Quadrilateral $ABCD$ is inscribed in a circle. It is known that $AC \perp BD$. Find the length of $BC$, if the distance from the center of the circle to side $AD$ is 2.
Draw the diameter $B B_{1}$. Then $C B_{1}=A D$.

## Solution

Let $B_{1}$ be the point diametrically opposite to point $B$. Since $B_{1} D \perp B D\left(B B_{1}\right.$ - the diameter of the circle) and $A C \perp B D$, then $B_{1} D \| A C$. Therefore, $C B_{1}=A D$, and since equal chords are equidistant from the center of the circle, the distance $O M$ from the center $O$ of the circle to the chord $B_{1} C$ is also 2.

Since $O M \perp B_{1} C, B C \perp B_{1} C$ and $O$ is the midpoint of $B B_{1}$, then $O M$ is the midline of the right triangle $B C B_{1}$. Therefore, $B C=2 \cdot O M=2 \cdot 2=4$.

![](https://cdn.mathpix.com/cropped/2024_05_06_96a7587f2ba0a7b3aee2g-08.jpg?height=604&width=646&top_left_y=1&top_left_x=709)

## Answer

4.

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