In a right angle triangle `Delta ABC` is which `/_ B = 90^@` a circle is drawn with AB diameter intersecting the hypotenuse AC at P.Prove that the tangent to the circle at PQ bisects BC.

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Please briefly explain why you feel this question should be reported.

Please briefly explain why you feel this answer should be reported.

Please briefly explain why you feel this user should be reported.

BQ=PQ-(1)

`/_APB=90^o`

`/_BPC=180-90=90^o`

`/_BPC=90^0`

`/_BPQ+/_QPC=90^o`

from equation 1

`/_BPQ=/_PBQ`

`/_PBQ+/_PCQ=90^o-(3)`

from equation 2 and 3

`/_BPQ+/_QPC=/_PBQ+/_PCQ`

`/_QPC=/_PCQ`

PQ=QC

From 1

BQ=QC

Q is mid point of BC.