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[Median line of a triangle
[ Trigonometric ratios in a right-angled triangle ]

In triangle $ABC$, the median $BM$ and the altitude $AH$ are drawn. It is known that $BM = AH$. Find the angle $MBC$.

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Drop a perpendicular from point $M$ to line $B C$.

## Solution

Drop the perpendicular $M K$ to line $B C$. Then $M K$ is the midline of triangle $A H C$. Therefore, $M K = 1/2 A H = 1/2 B M$, which means $\angle M B K = 30^{\circ}$. Consequently, $\angle M B C = 30^{\circ}$ or $150^{\circ}$.
![](https://cdn.mathpix.com/cropped/2024_05_06_96d74c2697cd905bd948g-01.jpg?height=272&width=962&top_left_y=2402&top_left_x=552)

## Answer

Inside the triangle, there are two points. The distances from one of them to the sides of the triangle are 1, 3, and 15, and from the other (in the same order) are 4, 5, and 11.

Find the radius of the inscribed circle of the given triangle.

## Hint

Let $M_{1}$ and $M_{2}$ be the given points. On the line $M_{1} M_{2}$, take a point $O$ such that $O M_{2} = M_{2} M_{1}$. Then $O$ is the center of the inscribed circle of the given triangle.

## Solution

Let $M_{1}$ and $M_{2}$ be the given points; $O$ be the point for which $M_{2}$ is the midpoint of $O M_{1}$. By the property of the midline of a trapezoid, the distances from point $O$ to the sides of the triangle are respectively $2 \cdot 4 - 1 = 7, 2 \cdot 5 - 3 = 7, 2 \cdot 11 - 15 = 7$.

Since the segment $O M_{1}$ cannot intersect any of the sides of the triangle, $O$ is the center of the inscribed circle, and its radius is 7.

![](https://cdn.mathpix.com/cropped/2024_05_06_96d74c2697cd905bd948g-02.jpg?height=829&width=491&top_left_y=886&top_left_x=792)

## Answer

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