Question

From an external point P, two tangents PA and PB are drawn to the circle with center O. Prove that OP is the perpendicular bisector of AB.

Answer 1

$△OPA\&△OPB:$

$PA=PB$ (tangents to circle drawn from same exterior point).

$PO=PO$ (Common).

$\angle PAO=\angle PBO={90}^0$ ($\because$ Tangent and normal make ${90}^0$ and normal passes through centre of circle).

$\therefore △OPA\cong ▽OPBbyR.H.S.$

$\Rightarrow \angle OPA=\angle OPB=0\left(let\right)$

$\Rightarrow Namein△DAP\&△DBP$

$\Rightarrow \angle OPA=\angle OPBor$

$\angle DPA=\angle DPB$

$PA=PB$

$PD=PD$

$△DAP\cong △DBP$

$\Rightarrow \angle ADP=\angle PDB=\propto \left(let\right).$

$\because$ Sum of angles on a straight line = ${180}^0$

$\Rightarrow 2\propto ={180}^0\Rightarrow \propto ={90}^0.$

Hence $OP$ is the perpendicular bisector of $AB$.