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In given figure , ABC is a triangle prod...

In given figure , ABC is a triangle produced meets the circumcircle of `Delta ABC` at `Q`, prove that `CP=CQ`

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In ` Delta ABP`,
`AB=AP` (given)
`angle1=angle2` (angles opposite to equal sides ) .......(1)
But `angle2 =angle3 ` (vertically opposite angle) ......(2)
`rArrangle1=angle3` [From (1) and (2)] .........(3)
But `angle1=angle4` (angles of same segment) .........(4)
`therefore angle3= angle4` [from (3)and (4)]
`rArrCP=CQ` (side opposite to equal of `DeltaCPQ`)
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