3. (15 points) In a convex quadrilateral $A B C D, A B=10, C D=15$. Diagonals $A C$ and $B D$ intersect at point $O, A C=20$, triangles $A O D$ and $B O C$ have equal areas. Find $A O$.

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# Solution:

From the equality of the areas of triangles $A O D$ and $B O C$ and the equality of angles $\angle A O D=\angle B O C$, it follows that $\frac{A O \cdot O D}{B O \cdot O C}=1$ (by the theorem on the ratio of areas of triangles with one equal angle). From this, we get $\frac{A O}{O C}=\frac{B O}{O D}$. Additionally, $\angle A O B=\angle D O C$ as vertical angles. Therefore, triangles $A O B$ and $C O D$ are similar. $\frac{A B}{B C}=\frac{A O}{O C}=\frac{2}{3}$, which means $A O=8$.

![](https://cdn.mathpix.com/cropped/2024_05_06_1ae08a41583c767d264cg-02.jpg?height=577&width=479&top_left_y=911&top_left_x=797)

Answer: 8.

Criteria:

| Points | Conditions for awarding |
| :---: | :--- |
| 15 points | Complete, justified solution |
| 12 points | The solution approach is correct, but a computational error has been made. |
| 10 points | Similarity of triangles $A O B$ and $C O D$ is proven, but further actions are not performed or are performed incorrectly. |
| 5 points | The theorem on the ratio of areas of triangles with one equal angle is correctly applied, but further actions are not performed or are performed incorrectly. |
| 0 points | The solution does not meet the criteria listed above |