A2. In triangle $A B C$, $\angle A C B=32^{\circ}$. On the extension of side $A B$ lie points $D$ and $E$, such that $|A D|=|A C|$ and $|B E|=$ $|B C|$ (see figure). How many degrees is the measure of angle $\angle D C E$?
(A) 90
(B) 96
(C) 100
(D) 106
(E) 116

![](https://cdn.mathpix.com/cropped/2024_06_07_95af3fbbcea2fb458cf3g-03.jpg?height=263&width=554&top_left_y=1011&top_left_x=1342)
A2. Let $\angle B A C=\alpha$. Then $\angle C B A=180^{\circ}-32^{\circ}-\alpha=148^{\circ}-\alpha$. Therefore, $\angle C A D=$ $180^{\circ}-\angle B A C=180^{\circ}-\alpha$ and $\angle E B C=180^{\circ}-\angle C B A=\alpha+32^{\circ}$. Triangles $C A D$ and $E B C$ are isosceles with vertices at $A$ and $B$, so $\angle D C A=\frac{180^{\circ}-\angle C A D}{2}=\frac{\alpha}{2}$ and $\angle B C E=\frac{180^{\circ}-\angle E B C}{2}=$ $74^{\circ}-\frac{\alpha}{2}$. From this, we calculate $\angle D C E=\angle D C A+\angle A C B+\angle B C E=32^{\circ}+74^{\circ}=106^{\circ}$.