The vertices at ΔABC are A(–1, 5), B(3, 1) and C(5, 7). D, E, F are the mid points of BC, CA and AB respectively. Let us find the area of triangular region ΔDEF and prove that ΔABC = 4 ΔDEF.

Let us find area of ΔABC

Area of triangle is given by formula

Area = × [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]

Where (x₁, y₁), (x₂, y₂) and (x₃, y₃) are vertices of triangle

Here,

A = (x₁, y₁) = (-1, 5)

B = (x₂, y₂) = (3, 1)

C = (x₃, y₃) = (5, 7)

Substituting values

⇒ Area(ΔABC) = × [(-1)(1 – 7) + 3(7 – 5) + 5(5 – 1)]

⇒ Area(ΔABC) = × [6 + 6 + 20]

⇒ Area(ΔABC) = × 32

⇒ Area(ΔABC) = 16 unit² … (i)

Let us now find the midpoints of BC, AC and AB i.e. points D, E and F respectively

Point F is the midpoint of AB

⇒ F =

⇒ F = (1, 3)

Point D is the midpoint of BC

⇒ D =

⇒ D = (4, 4)

Point E is the midpoint of AC

⇒ E =

⇒ E = (2, 6)

Let us find area of ΔDEF

Area = × [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]

Where (x₁, y₁), (x₂, y₂) and (x₃, y₃) are vertices of triangle

Here,

D = (x₁, y₁) = (1, 3)

E = (x₂, y₂) = (4, 4)

F = (x₃, y₃) = (2, 6)

Substituting values

⇒ Area(ΔDEF) = × [1(4 – 6) + 4(6 – 3) + 2(3 – 4)]

⇒ Area(ΔDEF) = × [-2 + 12 + (-2)]

⇒ Area(ΔDEF) = × 8

⇒ Area(ΔDEF) = 4 unit² … (ii)

From (i) and (ii) we can conclude that area(ΔABC) = 4 × area(ΔDEF)