Question
In a parallelogram $\text{ABCD}, E$ is the midpoint of $AB$ and $DE$ bisects $\angle D$. Prove that: $BC = BE.$
✓
Answer
$\angle CDE = \angle DEA ...($ALternate angles$)$
$\angle CDE = \angle EDA ...$(Given $DE$ bisects $\angle D)$
$\angle DEA = \angle EDA$
$\Rightarrow AD = AE ....(i)$(Sides opposite to equal angles are equal$)$
Now, $AD = BE ....(ii)($Opposite to equal angles are equal$)$
And, $AE = BE ....(iii)(E$ is the mid$-$point of $AB)$
$\Rightarrow BC = BE. ....($proved$)$
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