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 ∆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.