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Side $A B$ of triangle $A B C$ is a chord of a certain circle. Sides $A C$ and $B C$ lie inside the circle, the extension of side $A C$ intersects the circle at point $D$, and the extension of side $B C$ at point $E$, with $A B=A C=C D=2, C E=\sqrt{2}$. Find the radius of the circle.
Prove that angle $A$ is a right angle.

## Solution

By the theorem of the product of chord segments $B C=A C \cdot C D / C E=2 \sqrt{2}$.

From triangle $A B C$, we find that angle $A$ is a right angle. Therefore, $D B$ is the diameter of the circle, and $B D^{2}=A B^{2}+A D^{2}=$ 20.

![](https://cdn.mathpix.com/cropped/2024_05_06_9e7ee8b57660b9721937g-29.jpg?height=503&width=486&top_left_y=1929&top_left_x=795)

## Answer

$\sqrt{5}$.