In a convex quadrilateral $A B C D$, diagonals $A C$ and $B D$ are drawn. It is known that $A D=2, \angle A B D=\angle A C D=$ $90^{\circ}$, and the distance between the centers of the circles inscribed in triangles $A B D$ and $A C D$ is $\sqrt{2}$. Find

$B C$.
Points $A, D$ and the centers of the indicated circles lie on a circle, the center of which is located on the circumcircle of the quadrilateral $A B C D$.

## Solution

Let $O_{1}$ and $O_{2}$ be the centers of the indicated circles. Since

$$
\angle A O_{1} D=\angle A O_{2} D=90^{\circ}+\frac{1}{2} \angle A C D=90^{\circ}+45^{\circ}=135^{\circ}
$$

then points $A, O_{1}, O_{2}$ and $D$ lie on the same circle. Let $O$ be the center of this circle, and $R$ be its radius. Then

$$
A D=2 R \sin \angle A O_{1} D \text {, or } 2=2 R \sin 135^{\circ} .
$$

From which we find that $R=\sqrt{2}$.

In triangle $A O D$, the square of side $A D$ is equal to the sum of the squares of sides $O A$ and $O D\left((\sqrt{2})^{2}+(\sqrt{2})^{2}=4=2^{2}\right)$.

Therefore, $\angle A O D=90^{\circ}$. Since the segment $A D$ is seen from points $B, C$ and $O$ at a right angle, these points lie on a circle with diameter $A D$, i.e., on the circumcircle of the given quadrilateral $A B C D$.

Since $O_{1} O_{2}=\sqrt{2}=R$, then $\angle O_{1} O O_{2}=60^{\circ}$. Since $O$ is the midpoint of the arc $A D$ not containing point $B$, then $\mathrm{CO}$ and $B O$ are the angle bisectors of angles $A C D$ and $A B D$. Therefore, $B O_{1} O$ is a straight line and $\mathrm{CO}_{2} \mathrm{O}$ is a straight line. Since $\angle B O C=60^{\circ}$, then

$$
B C=A D \sin 60^{\circ}=2 \cdot \frac{\sqrt{3}}{2}=\sqrt{3}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_4bea0374a35e8bb7d318g-43.jpg?height=392&width=558&top_left_y=164&top_left_x=750)

## Answer