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@Reaper534 would you kindly help me?

sd|dw:1359901648939:dw|
can you now express

why

:) well done!

now what about b part? Any idea ?

why

i didnt say

but the question ask

asks*

can you tell how you got

ext. angle of triangle

@BAdhi thanks for the spot ..I cant believe I fell for that mistake ! sorry for that @kryton1212

so what is

|dw:1359966959882:dw|

for p, i saw triangle ABP
and for

where is p?

p=

thanks:)

but the question is asking

and why

RP is the angle bisector of

ok, i miss this-.-

BQR seems a bit complicated hmm

is BQR an isos. triangle? we need to prove that BQR is an isos. triangle

since the answer is also h+k..

|dw:1359969862634:dw|
I am not too sure what use we make of the tangent! :|

i know now,

BAC = QBP ? How come ?

PB is a tangent, and

|dw:1359970287522:dw|

P lies out side the circle, your logic is flawed.

but the theorem is this,....

Had P been any point on the circumference, then that'd been equal,

http://www.mathsrevision.net/gcse/pages.php?page=13

The condition is that AB should be a diameter.

Is it given In the ques ? That AB is diameter ?

not at all

Only then the theorem is valid.

Which is not the case here

Alternate Segment Theorem

As I said, those angles would be equal had AB been diameter.

Which is not our case.

are you sure? the alternate segment theorem is not about the diameter.

Wait, leme revise

Ohh I see !
It makes sense, I missed a step.
Yep thats correct.

thanks :)))