question_answer

Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that \[\angle PTQ=2\angle OPQ.\]

Answer:

Given, a circle with centre O, an external point T and two tangents TP and TQ.

Let \[\angle PTQ=\theta \].

To prove:           \[\angle PTQ=2\,\angle OPQ\].

Proof:                     \[TP=TQ\]

[Tangent from an external point]

So \[\Delta \,TPQ\] is an isosceles triangle

\[\angle TPQ=\angle TQP\]

[Angle opposite to equal sides of a \[\Delta \]]

So,                   \[\angle TPQ=\angle TQP=\frac{1}{2}(180-\theta )=90-\frac{\theta }{2}\]

But,                  \[\angle TPQ=90{}^\circ \]

[Angle between tangent and radius]

\[\therefore \]                  \[\angle OPQ=\angle OPT-\angle TPQ=90-\left( 90-\frac{\theta }{2} \right)\]

\[=\frac{\theta }{2}=\frac{1}{2}\angle PTQ\]

Or                     \[\angle PTQ=2\angle OPQ\]                                Hence Proved.