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

Given:

$\mathrm{D}, \mathrm{E}$ and $\mathrm{F}$ are respectively the mid-points of the sides $\mathrm{BC}, \mathrm{CA}$ and $\mathrm{AB}$ of a $\triangle \mathrm{ABC}$.

To do:

We have to show that

(i) BDEF is a parallelogram.
(ii) $\operatorname{ar}(\mathrm{DEF})=\frac{1}{4} \operatorname{ar}(\mathrm{ABC})$
(iii) $\operatorname{ar}(\mathrm{BDEF})=\frac{1}{2} \operatorname{ar}(\mathrm{ABC})$

Solution:

In $\triangle ABC$,

By mid point theorem,

$EF \| BC$

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

$BD = \frac{1}{2}BC$ ($D$ is the mid point of $BC$)

This implies,

$BD = EF$

Similarly,

$BF=DE$

$BF$ and $DE$ are parallel.

Here,

The pair of opposite sides are equal in length and parallel to each other.

Therefore,

$BDEF$ is a parallelogram.

(ii) Similarly, we can prove that both $DCEF$ and $AFDE$  are also parallelograms.

The diagonal of a parallelogram divides it into two triangles of equal area.

This implies,

In parallelogram $BDEF$,

$ar(\triangle BFD) = ar(\triangle DEF)$..........(i)

In parallelogram $DCEF$

$ar(\triangle DCE) = ar(\triangle DEF)$..........(ii)

In parallelogram $AFDE$,

$ar(\triangle AFE) = ar(\triangle DEF)$...........(iii)

From (i), (ii) and (iii), we get,

$ar(\triangle BFD) = ar(\triangle AFE) = ar(\triangle CDE) = ar(\triangle DEF)$

This implies,

$ar(\triangle BFD) +ar(\triangle AFE) +ar(\triangle CDE) +ar(\triangle DEF) = ar(\triangle ABC)$

$4 ar(\triangle DEF) = ar(\triangle ABC)$

$ar(\triangle DEF) = \frac{1}{4}ar(\triangle ABC)$

(iii) Area of parallelogram $BDEF = ar(\triangle DEF) +ar(\triangle BDE)$

$ar(BDEF) = ar(\triangle DEF) +ar(\triangle DEF)$

$ar(BDEF) = 2\times ar(\triangle DEF)$

$ar(BDEF) = 2\times \frac{1}{4}ar(\triangle ABC)$

$ar(BDEF) = \frac{1}{2}ar(\triangle ABC)$

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Updated on: 10-Oct-2022

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