Question

ABCD is a parallelogram. If E is the midpoint of BC and AE is the bisector of angle A. Prove that AB = 1/2 AD

Answer 1

let DE cut AB extended at F 

We have BE=EC

angle (BEF)= angle (DEC)

∠(FBE)=∠(ECD)

so triangle CED congruent to BEF ( case ASA)

so EF= ED and CD=BF=AB

∴ inΔ ADF, AE is the median and angle bisector of angle A so triangle ADF is isoceles.

so AD= AF=2.AB => AB= 1/2  AD