Question

How can we prove that "If two tangents AP and AQ are drawn to a circle with centre O from an external point A then ∠PAQ= 2∠OPQ= 2∠OQP ."

Answer 1

We are given a circle with center O ,

an external point A and two tangents

AP and AQ to the circle ,

Where P, Q are the points of contact.

To prove : <PAQ = 2<OPQ

Let <PAQ = x°

[ Now , By the theorem

The lengths of tangents drawn from an

external point to a circle are equal ]

AP = AQ ,

So , ∆APQ is an isosceles triangle.

Therefore ,

<APQ + <AQP + <PAQ = 180°

( Since , sum of three angles in a triangle )

<APQ = <AQP = 1/2 ( 180° - x° )

<APQ = <AQP = 90° - x/2

<OPA = 90°

So ,

<OPQ = <OPA - <APQ

[ Since , The tangent at any point of a

circle is perpendicular to the radius

through the point of contact . ]

<OPQ = 90° - [ 90° - x°/2 ]

<OPQ = x°/2 = ( <PAQ )/2

This gives

<PAQ = 2<OPQ ,

Similarly <PAQ = 2<OQP

I hope this helps you.

: )