The bisectors of angles $B$ and $C$ of parallelogram $A B C D$ intersect at point $O$. Find the area of the parallelogram if $\angle A=2 \arcsin \frac{2}{\sqrt{13}}, O A=2 \sqrt{10}, O D=5$. (Find all solutions).
Denote $B C=x, A B=a$ and use the cosine theorem for triangles $O C D$ and $O A B$ to form two equations in terms of $a$ and $x$. From these equations, $x$ can be eliminated.

## Solution

Let

$$
B C=A D=x, A B=C D=a, \angle B A D=\angle B C D=2 \alpha .
$$

Then

$$
\begin{gathered}
\angle B C O=\angle D C O=\alpha, \angle B O C=90^{\circ}, \angle A B O=\angle C B O=90^{\circ}-\alpha \\
B O=B C \sin \alpha=\frac{2 x}{\sqrt{13}}, O C=B C \cos \alpha=3 \times \sqrt{13}
\end{gathered}
$$

By the cosine theorem in triangle $O C D$, we find that

$$
O D^{2}=O C^{2}+C D^{2}-2 O C \cdot C D \cos \alpha \text {, or } 25=\frac{9 x^{2}}{13}+a^{2}-\frac{18 a x}{13}=\frac{9\left(x^{2}-2 a x\right)}{13}+a^{2} .
$$

From triangle $O A B$ similarly, we find that

$$
40=\frac{4\left(x^{2}-2 a x\right)}{13}+a^{2}
$$

Multiply the second equation by 9 and subtract from it the first equation, multiplied by 4. We get $260=5 a^{2}$. Therefore, $a=2 \sqrt{13}$.

Substituting the obtained value of $a$ into the second equation, we get a quadratic equation in terms of $x$:

$$
x^{2}-4 x \sqrt{13}+39=0
$$

Thus, $x=3 \sqrt{13}$ or $x=\sqrt{13}$. In the first case,

$$
S_{\mathrm{ABCD}}=a x \sin 2 \alpha=2 a x \sin \alpha \cos \alpha=72
$$

In the second case, $-S_{\mathrm{ABCD}}=24$.

![](https://cdn.mathpix.com/cropped/2024_05_06_1b164f828743b46f847cg-52.jpg?height=361&width=970&top_left_y=877&top_left_x=544)

## Answer

24 or 72.