VIII-3. Given a right triangle ABC ( $\angle \mathrm{C}=90^{\circ}$ ), whose median to the hypotenuse is equal to $20 \mathrm{~cm}$. From the midpoint $D$ of the hypotenuse, a perpendicular to the hypotenuse is drawn, intersecting one of the legs at point $E$, and $\overline{D E}=15 \mathrm{~cm}$. Calculate the legs of the triangle.
Solution: If $\overline{C D}=20 \mathrm{~cm}$, then $\overline{A B}=40 \mathrm{~cm}$.

From $\triangle A D E: \overline{\mathrm{AE}}^{2}=\overline{\mathrm{DE}}^{2}+\overline{\mathrm{AD}}^{2}$,

$\overline{A E}=\sqrt{15^{2}+20^{2}}=25 \mathrm{~cm}, \triangle A B C \sim \triangle A D E, \alpha$ is a common angle, so $\overline{\mathrm{AB}}: \overline{\mathrm{AE}}=\overline{\mathrm{AC}}: \overline{\mathrm{AD}} ; 40: 25=\overline{\mathrm{AC}}: 20$; $\overline{A C}=32 \mathrm{~cm}$, and $\overline{B C}=24 \mathrm{~cm}$.

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