The angle bisectors of angles $A$ and $B$ of trapezoid $A B C D$ ($B C \| A D$) intersect at point $O$. Find the sides $A B$ and $B C$, if $\angle A$

$=2 \arccos \sqrt{\frac{5}{6}}, O C=\sqrt{7}, O D=3 \sqrt{15}, A D=5 B C$
Let $A B=x, B C=a$ and use the cosine theorem for triangles $A O D$ and $B O C$ to form two equations in terms of $a$ and $x$. From these equations, $a$ can be eliminated.

## Solution

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

$$
\begin{gathered}
A D=5 a, \angle O A B=\angle O A D=\alpha, \angle A O B=90^{\circ} \\
\angle O B C=\angle O B A=90^{\circ}-\alpha, A O=A B \cos \alpha=\frac{x \sqrt{5}}{\sqrt{6}} \\
O B=A B \sin \alpha=\frac{x}{\sqrt{6}}
\end{gathered}
$$

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

$$
O D^{2}=A O^{2}+A D^{2}-2 A O \cdot A D \cos \alpha, \text { or } 135=\frac{5 x^{2}}{6}+25 a^{2}-\frac{25 a x}{3} .
$$

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

$$
7=\frac{x^{2}}{6}+a^{2}-\frac{a x}{3}
$$

Multiply the second equation by 25 and subtract the first from it. We get $40=\frac{10 x^{2}}{3}$. Therefore, $x^{2}=12$ and $A B=x=2 \sqrt{3}$. From the second equation, we find that

$$
a^{2}-\frac{2 a \sqrt{3}}{3}-5=0
$$

From this, it follows that $B C=a=\frac{5 \sqrt{3}}{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_e3b1ddf947c7dd2094eeg-10.jpg?height=334&width=880&top_left_y=-1&top_left_x=586)

## Answer

$2 \sqrt{3}, \frac{5 \sqrt{3}}{3}$