4. In triangle $A B C$, points $M$ and $N$ are the midpoints of sides $A C$ and $B C$ respectively. It is known that the point of intersection of the medians of triangle $A M N$ is the point of intersection of the altitudes of triangle $A B C$. Find the angle $A B C$.
Answer: $45^{\circ}$.

Solution. Triangles $E T N$ and $A T B$ are similar, therefore, $T N: T B=T E: T A=E N: A B=1: 4$. Therefore, $C T=1 / 2 B T$. Since $H$ is the intersection point of the medians of triangle $A M N$, $E H=1 / 3 A E=E T$. Therefore, $H T=1 / 2 A T$.

Thus, right triangles $C T H$ and $B T A$ are similar, and

$\square T C H=\square T B A$. But $C H$ is part of the height $C Q$ of triangle $A B C$, so these equal angles are acute angles

![](https://cdn.mathpix.com/cropped/2024_05_06_f77d0efe3862c4036ebag-1.jpg?height=403&width=485&top_left_y=1880&top_left_x=1431)
of the right triangle $C Q B$.