[ Pythagorean Theorem (direct and inverse). ]

In triangle $ABC$, angle $B$ is a right angle, medians $AD$ and $BE$ are perpendicular to each other. Find angle $C$.
Let $O$ be the point of intersection of the medians of triangle $ABC$. Denote $OE=a, OD=b$. Then $BO=2a, AO=2b, AB^{2}=OB^{2}+OA^{2}=4\left(a^{2}+b^{2}\right)$

$AD^{2}=DB^{2}+AB^{2}$, or $9b^{2}=b^{2}+4a^{2}+4a^{2}+4b^{2}$. Hence, $b=a\sqrt{2}$.

Therefore, $\operatorname{tg} \angle C=\operatorname{tg} \angle OBD=b / 2a=\frac{1}{\sqrt{2}}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_633abcb7b0e2bd041761g-04.jpg?height=494&width=345&top_left_y=2131&top_left_x=868)

## Answer

$\operatorname{arctg} \frac{1}{\sqrt{2}}$.