The median $A M$ of triangle $A B C$ is equal to half the side $B C$. The angle between $A M$ and the altitude $A H$ is $40^{\circ}$. Find the angles of triangle $A B C$.
Since the median of a triangle is equal to half the side to which it is drawn, the triangle is right-angled.

## Solution

Since the median of a triangle is equal to half the side to which it is drawn, the triangle is right-angled $\left(\angle A=90^{\circ}\right)$.

Let point $H$ lie on segment $B M$. Then $\angle A M B=90^{\circ}-40^{\circ}=50^{\circ}$ is the exterior angle of the isosceles triangle $A M C$, so $\angle C=1 / 2 \angle A M B=25^{\circ}$.

Therefore, $\angle B=90^{\circ}-25^{\circ}=65^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_4c124ad32a76c9faf91dg-19.jpg?height=300&width=666&top_left_y=656&top_left_x=702)

## Answer

$90^{\circ}, 25^{\circ}, 65^{\circ}$.

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