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 3^{rd} 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, \(\frac{Area\,of\,\triangle DEF}{Area\,of\triangle ABC}=\frac{1}{4}\)