Step 1: Make an imaginary figure by the given data

Given in ΔABC, D, E and F are midpoints of sides AB, BC and CA respectively.

BC = EC

Step 2: Recall the Midpoint theorem

**Midpoint Theorem**. The line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is congruent to one half of the third side.

Recall that the line joining the midpoints of two sides of a triangle is parallel to third side and half of it.

EXAMPLE: Form the figure

DF = \frac{1}{2} BC

\frac{DF}{BC} = \frac{1}{2} ............................(1)

Similarly, \frac{DE}{AC} = \frac{1}{2}...........................(2)

\frac{EF}{AB} = \frac{1}{2}..................................(3)

Step 3: Prove that the triangles are similar

From (1), (2) and (3) we have

\frac{DF}{BC} = \frac{DE}{AC} =\frac{EF}{AB} = \frac{1}{2}.

EXPLANATION: We know that, if in two triangles, sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar

Hence ΔABC ~ ΔEDF (since SSS similarity theorem)

Step 4: Find the areas ratio of the similar triangles

NOTE: The ratio of the areas of two similar triangles is equal to the ratio

of the squares of their corresponding sides

\frac{Area \triangle DEF}{Area \triangle ABC} = (\frac{DF}{BC})^2

(\frac{DF}{BC})^2 = (\frac{1}{2})^2

(\frac{DF}{BC})^2 = \frac{1}{4}

Hence area of ΔDEF : area of ΔABC = 1 : 4