# D, E and F are respectively the mid-points of sides AB, BC and CA of ▲ABC. Find the ratio of the areas of ▲DEF and ▲ABC.

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ΔABC & ΔDEF, mid-points of AB, BC & CA respectively.

To find:- $\frac{ar\triangle DEF}{ar\triangle ABC}$

Note:- Since we need to find ratio of area of ΔDEF and ΔABC.

We first need to prove these triangles are similar.

Solution:-

We know that, line joining mid-points of two sides of a triangle is parallel to the 3rd side.

In ΔABC,

D and F are mid-points of AB and AC resp,.

$\therefore$ DF || BC

So, DF || BE also   ........(1)

Similarly,

E and F are mid-points of BC and AC resp.

EF || AB

Hence, EF || DB .............(2)

From (1) & (2)

DF || BE & FE || DB

Therefore, opposite sides of quadrilateral is parallel.

$\therefore$ DBEF is a parallelogram.

$\because$ DBEF is a parallelogram

Now we know that,

in parallelogram, opposite angle are equal.

Hence, ∠DFE = ∠ABC ..........(3)

Similarity,

we can prove DECF is a parallelogram.

In a parallelogram, opposite angles are equal.

Hence, ∠EDF = ∠ACB  ..........(4)

Now, in ΔEDF = ΔABC

∠DFE = ∠ABC     (From (3))

∠EDF = ∠ACB     (From (4))

By using AA similarity criterion

ΔDEF $\sim$ ΔABC

We know that if two triangles are similar,

the ratio of their area is always equal to the square of the ratio of their corresponding side.

$\therefore \frac{ar\triangle DEF}{ar \triangle ABC}=\frac{DE^2}{AC^2}$

$\frac{ar\triangle DEF}{ar \triangle ABC}=\frac{FC^2}{AC^2}$

$\boxed{(\text{Since DEECF is a parallelogram,}\\\text{opposite sides are equal,}\, i.e, DE = FC.)}$

$\frac{Area \,of \triangle DEF}{Area \,of\,\triangle ABC}=\frac{\left(\frac{AC}{2}\right)^2}{(AC)^2}$   (As F is the mid-point of AC)

$\frac{Area\,of\,\triangle DEF}{Area\,of \,\triangle ABC}=\frac{\frac{(AC)^2}{4}}{(AC)^2}$

$\frac{Area\,of\,\triangle DEF}{Area\,of \,\triangle ABC}=\frac{\frac{1}{4}}{1}$

Hence, $\frac{Area\,of\,\triangle DEF}{Area\,of\triangle ABC}=\frac{1}{4}$

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