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Triangles between Same Parallels

D, E and F ...

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
(i) $B D E F$ is a parallelogram
(ii) $a r (D E F) = \frac{1}{4} a r (A B C)$
(iii) $a r (B D E F) = \frac{1}{2} a r (A B C)$

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Solution

In $Δ A B C$ ,
E and F are midpoints of side AC and AB

Therefore, $E F ∥ B C$ and $E F = \frac{1}{2} B C$
Similarly, $E D ∥ A B$ and $E D = \frac{1}{2} A B$
Since, $E F ∥ B C$ $\Rightarrow E F | | B D$ and $E D ∥ F B$
$∴ □ B D E F$ is a parallelogram.
(ii)Using above concept , we will have BDEF, CDFE,DFAE, are parallelogram ,

And as we know that diagonal of parallelogram divides it into two triangles of equal area
$∴ A (Δ B D F) = A (Δ D F E) = A (Δ C D E) = A (Δ A F E)$
$A (Δ B D F) + A (Δ D F E) + A (Δ C D E) + A (Δ A F E) = A (Δ A B C)$

$A (Δ D F E) = \frac{1}{4} A (Δ A B C)$
(iii) Using the relation in (ii), we have
$A (□ B D E F) = A (Δ D B F) + A (Δ D F E)$
and $A (Δ B D F) = A (Δ D F E) = A (Δ C D E) = A (Δ A F E) = \frac{1}{4} A (Δ A B C)$
$A (Δ B D E F) = A (Δ D B F) + A (Δ D F E) = \frac{1}{4} A (Δ A B C) + \frac{1}{4} A (Δ A B C) = \frac{1}{2} A (Δ A B C)$

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