From an external point `P ,` two tangents `P Aa n dP B` are drawn to the circle with centre `Odot` Prove that `O P` is the perpendicular bisector of `A Bdot`

Explanation

In `triangle PAC` and `triangle PBC`


`PA=PB` [length of tangents drawn from external point are equal]


`angle APC =angle BPC` [As, PA and PB are equally inclined to OP]


PC= PC [Common]


Therefore, by SAS criteria of similarity


`Delta PAC cong triangle PBC`


`Rightarrow AC =BC and angle ACP=angle BCP`


` angle ACP+angle BCP=180^{circ}`


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