7.1. In quadrilateral $A B C D$, $A B=B C=C D$. Let $E$ be the intersection point of $A B$ and $C D$ (with $B$ between $A$ and $E$). It turns out that $A C=C E$ and $\angle D B C=15^{\circ}$. Find $\angle A E D$.
Answer: $50^{\circ}$

Solution. We will solve the general problem. Let $\angle D B C=\alpha$, and $\angle B E C=\beta$. From the isosceles property of $\triangle D C B$, we get $\angle B D C=\alpha$. From the isosceles property of $\triangle A C E$, we get $\angle B E C=\angle B A C$, and from the isosceles property of $\triangle A B C$, we get $\angle B A C=\angle A C B=\beta$. Then $\angle A C D=2 \beta$ as an exterior angle. Therefore, the sum of the angles in triangle $B C D$ is $2 \alpha+3 \beta=180^{\circ}$. In our problem, $\alpha=15^{\circ}$, then $\beta=\frac{180^{\circ}-2 \cdot 15^{\circ}}{3}=50^{\circ}$.

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