3. (13 points) On the hypotenuse $A B$ of the right triangle $A B C$, a square $A B D E$ is constructed outwardly, with $A C=1, B C=4$. In what ratio does the angle bisector of angle $C$ divide the side $D E$?
Answer: $1: 4$.

Solution. Let $O$ be the center of the square, $P$ be the intersection point of lines $A B$ and $C O$, and $Q$ be the intersection point of lines $D E$ and $C O$.

![](https://cdn.mathpix.com/cropped/2024_05_06_f717c0793dcd8afe888ag-02.jpg?height=497&width=525&top_left_y=848&top_left_x=891)

In quadrilateral $A C B O$, the opposite angles are right angles, so it is a cyclic quadrilateral. Angles $\angle O A B$ and $\angle O C B$ subtend the same chord $O B$, hence they are equal, that is, $\angle O C B = \angle O A B = 45^{\circ}$. Therefore, $C P$ is the angle bisector of $\angle C$. By the angle bisector theorem, $\frac{A P}{P B} = \frac{1}{4}$. Then $\frac{D Q}{Q E} = \frac{1}{4}$ as well, since triangles $A P O$ and $D Q O$ are equal.

Evaluation criteria. Full solution - 13 points. It is shown that quadrilateral $A C B O$ is cyclic - 5 points. It is shown that $C P$ is the angle bisector of $\angle C$ - 3 more points, totaling 8 points. The ratio of $A P$ and $P B$, where $P$ is the intersection point of lines $A B$ and $C O$, is found - 2 more points, totaling 10 points.