The following diagram shows two parallelograms ABCD and BEFG.

Prove that :

Area of // gm ABCD = Area of // gm BEFG.

Given: ABCD and BEFG are two parallelograms.

To prove: Area of // gm ABCD = Area of // gm BEFG

Construction: Join AC and GE.

Proof: We know that if two triangles have the same base and height, their areas will be equal.

Since AG is a common base,

Area (Δ ACG) = Area (Δ AEG)

Adding Area (Δ ABG) on both sides,

⇒ ar.(Δ ABC) + ar.(Δ ABG) = ar.(Δ ABG) + ar.(Δ BGE)

⇒ ar.(Δ ABC) = ar.(Δ BGE) ……………….(1)

Since the area of a triangle is always half the area of a parallelogram when they share the same base and height, we can write:

ar.(Δ ABC) = $\dfrac{1}{2}$ ar.(∥gm ABCD)

ar.(Δ BGE) = $\dfrac{1}{2}$ ar.(∥gm BEFG)

From equation (1),

⇒ $\dfrac{1}{2}$ ar.(∥gm ABCD) = $\dfrac{1}{2}$ ar.(∥gm BEFG)

⇒ ar.(∥gm ABCD) = ar.(∥gm BEFG)

Hence, Area of // gm ABCD = Area of // gm BEFG.