2. The bisector of angle $B A D$ of a rectangular trapezoid $A B C D$ (with bases $A D$ and $B C, \angle B A D=90^{\circ}$) intersects the lateral side $C D$ at point $E$. Find the ratio $C E: E D$, if $A D+B C=A B$.

## Answer: $1: 1$.
# Solution:

Let's extend the bisector $A E$ until it intersects with line $B C$ (denote the intersection point as $F$). Since $\angle A B C$ is a right angle and $\angle F A B=45^{\circ}$, then $\angle A F B=45^{\circ}$, so $F A B$ is an isosceles right triangle, and $A B=B F$.

Therefore, $C F=B F-B C=A B-B C=A D$.

Thus, in quadrilateral $A C F D$, the opposite sides $A D$ and $C F$ are equal and parallel, so its diagonals $C D$ and $A F$ are bisected by their intersection point ($E$). Therefore, $C E: E D=1: 1$.

![](https://cdn.mathpix.com/cropped/2024_05_06_71062315d0492b8e06a2g-2.jpg?height=375&width=397&top_left_y=1600&top_left_x=1549)

## Criteria:

There is an additional construction leading to the solution - 1 point.

The equality of triangles AED and FEC is used but not proven - no more than 3 points.