Question

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

Answer 1

Given : A circle with centre O. Tangents TP  and TQ are drawn from a point T outside a  circle.

To Prove : OT is the right bisector of line  segment PQ.

Construction : Join OP & OQ

Proof : In ΔPTR and ΔQTR

In ΔOPT and ΔOQT

∠OPT = ∠OQT = 90°

OP = OQ (radius)

OT = OT (Common)

ΔOPT ≅  ΔOQT (By RHS congruence)

∠PTR = ∠QTR (cpct)

TP = TQ  (Tangents are equal)

TR = TR (Common)

∠PTR = ∠QTR  (OT bisects ∠PTQ)

ΔPTR ≅ ΔQTR (By SAS congruency)

PR = QR

∠PRT = ∠QRT

But ∠PRT+ ∠QRT = 180° (as PQ is line segment)

∠PRT = ∠QRT = 90°

Therefore TR or OT is the right bisector of  line segment PQ