${ }^{[}$The Law of Cosines Area of a triangle (through two sides and the angle between them). [Theorem on the lengths of a tangent and a secant; the product of the entire secant and its external part

In triangle $A B C$, medians $A M$ and $B P$ are drawn. It is known that $\angle A P B=\angle B M A, \cos \angle A C B=0.8, B P=1$.

Find the area of triangle $A B C$.

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Points $A, B, M$ and $P$ lie on the same circle, so $C M \cdot C B = C P \cdot C A$, or $2 C M^2 = 2 C P^2$. Therefore, $C M = C P, C A = C B$, which means triangle $A B C$ is isosceles.

Let $M C = x$. By the cosine rule, $1 = B P^2 = x^2 + 4 x^2 - 2 x \cdot 2 x \cdot 4 / 5$, so $x^2 = 5 / 9$. Thus, $S_{A B C} = 1 / 2 A C \cdot B C \sin \angle C = 1 / 2 (2 x)^2 \cdot 3 / 5 = 6 / 5 x^2 = 2 / 3$.

![](https://cdn.mathpix.com/cropped/2024_05_06_14311972f49418f31424g-19.jpg?height=529&width=352&top_left_y=1891&top_left_x=859)

## Answer

$2 / 3$.