4. (3 points) Point $C$ is located on the segment $A E$. On one side of the line $A E$, points $B$ and $D$ are marked such that $A B C$ is an equilateral triangle, and $C D E$ is an isosceles right triangle with a right angle at $D$. It turns out that triangle $B C D$ is isosceles with base $B C$. Find the angle $A D E$.
# Answer: $105^{\circ}$

Solution: The acute angles of triangle $C D E$ are each $45^{\circ}$, and the angles of triangle $A B C$ are each $60^{\circ}$. Therefore, angle $B C D$ is $75^{\circ}$. Consequently, $\angle C B D=75^{\circ}, \angle C D B=30^{\circ}$.

The altitudes of triangles $A B C$ and $B C D$, dropped to side $B C$, fall at its midpoint, meaning they are extensions of each other. Since they are also angle bisectors, $\angle A D C=\frac{\angle B D C}{2}=15^{\circ}$.

Thus, $\angle A D E=\angle A D C+\angle C D E=15^{\circ}+90^{\circ}=105^{\circ}$.