10.422 In a right-angled triangle, the bisector of the right angle cuts off segments of length $a$ and $b$ on the hypotenuse. Find the area of the square whose side is this bisector.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-254.jpg?height=372&width=931&top_left_y=1065&top_left_x=103)

Let $AD$ be the bisector of the right angle of the given $\triangle ABC$, $BD=a$, $DC=b$. Drop perpendiculars $DM$ and $DN$ to the legs $AB$ and $AC$. Then $AMDN$ is a square with side $z$. Let $BM=x$, $NC=y$.

$$
\begin{gathered}
\triangle BDM \sim \triangle DCN \Rightarrow \frac{x}{z}=\frac{a}{b}, \frac{z}{y}=\frac{a}{b} \Rightarrow \\
\Rightarrow x^{2}=\frac{a^{2}}{b^{2}} z^{2}, y^{2}=\frac{b^{2}}{a^{2}} z^{2} .
\end{gathered}
$$

By the Pythagorean theorem for $\triangle BDM$, $\triangle DCN$ we have:

$$
\begin{gathered}
\left\{\begin{array}{l}
x^{2}+z^{2}=a^{2}, \\
y^{2}+z^{2}=b^{2}
\end{array} \Rightarrow\left(x^{2}+y^{2}\right)+2 z^{2}=a^{2}+b^{2} \Rightarrow z^{2}\left(\frac{a^{2}}{b^{2}}+\frac{b^{2}}{a^{2}}+2\right)=a^{2}+b^{2} \Rightarrow\right. \\
\Rightarrow z^{2}=\frac{a^{2} b^{2}}{a^{2}+b^{2}} \Rightarrow AD^{2}=2 z^{2}=\frac{2 a^{2} b^{2}}{a^{2}+b^{2}}
\end{gathered}
$$

Answer: $\frac{2 a^{2} b^{2}}{a^{2}+b^{2}}$.