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.

#### Solution

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 ∆OPT and ∆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)

∴ ∆OPT ≅ ∆OQT (By SSS congruence rule)

So, ∠PTO = ∠QTO (By CPCT) .....(1)

Now, in ∆PRT and ∆QRT,

TP = TQ (Lengths of tangents drawn from an external point to a circle are equal.)

∠PTO = ∠QTO [From (1)]

RT = RT (Common sides)

∴ ∆PRT ≅ ∆QRT (By SAS congruence rule)

So, PR = QR .....(2) (By CPCT)

And, ∠PRT = ∠QRT (By CPCT)

Now,

∠PRT + ∠QRT = 180° (Linear pair)

⇒ 2∠PRT = 180°

⇒ ∠PRT = 90°

∴ ∠PRT = ∠QRT = 90° .....(3)

From (2) and (3), we can conclude that

OT is the right bisector of the line segment PQ.