Question

PA and PB are tangent segments to a circle with centre O from an external point P. if OP intersects the circle in M, prove that AM bisects angle PAB.

Answer 1

Whats your progress?

Answer 2

I wouldn'd say PA=PB, but angle OAP=angle OBP=90 degrees (tangents)

Answer 3

Well why wouldn't you say that but? :P

Answer 4

I suppose you are joking?

Answer 5

I am just saying you said "You won't say PA=PB" , I mean, why not ? They are equal anyways.

Answer 6

Anyways, thats not important here, let us focus at the problem.

Answer 7

It could be the problem.
I use the right angle between a radius and a tangent.
You claim line segments made by intersecting tangents are the same length.
I am not familiar with this. Why would this be true?

Answer 8

OAP and OBP are congruent triangles.

Answer 9

But this is just what you have to prove. After that, several things follow, among them PA=PB.

Answer 10

@shubhamsrg
am i right
in triangle AOP & BOP,
OP= OP (Common)
OA=OB (Radii of same circle)
PA=PB (Tangents from external cirlce)

Answer 11

thus triangles AOT & BOT are congruent

Answer 12

vat after that

Answer 13

@ZeHanz we don't have to prove that.(Though we can) , main problem is to prove AM bisects angle PAB.

Answer 14

@singhmmm: (my 3rd property is different)

in triangle AOP & BOP,
OP= OP (Common)
OA=OB (Radii of same circle)
angle OAP=angle OBP=90 degrees (tangents)
So the triangles are congruent

And therefore angle APO=angle BPO, so PO is bisector.