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[ $\underline{\text { Segment visible from two points at the same angle }}$

The diagonals of a convex quadrilateral $A B C D$ intersect at point $E, A B=B C, D B$ is the bisector of angle $D, \angle A B C=100^{\circ}, \angle B E A=70^{\circ}$. Find the angle $\angle C A D$.

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Let $F$ be the point symmetric to vertex $A$ with respect to the line $B D$. Then the points $C, F, B$, and $E$ lie on the same circle.

## Solution

Triangle $A B C$ is isosceles, so $\angle C A B = \angle A C B = 40^{\circ}$. Therefore,

$$
\angle A B E = 180^{\circ} - 70^{\circ} - 40^{\circ} = 70^{\circ}, \angle C B E = \angle C B A - \angle A B E = 100^{\circ} - 70^{\circ} = 30^{\circ}.
$$

Let $F$ be the point symmetric to vertex $A$ with respect to the line $B D$. Since $D B$ is the angle bisector of $\angle A D C$, point $F$ lies on the ray $D C$, and $\angle B F E = \angle B A E = 40^{\circ} = \angle B C E$. Thus, segment $B E$ is seen from points $C$ and $F$ at the same angle, and points $C$ and $F$ lie on the same side of line $B E$. Therefore, points $C, F, B$, and $E$ lie on the same circle. Hence,

$$
\angle C F E = \angle C B E = 30^{\circ}
$$

Therefore,

$$
\angle C A D = \angle D F E = \angle C F E = 30^{\circ}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_1b164f828743b46f847cg-44.jpg?height=531&width=512&top_left_y=1221&top_left_x=773)

## Answer

$30^{\circ}$. 

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