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$ABC$ is a triangle. $D$ is a point on $AB$ such that $AD = \frac{1}{4}AB$ and $E$ is a point on $AC$ such that $AE = \frac{1}{4}AC$. Prove that $DE =\frac{1}{4}BC$.
Given:
$ABC$ is a triangle. $D$ is a point on $AB$ such that $AD = \frac{1}{4}AB$ and $E$ is a point on $AC$ such that $AE = \frac{1}{4}AC$.
To do:
We have to prove that $DE =\frac{1}{4}BC$.
Solution:
Join $DE$.
Let $P$ and $Q$ be the mid points of $AB$ and $AC$ and join them.
This implies,
$\mathrm{PQ}=\frac{1}{2} \mathrm{BC}$
Similarly,
In $\triangle \mathrm{APQ}$,
$\mathrm{D}$ and $\mathrm{E}$ are mid-points of $\mathrm{AP}$ and $\mathrm{AQ}$
$\mathrm{DE}=\frac{1}{2} \mathrm{PQ}$
$=\frac{1}{2}(\frac{1}{2} \mathrm{BC})$
$=\frac{1}{4} \mathrm{BC}$
Hence proved.
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