x$, then $B E=2 x$. Since the right triangles $A B E$ and $C D F$ are equal, $F D=2 x$, $B D=5 x$.

The height $B K$ of the parallelogram is $\sqrt{2}$. By the Pythagorean theorem, $D K^{2}=25 x^{2}-2, D A^{2}-9 x^{2}=A E^{2}=4-4 x^{2}$, so $A D^{2}=5 x^{2}+4$. Moreover, from the similarity of triangles $E D K$ and $A D B$, it follows that $D A \cdot D K=D B \cdot D E=15 x^{2}$.

Therefore, $(5 D A-D K)^{2}=25 D A^{2}-10 D A \cdot D K+D K^{2}=100-2=98$, from which $5 D A-D K=7 \sqrt{2}$. Since $D A-D K=\sqrt{2}$, then $4 D A=6 \sqrt{2}$, and

$S_{A B C D}=A D \cdot B K=3$.

## Answer

3.