and are the tangents drawn to ⨀ (O, r) from point P lying in the exterior of the circle and T and R are their points of contact respectively. Prove that m∠TPR = 2m∠OTR.

Given that PT and PR are tangents drawn to circle (O, r) from point P lying in the exterior of the circle and T and R are their points of contact respectively.

We have to prove that m∠TPR = 2m∠OTR

Proof:

By theorem,

PT ≅ PR

We know that angles opposite to congruent sides are equal.

∴ ∠PTR = ∠PRT

We know that sum of all angles in a triangle is 180°.

In ΔPTR,

⇒ ∠PTR + ∠PRT + ∠TPR = 180°

⇒ ∠PTR + ∠PTR + ∠TPR = 180°

⇒ 2∠PTR + ∠TPR = 180°

⇒ 2∠PTR = 180° – ∠TPR

⇒ ∠PTR = 90° – 1/2 ∠TPR

∴ 1/2 ∠TPR = 90° – ∠PTR … (1)

Then, OT ⊥ PT,

⇒ ∠OTP = 90°

⇒ ∠OTR + ∠PTR = 90°

⇒ ∠OTR = 90° – ∠PTR … (2)

From (1) and (2),

⇒ 1/2 ∠TPR = ∠OTR

⇒ ∠TPR = 2∠OTR

∴ m∠TPR = 2m∠OTR

Hence proved.