Question
In the adjoining figure, $ABCD$ is a parallelogram in which $AB$ is produced to $E$ so that $BE = AB$. Prove that $ED$ bisects $BC$. 
✓
Answer
Given: $ABCD$ is a parralegram in which $AB$ is produced to $E$ such that $BE = AB$. $DE$ is joined which cuts $BC$ at $O$.
To Prove: $\text{OB = OC}$
Proof: In $\triangle\text{OCD}$ and $\triangle\text{OBE},$ we have,
$\angle\text{DOC}=\angle\text{EOB}$ [vertically opposite angles are equal]
$\angle\text{OCD}=\angle\text{OBE}$ [AB || CD, BC is a transversal thus, alternate angles are equal]
$\text{DC = BE}$ $[AB = CD$ and $BE = AB]$
Thus, by Angle-Angle-Side criterion of congruence, we have
$\therefore\triangle\text{OCD}\cong\triangle\text{OBE}$ [by AAS]
The corresponding parts of the congruent triangles are equal.
$\therefore\text{OC = OB}$
Hence, $ED$ bisect $BC$.
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