The medians of a triangle ABC intersect each other at point G. If one of its medians is AD, prove that :

(i) Area (△ ABD) = 3 × Area (△ BGD)

(ii) Area (△ ACD) = 3 × Area (△ CGD)

(iii) Area (△ BGC) = $\dfrac{1}{3}$ × Area (△ ABC)

(i) Given,

Medians of a triangle ABC intersect each other at point G.

We know that,

Medians intersect at centroid, also the centroid divides medians in the ratio 2 : 1.

∴ AG : GD = 2 : 1.

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.

$\Rightarrow \dfrac{\text{Area of △ AGB}}{\text{Area of △ BGD}} = \dfrac{AG}{GD} \\[1em] \Rightarrow \dfrac{\text{Area of △ AGB}}{\text{Area of △ BGD}} = \dfrac{2}{1} \\[1em] \Rightarrow \text{Area of △ AGB} = 2 \text{ Area of △ BGD}.$

From figure,

⇒ Area of △ ABD = Area of △ AGB + Area of △ BGD

⇒ Area of △ ABD = 2 Area of △ BGD + Area of △ BGD

⇒ Area of △ ABD = 3 Area of △ BGD.

Hence, proved that area of △ ABD = 3 area of △ BGD.

(ii) We know,

AG : GD = 2 : 1

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

$\Rightarrow \dfrac{\text{Area of △ AGC}}{\text{Area of △ CGD}} = \dfrac{AG}{GD} \\[1em] \Rightarrow \dfrac{\text{Area of △ AGC}}{\text{Area of △ CGD}} = \dfrac{2}{1} \\[1em] \Rightarrow \text{Area of △ AGC} = 2 \text{ Area of △ CGD}.$

From figure,

⇒ Area of △ ACD = Area of △ AGC + Area of △ CGD

⇒ Area of △ ACD = 2 Area of △ CGD + Area of △ CGD

⇒ Area of △ ACD = 3 Area of △ CGD.

Hence, proved that area of △ ACD = 3 area of △ CGD.

(iii) From part (i),

⇒ Area of △ ABD = 3 Area of △ BGD ……..(1)

From part (ii),

⇒ Area of △ ACD = 3 Area of △ CGD ……….(2)

Adding equations (1) and (2), we get :

⇒ Area of △ ABD + Area of △ ACD = 3 Area of △ BGD + 3 Area of △ CGD

⇒ Area of △ ABC = 3(Area of △ BGD + Area of △ CGD)

⇒ Area of △ ABC = 3 Area of △ BGC

⇒ Area of △ BGC = $\dfrac{1}{3}$ Area of △ BGC

Hence, proved that area of △ BGC = $\dfrac{1}{3}$ area of △ BGC.