Diagonals of a convex quadrilateral $A B C D$ intersect at point $E, A B=A D, C A-$ is the bisector of angle $C$,

$\angle B A D=140^{\circ}, \angle B E A=110^{\circ}$.

Find the angle $C D B$.

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Continue sides $B C$ and $A D$ until they intersect at point $F$ and prove that triangle $C D F$ is isosceles.

## Solution

The angles at the base $B D$ of the isosceles triangle $B A D$ are each $20^{\circ}$. Therefore, $\angle C A D = \angle A E B - \angle A D E = 90^{\circ}$.

Extend sides $B C$ and $A D$ until they intersect at point $F$. Since the bisector $C A$ of triangle $C D F$ is also its altitude, triangle $C D F$ is isosceles. Therefore, $F A = A D = A B$.

Since the median $A B$ of triangle $B F D$ is half the length of side $D F$, $\angle D B F = 90^{\circ}$. Therefore, $\angle C D F = \angle B F D = 90^{\circ} - \angle B D F = 70^{\circ}$.

Thus, $\angle C D B = \angle C D F - \angle B D A = 50^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_8c5c78c6462ff090b828g-33.jpg?height=574&width=440&top_left_y=342&top_left_x=815)

Answer

$50^{\circ}$.