Question

E is the midpoint of the side BC of parallelogram ABCD and AE is the bisector of ∠A. Prove that 2AB = 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