$3+$ [ Law of Sines $\quad]$

Quadrilateral $A B C D$ is inscribed in a circle, $A C=a, B D=b, A B \perp C D$. Find the radius of the circle.
Let $R$ - radius of the circle, $\angle B A D=\alpha$, then $\angle C D A=90^{\circ}-\alpha$ (see fig.). By the corollary of the sine theorem $B D=2 R \sin \alpha, A C=2 R \cos \alpha$. Therefore, $B D^{2}+A C^{2}=4 R^{2}$, hence, $R=\frac{\sqrt{a^{2}+b^{2}}}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9fc5d17caeb61bf6ddabg-03.jpg?height=489&width=486&top_left_y=842&top_left_x=795)

Answer

$\frac{\sqrt{a^{2}+b^{2}}}{2}$.