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

PT

In ΔOPT and ΔOQT
∠OPT = ∠OQT = 90°

OP = OQ (radius)

OT = OT (common side)

∴ ΔOPT and ΔOQT are congruent by RHS.

So, ∠ PTO = ∠QTO (by CPCT) ….(1)

Also, PT = QT (by CPCT) …(2)

Now,
In ΔPCT and Δ QCT,
PT = QT (From eq (2))

∠ PTC = ∠ QTC (From eq(1))

CT = CT (common side)

∴ ΔPCT and Δ QCT are congruent by SAS property.

So, ∠ PCT = ∠ QCT (by CPCT) …..(3)

As PQ is a line segment,
∠ PCT + ∠ QCT = 180°

⇒ ∠ 3 + ∠ 4 = 180°

⇒ ∠ 3 + ∠ 3 = 180°

⇒ 2∠3 = 180°

⇒ ∠ 3 = 90°

Therefore, ∠3 = ∠4 = 90°

Hence proved that OT is the right bisector of PQ.