Question

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

Given: A circle with centre O and an external point T from which tangents TP and TQ are drawn to touch the circle at P and Q.

To prove: ∠PTQ = 2∠OPQ.

Proof: Let ∠PTQ = xº. Then,

∠TQP + ∠TPQ + ∠PTQ = 180º

[∵ sum of the ∠s of a triangle is 180º]

⇒ ∠TQP + ∠TPQ = (180º – x) ….(i)

We know that the lengths of tangent drawn from an external point to a circle are equal.

So, TP = TQ.

Now, TP = TQ

⇒ ∠TQP = ∠TPQ

`=\frac{1}{2}(180^{0}-x)=( 90^\text{o}-\frac{x}{2})`

∴ ∠OPQ = (∠OPT–∠TPQ)

`=90^{0}-( 90^\text{o}-\frac{x}{2})=\frac{x}{2} `

`⇒ ∠OPQ = \frac { 1 }{ 2 } ∠PTQ`

⇒ 2∠OPQ = ∠PTQ

Answer 2

Given: TP and TQ are two tangents of a circle with centre O and P and Q are points of contact

To prove: ∠PTQ = 2∠OPQ

Suppose ∠PTQ = θ.

Now by theorem, "The lengths of a tangents drawn from an external point to a circle are equal".

So, TPQ is an isoceles triangle.

Therefore, ∠TPQ = ∠TQP = `1/2`(180^°− θ) = 90^°−`θ/2`.

Also by theorem "The tangents at any point of a circle is perpendicular to the radius through the point of contact" ∠OPT = 90°.

Therefore,

∠OPQ = ∠OPT − ∠TPQ

`= 90^@ - (90^@ -  1/2theta)`

`= 1/2 theta`

= `1/2` ∠PTQ

Hence, 2∠OPQ = ∠PTQ.