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Parallelograms on the Same Base and between the Same Parallels Are Equal in Area

D,E and F are...

Question

D,E and F are respectively the mid-points of the sides BC, CA and AB of a $△$ ABC show that
1) BDEF is a parallelogram.
2) ar (DEF)= $\frac{1}{4}$ ar (ABC)
3) ar (BDEF)= $\frac{1}{2}$ ar (ABC)

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Solution

D and E are mid-points of sides BC and AC respectively.
So, $D E ∥ B A ⟹ D E ∥ B F$

Similarly, $F E ∥ B D$ . So, BDEF is a parallelogram.

Similarly, DCEF and AFDE are parallelograms.

Now, DF is a diagonal of parallelogram BDEF.

Therefore,

$a r (△ B D F) = a r (△ D E F)$ .....(1)

DE is a diagonal of parallelogram DCEF.

So, $a r (△ D C E) = a r (△ D E F)$ .....(2)

FE is a diagonal of parallelogram AFDE.

$a r (△ A F E) = a r (△ D E F)$ ....(3)

From 1,2 &3, we have,

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

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

So, $4 a r (△ D E F) = a r (△ A B C)$

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

Now, $a r (∥^{g m} B D E F) = 2 a r (△ D E F)$

$a r (∥^{g m} B D E F) = 2 \times \frac{1}{4} a r (△ A B C) = \frac{1}{2} a r (△ A B C)$

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D,E and F are respectively the mid-points of the sides BC, CA and AB of a △ ABC show that 1) BDEF is a parallelogram.2) ar (DEF)=14ar (ABC)3) ar (BDEF)=12 ar (ABC)

D,E and F are respectively the mid-points of the sides BC, CA and AB of a $△$ ABC show that
1) BDEF is a parallelogram.
2) ar (DEF)= $\frac{1}{4}$ ar (ABC)
3) ar (BDEF)= $\frac{1}{2}$ ar (ABC)