Question

From a point T outside a circle of centre O, tangents TP and TQ are drawn to the circle. Prove that OT is the right bisector of line segment PQ.

Answer 1

TP and TQ are tangents drawn from an external point T to the circle. O is the centre of the circle.
Suppose OT intersect PQ at point R.
In $\Delta$ OPT and $\Delta$ OQT,
OP = OQ (Radii of the circle)
TP = TQ (Lengths of tangents drawn from an external point to a circle are equal)
OT = OT (common sides)
$\therefore \Delta OPT\cong \Delta OQT$ (By SSS congruence rule)
So, $\angle PTO=\angle QTO$ (By CPCT) ......(1)
Now, in $\Delta$ PRT and $\Delta$ QRT,
TP = TQ (Lengths of tangents drawn from an external point to a circle are equal)
$\angle PTO=\angle QTO$ [From(1)]
RT = RT (Common sides)
$\therefore \Delta PRT\cong \Delta QRT$ (By SSS congruence rule)
So, PR = QR ....(2) (By CPCT)
And, $\angle PRT=\angle QRT$ (By CPCT)
Now,
$\angle PRT+\angle QRT={180}^{\circ }$ (Linear Pair)
$\Rightarrow 2\angle PRT={90}^{\circ }$
$\therefore \angle PRT=\angle QRT={90}^{\circ }$ ....(3)
From (2) and (3) we can conclude that OT is the right bisector of the line segment PQ.