$ABC$ 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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Updated on: 10-Oct-2022

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