Question

In parallelogram ABCD, E is the mid point of side AB and CE bisects angle BCD. Prove that AE-AB, DE bisects angle ADC and angle DEC is a right angle.

Answer 1

(angles)a+b+c+d=360°
(angles)d+c=180°
in ∆ecd
(angles)edc+dce=90°
but sum of all angles of∆=180°
hence,angle dec =180°-(angle edc+angle dce)
=180°-90°
=90°
PROVED