8.3. In trapezoid $A B C D$ with base $A D$, it is known that $A B=B C, A C=C D$ and $B C+C D=A D$. Find the angles of the trapezoid.
Answer: $\quad \angle A=72^{\circ}, \quad \angle B=108^{\circ}$, $\angle C=144^{\circ}, \angle D=36^{\circ}$.

Mark a point $K$ on the base $A D$ of the trapezoid such that $A K=B C$. Then $K D=C D$ and $A B C K-$ is a rhombus. Let the angle $\angle C A K$ be $\alpha$. We get, $\angle C A K=$

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$\angle A C K=\angle A D C=\alpha$. Then $\angle C K D=\angle K C D=2 \alpha$, and since the sum of the angles in triangle $A C D$ is $180^{\circ}$, we get that $5 \alpha=180^{\circ}$, from which $\alpha=36^{\circ}$. Therefore, the angles of the trapezoid are: $\angle A=72^{\circ}, \angle B=108^{\circ}, \angle C=144^{\circ}, \angle D=36^{\circ}$.