1.

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.

Answer» 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.


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