Question

The given figure, two tangents TP and TQ are drawn to a circle with centre O from an external point T.
Prove that ∠PTQ=2∠OPQ.

Answer 1

Solution:

Construction: Join OQ and join OT.

In ΔOPT, since TP is a tangent; TP will be perpendicular to the radius of the circle. So, OP ⟂ TP.
In ΔOQT, since TQ is a tangent; TQ will be perpendicular to the radius of the circle. So, OQ ⟂ TQ.

Consider ΔOPT and ΔOQT
∠OPT = ∠OQT = 90° (Right angle)
TO = TO (Hypotenuse)
OP = OQ (Radii of the same circle)
ΔOPT ≅ ΔOQT (RHS congruence criterion)

TP = TQ
∠TQP = ∠TPQ.........(i)
(Angle opposite to equal sides are also equal.)

Tangent at any point of circle is perpendicular to the radius through point of contact.
(∴ OP⊥TP)

∴ ∠OPT = 90°
∠OPQ + ∠TPQ = 90°
∠TPQ = 90° − ∠OPQ.........(ii)

In △PTQ,
∠TPQ + ∠PQT + ∠PTQ = 180°
(∴ Sum of angles triangle is 180°)
90° − ∠OPQ + ∠TPQ + ∠PTQ = 180°
2(90° − ∠OPQ) + ∠PTQ = 180°
[from equation (i) and equation (ii)]
180° − 2∠OPQ + ∠PTQ = 180°
∴ 2∠OPQ=∠PTQ
Hence proved.