ABCD is a cyclic quadrilateral. The two sides AB and CD are produced to meet in the point P and other two sides AD and BC are produced to meet in the point R . The two circumcircle of `DeltaBCPandDeltaCDR` intersect at the point T . Prove that points P , T , R are collinear.

Explanation

ABCD is a cyclic quadrilateral . The two sides AB and DC are produced to meet in the point P and other two sides AD and BC are produced to meet in the points R . The two circumcircle of `DeltaBDRandDeltaCDR` intersect at the point T.
To prove : Three points P ,T , R are collinear
Two angles in circle produced by the arc CP are `angleCBPand angleCTP`
`thereforeangleCBP=angleCTP` .........(1)
Again , `angleCDRand angleCTR` are two angles in circle produced by the chord CR.
`therefore" " angleCDR=angleCTR` ........(2) Now , ABCD is a cyclic quadrilateral,
`thereforeangleABC+angleADC=180^(@) ` [ by theorem 38]
or, `180^(@)-angleCBP+180^(@)-angleCDR=180^(@)[becauseangleABC+angleCBP=180^(@)andangleADC+angleCDR=180^(@)]`
or , `angleCBP+angleCDR=180^(@)`
or , `angleCTP+angleCTP=180^(@)` [from(1) and (2)] `thereforeanglePTR=1` straight angle
Hence the three points P , T ,R are collinear