$17 \cdot 9$ In triangle $ABC$, $AB=AC$, $\angle A=40^{\circ}$, point $O$ is inside the triangle, and $\angle OBC=\angle OCA$. Then the measure of $\angle BOC$ is
(A) $110^{\circ}$.
(B) $35^{\circ}$.
(C) $140^{\circ}$.
(D) $55^{\circ}$.
(E) $70^{\circ}$.
(5th American High School Mathematics Examination, 1954)
[Solution] Extend $B O$ to intersect $A C$ at $D$. By the exterior angle property of a triangle, we have
$$
\begin{aligned}
\angle B O C & =\angle O C D+\angle O D C \\
& =\angle O C D+(\angle O A D+\angle A O D) \\
& =\angle O C D+\angle O A D+\angle O A B+\angle A B O \\
& =\angle B A C+\angle A B C,
\end{aligned}
$$

Noting that $\angle O B C=\angle O C A$.
Given $\angle B A C=40^{\circ}$, then
$$
\angle A B C=\frac{1}{2}\left(180^{\circ}-40^{\circ}\right)=70^{\circ} \text {, }
$$

Therefore, $\angle B O C=40^{\circ}+70^{\circ}=110^{\circ}$.
Hence, the answer is $(A)$.