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 [email protected]

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

Answer 2

since AE is the bisector of angle A

Therefore

angle 1 = 1/2 angle A ..... (1)

since ABCD is a parallelogram is parallelogram

Therefore , AD || BC and AB intersects them.

=> angle A + angle B = 180 ( sum of interior angles is 180)

=> angle B = 180 - angle A

In tri ABE , we have

angle 1 + angle 2 + angle B = 180

=> 1/2 angle A + angle 2 + 180- angle A = 180

angle 2 - 1/2 angle A= 0

angle 2 = 1/2 angle A ..... (2)

from (1) and (2)

angle 1 = angle 2

In tri ABE we have

BE = AB ( sides opposite to equal angles are equal)

2 BE = 2 AB ( multiplying by 2 both sides )

BC = 2 AB ( E is the mid point of BC )

AD = 2AB ( ABCD is a parallelogram therefore AD = BC )

AB= 1/2 AD