$\underline{53619}$ topics:

In an isosceles triangle $ABC$ with base $AB$, the bisector $BD$ is drawn. A point $E$ is taken on the line $AB$ such that $\angle EDB=90^{\circ}$.

Find $BE$, if $AD=1$.
Connect point $D$ with the midpoint of segment $C E$.

## Solution

Let $M$ be the midpoint of $B E$. Then $D M$ is the median of the right triangle $E D C$, drawn to the hypotenuse $E B$, so $D M=1 / 2 B E=B M$. Therefore,

$\angle D M A=2 \angle M B D=\angle B=\angle A=\angle D A M$. Hence, $D M=D A=1, B E=2 D M=2$.

![](https://cdn.mathpix.com/cropped/2024_05_06_f7f9d90261bfc416bd71g-10.jpg?height=346&width=969&top_left_y=1519&top_left_x=545)

## Answer

2.