$17 \cdot 10$ As shown in the figure, given that $O$ is the intersection of the perpendicular bisectors of the three sides of acute $\triangle A B C$, $\angle A=50^{\circ}$. Then the degree measure of $\angle B O C$ is
(A) $90^{\circ}$.
(B) $95^{\circ}$.
(C) $100^{\circ}$.
(D) $105^{\circ}$.
(China Sichuan Province Junior High School Mathematics Competition, 1991)
[Solution] Connect $AO$ and extend it to intersect $BC$ at $D$, then $\angle OAB = \angle OBA, \angle OAC = \angle OCA$.
Thus, $\quad \angle BOC = \angle BOD + \angle DOC$ $= \angle OAB + \angle OBA + \angle OAC + \angle OCA$ $= 2 \angle OAB + 2 \angle OAC = 2 \angle A = 100^{\circ}$.
Therefore, the answer is $(C)$.