ABCD is a cyclic quadrilateral. The side BC of it is extended to E . Prove that the two bisectors of `angleBADandangleDCE` meet on the circumferncee of the circle .

Explanation

Let ABCD is a cyclic quadrilateral in the circle with centre at O . The side BC of it is extended to E .
To prove : The two bisectors of `angleBADandangleDCE` meet on the circumference of the circle.
Construction : Let us construct the bisector AP of `angleBAD` , which intersect the circle at P . Let us join P , C and PC is extended to Q
Proof : `angleBCP=angleQCE" " [because` opposite angles] ........(1)
Now , ADCP is a cyclic quadrilateral.
`anglePAD+anglePCD=180^(@)`...........(2)
But in the figure , `anglePCD+angleDCQ=180^(@)` ...........(3)
from (2) and (3)n we get, `anglePAD+anglePCD=anglePCD+angleDCQ" or " anglePAD=angleDCQ`
or, `anglePAB=angleDCQ..........(4)[because"AP is the bisector of"angleBAD]`
or, `anglePAB =angleDCQ" " [because"both"anglePABand anglePCB` are angles in circle produced by the arc BP.]
or, `angleQCE=angleDCQ` [from (1) or , `angleECQ=angleDCQ`
`therefore` CQ is the bisector of `angleDCE`
i.e PQ is the bisector of `angleDCE` .
Hence the two bisectors of `angleBADandangleDCE` meet on the circumference of the circle