In triangle $A B C$, a point $P$ is taken on side $B C$, and a point $M$ is taken on side $A C$. Segments $A P$ and $B M$ intersect at point $O$. It is known that triangles $B O P, A O M$, and $B O A$ are similar, $B M=1, \cos \angle B=0.6$. Find the area of triangle $A B C$.

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Prove that triangle $ABC$ is a right triangle.

## Solution

From the similarity of triangles $BOP$, $AOM$, and $BOA$, it follows that the angles $\angle BOP$ and $\angle BOA$ are equal. Since these are adjacent angles, $\angle BOP = \angle BOA = 90^\circ$.

Moreover, $\angle PBO = \angle ABO$, which means $BM$ is the bisector of angle $B$. From the equality of angles $\angle MAO$ and $\angle ABO$, it follows that $\angle A = 90^\circ$.

Let $\angle B = 2\beta$. Then $\angle ABM = \beta$, $AB = BM \cos \beta = \cos \beta$, $AC = AB \tan 2\beta = \cos \beta \tan 2\beta$, $S_{ABC} = \frac{1}{2} AB \cdot AC = \frac{1}{2} \cos^2 \beta \tan 2\beta = \frac{1}{4} (1 + \cos 2\beta) \tan 2\beta = \frac{1}{4} \cdot \frac{8}{5} \cdot \frac{4}{3} = \frac{8}{15}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_36c4956010bd91c4b4eeg-37.jpg?height=397&width=720&top_left_y=133&top_left_x=675)

## Answer

$\frac{8}{15}$.

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