Question

A, B are the points on ⦿ (O,r) such that tangents at A and B intersect in P. Prove that OP is the bisector of ∠AOB and PO is the bisector of ∠APB.

Answer 1

For prooving the above situation we need atleast three condition

Now,proceed according to image attached

In ΔAOP and ΔOBP

1) AP = PB (Tangent to the circle)

2) ∠OAP = ∠OBP =90° (tangent ot any circle makes 90° with radius)

3) OA = OB (Radius of circle)

4) OP = OP ( Common side of triangle)

5) Hence,∠AOP=∠BOP (which shows OP is bisector of ∠APB)

Answer 2

Let PA and PB be two tangents drawn

from an external point P to a circle

with center O .

We have to prove that OP is the bisector

of <APB and <AOB

In right triangles OAP and OBP

PA = PB

[ since Tangents drawn from an external

point are equal ]

OA = OB ( radii of same circle )

and , OP = OP ( common)

By SSS - criterion of congruence ,

∆OAP is congruent to ∆OBP

<OPA = <OPB

and

<AOP = <BOP

Therefore ,

OP is bisector of <APB and <AOB

I hope this helps you.

: )