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# D, E and F are respectively the mid-points of sides AB, BC and CA of $âˆ†ABC$. Find the ratio of the areas of $âˆ†DEF$ and $âˆ†ABC$.

Given:

D, E and F are respectively the mid-points of sides AB, BC and CA of $∆ABC$.

To do:

We have to find the ratio of the areas of $∆DEF$ and $∆ABC$.

Solution:

$D$ and $E$ are the mid-points of sides $AB$ and $BC$ respectively.

We know that,

The line segment joining the mid-points of two sides of a triangle is parallel to the third side and half of it.

This implies,

$EF=\frac{1}{2}AB$

$DF=\frac{1}{2}BC$

$DE=\frac{1}{2}AC$

Therefore,

$\frac{\mathrm{DE}}{\mathrm{AC}}=\frac{\mathrm{EF}}{\mathrm{AB}}=\frac{\mathrm{DF}}{\mathrm{BC}}=\frac{1}{2}$

Therefore, by SSS criterion,

$\Delta \mathrm{DEF} \sim \triangle \mathrm{ABC}$

This implies,

$\frac{\operatorname{ar} \triangle \mathrm{DEF}}{\text { ar } \triangle \mathrm{ABC}}=\frac{\mathrm{DE}^{2}}{\mathrm{AC}^{2}}$

$=(\frac{1}{2})^{2}$

$=\frac{1}{4}$

The ratio of the areas of $∆DEF$ and $∆ABC$ is $1: 4$.

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