Question

abcd is a parallelogram . if e is midpoint of bc and ae is the bisector of angle a.prove that ab=1/2ad

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