Question

In the figure two tangents TP and TQ are drawn to a circle with centre O from an
external point P. Prove that angel PTQ=2 angel OPQ

Answer 1

Given :-

• TP and TQ are tangents.
• OP and OQ are radius
• O is the centre of the circle

To prove :-

➟ ∠PTQ = 2 ∠OPQ

Solution :-

In Δ PTQ,

Since, tangents drawn from an exterior point are equal,

TP = TQ

We know that;

Angles opposite to equal sides are equal.

So, ∠TPQ = ∠TPQ    ------- {Equation 1}

Now, By Tangent Perpendicular Theorem,

∠OPT = 90°

→ ∠OPQ + ∠QPT = 90°

→ ∠QPT = 90° - ∠OPQ      ------- [Equation 2]

In Δ PQT,

By Angle Sum Property,

∠TPQ + ∠PQT + ∠PTQ = 180°

➥ ∠PQT + ∠PQT + ∠PTQ = 180°    [From Equation 1]

➝ 2 ∠TPQ + ∠PTQ = 180°

➝ 2 ∠TPQ = 180° - ∠PTQ

➝ 2(90° - ∠OPQ) = 180° - ∠PTQ      [From Equation 2]

➝ 180° - 2 ∠OPQ = 180° - ∠PTQ

⇒ - 2∠OPQ = - ∠PTQ

∴ ∠PTQ = 2 ∠OPQ

✬ HENCE PROVED ✬