Question

How can we prove that "The centre of a circle lies on the bisector of the angle between two tangents drawn from a point outside it"?

Answer 1

Given : A circle C( O , r ) and a point A

outside the circle such that AP and AQ

are the tangents drawn to the circle

from point A .

To Prove :

i ) <AOP = <AOQ

ii ) <OAP = <OAQ

Proof :

In right triangles OAP and OAQ , we have

AP = AQ

( tangents from an external point

are equal )

OP = OQ ( radii of a Circle )

OA = OA ( common side )

∆OAP congruent to ∆OAQ

=> <AOP = <AOQ and

<OAP = <OAQ

Hence proved.

: )

Answer 2

$\textbf{Answer is in Attachment !}$