Moskvitin N.A.

In triangle $A B C$, $A B = B C$. A perpendicular $E D$ is dropped from point $E$ on side $A B$ to $B C$. It turns out that $A E = E D$. Find the angle $D A C$.
According to the exterior angle theorem, $\angle A E D=90^{\circ}+\angle B=270^{\circ}-2 \angle A$ (see figure). Therefore, $\angle E A D=1 / 2\left(180^{\circ}-\right.$ $\angle A E D)=\angle A-45^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9617f685748a8dfcd82ag-32.jpg?height=1106&width=849&top_left_y=167&top_left_x=633)

## Answer

$45^{\circ}$.