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The Mid-Point Theorem

D, E and F ar...

Question

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

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Solution

$1]$ 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.

$2]$ 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, $4 a r (△ D E F) = a r (△ A B C)$
$a r (△ D E F) = \frac{1}{4} a r (△ A B C)$

$3]$ 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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Similar questions

D, E and F are respectively the mid-points of the sides BC, CA and AB of a ΔABC. Show that

(i) BDEF is a parallelogram.

(ii) ar (DEF) = ar (ABC)

(iii) ar (BDEF) = ar (ABC)

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)

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

Q. If

D, E, F

are the mid-points of the sides

B C, C A

and

A B

respectively of

△ A B C

, prove that

B D E F

is a parallelogram. And also show that

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

Q. In the figure

Δ A B C

, D, E , F are the midpoints of side BC, CA and AB respectively. Show that
(i) BDEF is a parallelogram
(ii)

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

(ii)

a r (B D E F) = \frac{1}{2} a r (Δ 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 (i) BDEF is a parallelogram (ii) ar (DEF) = 14 ar (ABC)(iii) ar BDEF) 12 ar (ABC)

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