8,9 |  |

In a right triangle $A B C$, the hypotenuse $A B$ is equal to $c$ and $\angle B=\alpha$. Find all medians of this triangle.
As is known, the median of the right triangle is equal to $c / 2$.

Let $M$ be the midpoint of $B C$. Then $C M=1 / 2 B C=1 / 2 c \sin \alpha$.

By the Pythagorean theorem, $A M^{2}=A C^{2}+C M^{2}=1 / 4\left(4 \cos ^{2} \alpha+\sin ^{2} \alpha\right)=1 / 4\left(1+3 \cos ^{2} \alpha\right)$.

Similarly, we find the third median.

## Answer

\section*{[ Perpendicular bisector of a segment (GMT). ] The perimeter of triangle [ Properties and criteria of isosceles triangles. ] Problem 54503 Topics:

![](https://cdn.mathpix.com/cropped/2024_05_06_c1015ea0b2d2805d8876g-04.jpg?height=94&width=21&top_left_y=289&top_left_x=1443) Classes: 8,9}

In triangle $A B C$, it is known that $A B=B C, A C=10$. From the midpoint $D$ of side $A B$, a perpendicular $D E$ to side $A B$ is drawn until it intersects side $B C$ at point $E$. The perimeter of triangle $A B C$ is 40. Find the perimeter of triangle $A E C$.

## Hint

Apply the theorem about the perpendicular bisector of a segment.

## Solution

Since triangle $A B C$ is isosceles, then $A B=B C=(40-10): 2=15$.

Since $D E$ is the perpendicular bisector of segment $A B$, then $A E=B E$. Therefore, $A C+C E+A E=A C+C E+$ $B E=A C+B C=10+15=25$.

## Answer

25.