In the following figure,if $AB = 10$,then $AC = \ldots$
(A) In the given figure,$OP$ is the line segment joining the center $O$ to the external point $P$. The line $OP$ is the perpendicular bisector of the chord $AB$.
Since $OP \perp AB$ and $OP$ bisects $AB$,we have $AC = CB = \frac{AB}{2}$.
Given $AB = 10$,therefore $AC = \frac{10}{2} = 5$.
Since $OP \perp AB$ and $OP$ bisects $AB$,we have $AC = CB = \frac{AB}{2}$.
Given $AB = 10$,therefore $AC = \frac{10}{2} = 5$.
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