Answer: 9.

## Solution.

From the fact that $A D$ is the diameter of the circle circumscribed around the trapezoid, it follows that $A O = B O = C O = D O$ as radii. Therefore, triangle $B O D$ is isosceles. Hence, $\angle O B D = \angle O D B = \angle D B C$ (the last equality is due to $A D \parallel B C$). Thus, $B D$ is the bisector of angle $O B C$. Similarly, $A C$ is the bisector of angle $O C B$. Since the center of the inscribed circle in triangle $B O C$ coincides with the intersection of the angle bisectors, we have: $I = A C \cap B D$. Further, due to the similarity of triangles $A I D$ and $B I C$, the areas are proportional to the square of the similarity coefficient: $\left(\frac{A D}{B C}\right)^{2} = \left(\frac{15}{5}\right)^{2} = 9$.

![](https://cdn.mathpix.com/cropped/2024_05_06_6a3c59b5c78061150e51g-4.jpg?height=485&width=523&top_left_y=598&top_left_x=839)
5. 5.1. An isosceles trapezoid \(ABCD\) is inscribed in a circle with diameter \(AD\) and center at point \(O\). A circle is inscribed in triangle \(BOC\) with center at point \(I\). Find the ratio of the areas of triangles \(AID\) and \(BIC\), given that \(AD=15, BC=5\).