5. Given a triangle $A B C, \angle B=90^{\circ}$. On the sides $A C, B C$ points $E$ and $D$ are chosen respectively, such that $A E=E C, \angle A D B=\angle E D C$. Find the ratio $C D: B D$.
Answer. $2: 1$

Solution. Construct triangle $A^{\prime} B C$, symmetric to the given one with respect to side $B C$. Points $A^{\prime}, D, E$ lie on the same line, since $\angle A^{\prime} D B=\angle E D C$. Therefore, $D$ is the point of intersection of the medians $A^{\prime} E$ and $C B$ of triangle $A A^{\prime} C$, and divides them in the ratio $2: 1$ from the vertex.

![](https://cdn.mathpix.com/cropped/2024_05_06_1175bac21bd43d8433a0g-6.jpg?height=854&width=893&top_left_y=704&top_left_x=593)

Criteria.

$(-)$

$(-$.

(-/+) Correct additional constructions, there is progress in the solution.

(+/2) Solution matching the one provided above. It is not proven that points $D, E, A^{\prime}$ lie on the same line.

(+/-) Generally correct solution with some inaccuracies

$(+$.

$(+)$