$17 \cdot 36$ As shown in the figure, in $\triangle A B C$, the external angle bisectors $A D, B E$ of $\angle A, \angle B$ intersect the extensions of the opposite sides at points $D, E$, respectively, and $A D=A B=B E$. Then the degree measure of $\angle A$ is
(A) $10^{\circ}$.
(B) $11^{\circ}$.
(C) $12^{\circ}$.
(D) None of the above.
(China Shanghai Junior High School Mathematics Competition, 1986)
[Solution] Let $\angle A=\angle E=\alpha, \angle D=\angle A B D=\beta$, $\angle C B E=\gamma, \angle A C B=\delta$. According to the problem, we have $\beta=2 \gamma$, and $\beta=\alpha+\delta, \quad \delta=\gamma+\alpha$.
Therefore,
$$
\beta=\alpha+(\gamma+\alpha)=2 \alpha+\gamma,
$$

Also, $2 \gamma=\beta=2 \alpha+\gamma$, so $\gamma=2 \alpha$.

Thus,
$$
\beta=2 \gamma=4 \alpha \text {. }
$$
$$
\begin{array}{cc}
\because & \frac{1}{2}\left(180^{\circ}-\alpha\right)+2 \beta=180^{\circ}, \\
\therefore & 4 \beta-\alpha=180^{\circ} . \\
\text { Then } & 4 \cdot 4 \alpha-\alpha=180^{\circ} .
\end{array}
$$
Then
we get $\alpha=12^{\circ}$, i.e., $\angle A=12^{\circ}$. Therefore, the answer is $(C)$.