Area of a triangle = $$frac{1}{2} imes base imes height$$
Given, D and E are the mid-points of AB and AC of ΔABC
∴ $$frac{AE}{AC}=frac{AE}2frac{AC}2=frac{AD}{AB}$$
ΔABC and ΔADE are similar by SAS (Side, Angle and Side) as $$frac{AE}{AC}=frac{AD}{AB}$$ and common angle ∠A
∴$$frac{AE}{AC}=frac{DE}{BC}$$
As,E is mid-point of AC
∴AC = 2AE
DE = BC/2 --------equ.(1)
Now, the height of triangle ABC is AF.
Now, AT will be half of AF as ΔADE is in a proportion of 1: 2 with ΔABC.
QP = TF as both are the perpendicular distances between same parallel lines.
∴ QP = AF/2--------equ(2)
Area of triangle PED = $$frac{1}{2} imes QP imes DE$$
From equation1 and 2 ....
Area of triangle PED = $$frac{1}{2}frac{AF}{2}frac{BC}{2}$$ ---------equ (3)
Area of triangle ABC = $$frac{1}{2} imes{AF} imes{BC}$$ ----------equ(4)
Dividing equation3 and 4, we have
$$frac{AreaoftrianglePED}{AreaoftriangleABC}$$ = $$frac{frac{AreaoftrianglePED}{AreaoftriangleABC}}{frac{1}{2} imes{AF} imes{BC}}$$
ΔPED=1/4 ΔABC