The sides $AB$ and $CD$ of a parallelogram $ABCD$ are bisected at $E$ and $F$. Prove that $EBFD$ is a parallelogram.
Given:
The sides $AB$ and $CD$ of a parallelogram $ABCD$ are bisected at $E$ and $F$.
To do:
We have to prove that $EBFD$ is a parallelogram.
Solution:
Join $DE$, $BF$ and $EF$.
$ABCD$ is a parallelogram
This implies,
$AB = CD$
$AB \parallel CD$ (Opposite sides of a parallelogram are equal and parallel)
$EB \parallel DF$
$EB = DF$ ($E$ and $F$ are mid points of $AB$ and $CD$)
Therefore,
$EBFD$ is a parallelogram.
Hence proved.
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