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In the figure, PB is a tangent to the circle at B. R is a point on AB such that RP is the angle bisector of

Mathematics
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|dw:1359898971161:dw|
@Reaper534 would you kindly help me?
sd|dw:1359901648939:dw| can you now express

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Other answers:

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 :)))

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