Problem 3. Point $M$ is located on side $CD$ of a square such that $CM: MD=1: 3$. Line $AM$ intersects the circumcircle of the square at point $E$. The area of triangle $ACE$ is 14. Find the side length of the square.
Answer: 10.

Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_12a9b3baa17b6febd8d6g-2.jpg?height=477&width=506&top_left_y=2017&top_left_x=638)

Triangles $A M D$ and $C M E$ are similar with a similarity coefficient $k=5$. Then

$$
C E=\frac{4 x}{5}, M E=\frac{3 x}{5} \rightarrow A E=5 x+\frac{3 x}{5}=\frac{28}{5} x
$$

Triangle $A C E$ is a right triangle, therefore:

$$
S_{A C E}=\frac{1}{2} A E \cdot C E=\frac{1}{2} \cdot \frac{28}{5} \cdot \frac{4}{5} x^{2}=14 \rightarrow x^{2}=\frac{25}{4} \rightarrow x=\frac{5}{2} \rightarrow A D=4 x=10
$$