In parallelogram $A B C D$, the longer side $A D$ is equal to 5. The angle bisectors of angles $A$ and $B$ intersect at point $M$.

Find the area of the parallelogram if $B M=2$, and $\cos \angle B A M=\frac{4}{5}$.
Prove that triangle $A M B$ is a right triangle.

## Solution

Since $\angle B A D+\angle A B C=180^{\circ}$, triangle $A M B$ is a right triangle. Therefore,

$$
A B=\frac{B M}{\sin \angle B A M}=2 \cdot \frac{5}{3}=\frac{10}{3}
$$

Using the formula $\sin 2 \boldsymbol{\alpha}=2 \sin \boldsymbol{\alpha} \cos \boldsymbol{\alpha}$, we find that

$$
\sin \angle B A D=3 \cdot \frac{3}{5} \cdot \frac{4}{5}=\frac{24}{25}
$$

Thus,

$$
S_{\mathrm{ABCD}}=A D \cdot A B \sin \angle B A D=5 \cdot \frac{10}{3} \cdot \frac{24}{25}=16
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_b9b3326e3456e69a696eg-29.jpg?height=271&width=654&top_left_y=1831&top_left_x=702)

Answer

16.

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