$\left.\begin{array}{ll}\text { [Mean proportionals in a right triangle] } \\ {[\quad \text { Pythagorean Theorem (direct and inverse). }}\end{array}\right]$

In a right triangle $ABC$, a height $CD$ is dropped from the vertex of the right angle $C$. The projection of segment $BD$ onto the leg $BC$ is $l$, and the projection of segment $AD$ onto the leg $AC$ is $m$. Find the hypotenuse $AB$.
Let $E$ be the projection of point $D$ onto the leg $A C$, and $F$ be the projection of point $D$ onto the leg $B C$. Denote $C F=a, C E=b, \angle C B A = \alpha$. Since $C E D F$ is a rectangle, $D E = C F = a, D F = C E = b$. Then

$$
\begin{aligned}
& a 2 = b m, \quad b 2 = a l, \quad \frac{a^{2}}{b^{2}} = \frac{b}{a} \cdot \frac{m}{l}, \quad \left(\frac{a}{b}\right)^3 = \frac{m}{l}, \quad \operatorname{tg} \alpha = \frac{a}{b} = \left(\frac{m}{l}\right)^{\frac{1}{3}} \\
& \cos \alpha = \frac{1}{\sqrt{1 + l^{2} g^{2}}} = \frac{1}{\sqrt{1 + \left(\frac{m}{T}\right)^{2}}} = \frac{1 l^{\frac{1}{3}}}{\sqrt{l^{2} + m^{\frac{2}{3}}}} \\
& \sin \alpha = \operatorname{tg} \alpha \cos \alpha = \frac{m^{\frac{1}{3}}}{\frac{1}{t^{3}}} \cdot \frac{t^{\frac{1}{3}}}{\sqrt{l^{2} + m^{2}}} = \frac{m^{\frac{1}{3}}}{\sqrt{t^{\frac{2}{3}} + m^{2}}}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_d69082500b56170eb8ddg-24.jpg?height=103&width=920&top_left_y=1432&top_left_x=569)

Therefore,

$$
A B = B D + A D = l^{\frac{2}{3}} \sqrt{l^{\frac{2}{3}} + m^{\frac{2}{3}}} + m^{\frac{2}{3}} \sqrt{l^{\frac{2}{3}} + m^{\frac{2}{3}}} = \left(l^{\frac{2}{3}} + m^{\frac{2}{3}}\right) \sqrt{l^{\frac{2}{3}} + m^{\frac{2}{3}}}
$$

## Answer

$\left(l^{\frac{2}{3}} + m^{\frac{2}{3}}\right) \sqrt{l^{\frac{2}{3}} + m^{\frac{2}{3}}}$