D and E are the mid­points of AB and AC of ΔABC; BC is produced to any point P; DE, DP and EP are joined. Then,

Area of a triangle = $$\frac{1}{2}\times base\times height$$
Given, D and E are the mid-points of AB and AC of ΔABC
∴ $$\frac{AE}{AC}=\frac{AE}2\frac{AC}2=\frac{AD}{AB}$$
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 = $$\frac{1}{2}\times QP\times DE$$
From equation1 and 2 ....
Area of triangle PED = $$\frac{1}{2}\frac{AF}{2}\frac{BC}{2}$$ ---------equ (3)
Area of triangle ABC = $$\frac{1}{2}\times{AF}\times{BC}$$ ----------equ(4)
Dividing equation3 and 4, we have
$$\frac{AreaoftrianglePED}{AreaoftriangleABC}$$ = $$\frac{\frac{AreaoftrianglePED}{AreaoftriangleABC}}{\frac{1}{2}\times{AF}\times{BC}}$$
ΔPED=1/4 ΔABC