Question

ABCD is a parallelogram having AB>AD , from AB cutoff  AE = AD. 
Prove that DC bisects Angle ADC

Answer 1

AB parallel to CD.  So  B+C = 180 degrees
AE = AD => in ΔAED,    angle AED  = angle EDA  = let us say  Ф
So angle BED = 180 - Ф                    angle EDC = ?
In quadrilateral  BEDC, sum all angles = 360.  angle B +C = 180 .  So angles BED + EDC = 180.
So angle EDC = 180 - (180 - Ф) = Ф
So DE bisects ADC.

Answer 2

AB II to CD. 
So  B+C = 180 (Linear Pair)
AE = AD (given) I'm not really sure
 There fore,  AED  = angle EDA  = let us say  x
So angle BED = 180 - x                   angle EDC = ?
In quadrilateral  BEDC, sum all angles = 360.  angle B +C = 180 .  So angles BED + EDC = 180.     hope it helps you out
So angle EDC = 180 - (180 - x) = x

So DE bisects ADC. Yay!! Proven :P