In ΔABC, D and E are mid-points of AB and AC respectively.
Find: ar(ΔADE) : ar(ΔABC).

We know that,

The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and is half of the third side.

Given,

In Δ ABC,

D is mid-point of side AB and E is the mid-point of the side AC.

∴ DE ∥ BC and DE = $\dfrac{1}{2}$ BC

D is mid-point of side AB.

∴ AB = 2AD

Let us consider ΔADE and ΔABC

∠DAE = ∠BAC [Common angle]

∠ADE = ∠ABC [Corresponding angle are equal]

∴ ΔADE ∼ ΔABC by AA similarity.

$\Rightarrow \dfrac{\text{area(ΔADE)}}{\text{area(ΔABC)}} = \dfrac{AD^2}{AB^2} \\[1em] \Rightarrow \dfrac{\text{area(ΔADE)}}{\text{area(ΔABC)}} = \dfrac{AD^2}{(2AD)^2} \\[1em] \Rightarrow \dfrac{\text{area(ΔADE)}}{\text{area(ΔABC)}} = \dfrac{1}{4}.$

Hence, ar(ΔADE) : ar(ΔABC) = 1 : 4.