Question

$ABCD$ is a parallelogram. If $E$ is mid-point of $BC$ and $AE$ is the bisector of $\angle A$, prove that $AB=\frac{1}{2}AD$.

Answer 1

$\Rightarrow$ $ABCD$ is a parallelogram. Mid-point of $BC$ is $E$

$\Rightarrow$ Join $A$ to $E$, $\angle EAD=\angle EAB$ ----- ( 1 )

$\Rightarrow$ $AD$ is parallel to $BC$ and $AE$ is a transversal

$\therefore$ $\angle EAD=\angle BEA$ ----- ( 2 ) [Alternate angle]

$\Rightarrow$ $\angle EAB=\angle BEA$ [From ( 1 ) and ( 2 )]

$\therefore$ $AB=BE$ ---- ( 3 )

$\Rightarrow$ But $BE=\frac{BC}{2}=\frac{AD}{2}$ ---- ( 4 )

$\Rightarrow$ Put value of BE in equation ( 3 )

$\therefore$ $AB=AD\frac{1}{2}$ [Proved]