The base BC of triangle ABC is divided at D so that BD = $\dfrac{1}{2}$ DC.

Prove that the area of Δ ABD = $\dfrac{1}{3}$ of the area of Δ ABC.

In △ ABC,

⇒ BD = $\dfrac{1}{2}$ DC

⇒ $\dfrac{BD}{DC} = \dfrac{1}{2}$

We know that,

Ratio of the area of triangles with same vertex and bases along the same line is equal to the ratio of their respective bases.

$\Rightarrow \dfrac{\text{Area of Δ ABD}}{\text{Area of Δ ADC}} = \dfrac{BD}{DC} \\[1em] \Rightarrow \dfrac{\text{Area of Δ ABD}}{\text{Area of Δ ADC}} = \dfrac{1}{2} \\[1em] \Rightarrow \text{Area of Δ ADC} = \text{2 Area of Δ ABD}.$

From figure,

⇒ Area of Δ ABC = Area of Δ ABD + Area of Δ ADC

⇒ Area of Δ ABC = Area of Δ ABD + 2 Area of Δ ABD

⇒ Area of Δ ABC = 3 Area of Δ ABD

⇒ Area of Δ ABD = $\dfrac{1}{3}$ Area of Δ ABC.

Hence, proved that area of Δ ABD = $\dfrac{1}{3}$ area of Δ ABC.