Question

$ABCD$ is a parallelogram in which $AB$ is produced to $E$. Show that $BE=AB$. Prove that $ED$ bisects $BC$

Answer 1

Given : $ABCD$ is a parallelogram

$BE=AB$

To prove : $ED$ bisects $AB$

Proof :

$AB=BE$ (given)

$AB=CD$ (opposite sides of a parallelogram

$therefore$ $BE=CD⟶\left(1\right)$

Let $DE$ intersects $BC$ at $F$

Now in $\Delta CDF$ & $\Delta BEF$

$\angle DCF=\angle EBF$ ($\because$ $AB\parallel CD$)

$\angle DFC=\angle EFB$ (vertically opposite angles)

$BE=CD$ (proved in equation $\left(1\right)$

Thus $BF=FC$ (by $CPCT$)

Thus, $ED$ bisect $BC$

Hence proved.