On the hypotenuse $A B$ of the right triangle $A B C$, a square $A B D E$ is constructed outward. It is known that $A C=1, \quad B C=3$.

In what ratio does the angle bisector of angle $C$ divide the side $D E$?
Let $O$ be the center of the given square. Prove that $C O$ is the bisector of angle $C$.

## Solution

Let $O$ be the center of the given square. Points $C$ and $O$ lie on the circle with diameter $A B$ (see the left figure). From the equality of chords $A O$ and $B O$, it follows that the inscribed angles $A C O$ and $B C O$ are equal, so the ray $C O$ is the bisector of angle $C$. Therefore, the line $C O$ divides the side $A B$ of the square into segments whose ratio is equal to the ratio of the sides $A C$ and $B C$ of triangle $A B C$, which is $1: 3$. Since this line passes through the center of the square, it also divides the opposite side $D E$ in the same ratio.

![](https://cdn.mathpix.com/cropped/2024_05_06_60e93adedf73503f106cg-07.jpg?height=586&width=1357&top_left_y=633&top_left_x=379)

## Answer

$1: 3$.

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