# 5. CONDITION

A right triangle $ABC$ with hypotenuse $AB$ is inscribed in a circle. A point $D$ is taken on the larger leg $BC$ such that $AC = BD$, and point $E$ is the midpoint of the arc $AB$ containing point $C$. Find the angle $DEC$.
Solution. Point $E$ is the midpoint of arc $AB$, so $AE = BE$. Moreover, inscribed angles $CAE$ and $EBD$, subtending the same arc, are equal. Given that $AC = BD$, triangles $ACE$ and $BDE$ are congruent, which implies that angle $CEA$ is equal to angle $BED$. Therefore, angle $DEC$ is equal to angle $BEA$ and both are $90^{\circ}$, since angle $BEA$ is equal to angle $BCA$.

![](https://cdn.mathpix.com/cropped/2024_05_06_a178f1cf7922fb904239g-2.jpg?height=706&width=714&top_left_y=1880&top_left_x=708)

## Answer. $90^{0}$.