ABCD is a cyclic quadrilateral such that AC bisects `angleBAD` . AD is produced to E in such a way that DE =AB . Prove that CE = CA .

Explanation

Given : ABCD is a cyclic quadrilateral . AC bisects `angleBAD` . AD is produced to E in such a way that DE =AB .
To Prove : CE=CA
Proof : ABCD is a cyclic qudrilateral.
`thereforeangleABC+angleADC=180^(@)` ……….(1)
Again , `angleADC+angleCDE=1 " straight angle" =180^(@)` ..........(2)
From (1) and (2) we get , `angleABC+angleADC=angleADC+angleCDE`
`rArrangleABC=angleCDE` ............(3)
Now , in `DeltaABCandDeltaCDE` DE =AB (given)
BC=CD [`because` the angle in circle `angleBAC` produced by the chord BC and the angle in circle `angleCAD` produced by the chord CD are equal , `therefore` BC=CD .] and included `angleABC ="includedangleCDE` [from (3)]
`thereforeDeltaABCcongDeltaCDE` [by the S-A-S condition of congruncy ]
`thereforeCA=CE[because` similar sides of conguent triangles]
Hence CA=CE .