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

Answers (1)

AB parallel to CD. So B+C=180^circ
AE=AD
In Traingle AED, angle AED=angle EDA= let us say phi
So angle BED = 180 - phi 
Angle EDC=?
In Quadrilateral BEDC,sum all angles=360.
Angle B+Angle C=180.
Angle BED+ Angle EDC=180
By above,Angle EDC=180-(180-phi)=phi
Which means that DE bisects ADC

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