Question

$D, E$ and $F$ are respectively the mid-points of the sides $B C, C A$ and $A B$ of a $△ A B C$ . Show that

$a r (D E F) = \frac{1}{4} a r (A B C)$

Accepted Answer

In $△ A B C,$

$E$ and $F$ are the mid-points of side $A C$ and $A B$ respectively.

From mid-point theorem, we have $E F ∥ B C$ and $E F = \frac{1}{2} B C$

Also, $B D = \frac{1}{2} B C$ ( $D$ is the mid-point of $B C$ )

So, $B D = E F$ and $B D ∥ E F$

Therefore, $B D E F$ is a parallelogram.

Similarly, $DCEF, AFDE$ are parallelogram.

From above results, it can be said that quadrilaterals $B D E F, D C E F, A F D E$ are parallelograms.

We know, the diagonal of a parallelogram divides it into two triangles of equal area.

We have, $a r (△ B F D) = a r (△ D E F)$ (For parallelogram $B DEF$ )

$ar (△ C D E) = a r (△ D E F)$ (For parallelogram $D C E F$ )

$a r (△ A F E) = a r (△ D E F)$ (For parallelogram $A F D E$ )

$∴ a r (△ A F E) = a r (△ B F D) = a r (△ C D E) = a r (△ D E F)$

Also,

$a r (△ A F E) + a r (△ B D F) + a r (△ C D E) + a r (△ D E F) = a r (△ A B C)$

$\Rightarrow ar (△ DEF) + ar (△ DEF) + ar (△ DEF) + ar (△ DEF) = ar (△ ABC)$

$\Rightarrow 4 a r (△ D E F) = a r (△ A B C)$

$\Rightarrow a r (△ D E F) = \frac{1}{4} a r (△ A B C)$

Hence proved

EXERCISE

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