If ABCD is a parallelogram, E is the mid-point of AB and CE bisects `/_BCD`, then `/_DEC` is

ABCD is a parallelogram, E is the mid-point of AB and CE bisects angleBCD . Then angleDEC is

ABCD is a parallelogram E is the mid point of AB and CE bisects angleBCD . Prove that: DE bisects angleADC

ABCD is a parallelogram E is the mid point of AB and CE bisects angleBCD . Prove that: AE=AD?

ABCD is a parallelogram E is the mid point of AB and CE bisects angleBCD . Prove that: angleDEC=90^@

ABCD is a parallelogram E is mid-point of AB and DE bisects angle D. Prove that BC= BE

ABCD is a parallelogram E is mid-point of AB and DE bisects angle D. Prove that CE bisects angle C

ABCD is a parallelogram E is mid-point of AB and DE bisects angle D. Prove that angleDEC= 90^(@)

ABCD is a parallelogram. E is the mid-point of AB and F is the mid-point of CD. GH is any line that intersects AD, EF and BC at G, P and H respectively. Prove that : GP=PH

In parallelogram ABCD, E is the mid-point of side AB and CE bisect angle BCD. Prove that : (i) AE=AD (ii) DE bisect angleADC and (iii) Angle DEC is a right angle.

In the given figure,ABCD is a parallelogram, in which is the mid-point of AD and O is a point on AC such that AO=(1)/(4)AC. When EO produced,meets AB at Frove that F is the mid-point of AB.