8,9 In triangle $A B C$ angle $A$ is equal to $\arccos \frac{7}{8}, B C=a$, and the height dropped from vertex $A$ is equal to the sum of the other two heights. Find the area of triangle $A B C$.
Let $A C=x, A B=y$, form a system of equations and express $xy$ from it.

## Solution

Since $\cos \angle A=\frac{7}{8}$, then

$$
\sin \angle A=\sqrt{1-\left(\frac{7}{8}\right)^{2}}=\frac{\sqrt{15}}{8}
$$

Let $A M, C D$ and $B E$ be the altitudes of triangle $A B C$. Denote $A C=x, A B=y$. By the cosine rule in triangle $A B C$, we find that

$$
B C^{2}=A C^{2}+A B^{2}-2 A C \cdot A B \cos \angle A, \text { or } a^{2}=x^{2}+y^{2}-\frac{7 x y}{4} \text {. }
$$

Moreover,

$$
S_{\triangle \mathrm{ABC}}=\frac{1}{2} a \cdot A M=\frac{1}{2} x \cdot B E=\frac{1}{2} y \cdot C D, C D+B E=A M
$$

From these equalities, it follows that

$$
\frac{2 S_{\triangle A B C}}{x}+\frac{2 S_{\triangle A B C}}{y}=\frac{2 S_{\triangle A B C}}{a}
$$

or

$$
\frac{1}{x}+\frac{1}{y}=\frac{1}{a} \Leftrightarrow a(x+y)=x y
$$

From the system of equations

$$
\left\{\begin{array}{l}
x^{2}+y^{2}-\frac{7 x y}{4}=a^{2} \\
a(x+y)=x y
\end{array}\right.
$$

we find that $x y=4 a^{2}$. Therefore,

$$
S_{\triangle \mathrm{ABC}}=\frac{1}{2} x y \sin \angle A=\frac{a^{2} \sqrt{15}}{4}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_f7f9d90261bfc416bd71g-26.jpg?height=343&width=780&top_left_y=480&top_left_x=639)

Answer

$\frac{a^{2} \sqrt{15}}{4}$.