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.
You need to login to perform this action.
You will be redirected in
3 sec