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Question
If PA and PB are two tangents drawn from a point P to a circle with center C touching it A and B, prove that CP is the perpendicular bisector of AB.
Solution
We shall prove that ∠ACP = ∠BCP = 90°
and AC = BC
Now, ∠APC = ∠BPC
Since O lies on the bisector of ∠APB.
Δs ACP and BCP are congruent triangles by SAS congruence criterion,
∴ AC = BC
and ∠ ACP = ∠ BCP
Since ∠ ACP + ∠ BCP = 180°
2 ∠ ACP = 180°
∠ ACP = 90°
∠ACP = ∠BCP = 90°
Hence proved.
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