Show that :

(i) a diagonal divides a parallelogram into two triangles of equal area.

(ii) the ratio of the areas of two triangles of the same height is equal to the ratio of their bases.

(iii) the ratio of the areas of two triangles on the same base is equal to the ratio of their heights.

(i) Suppose ABCD is the parallelogram and diagonal AC divides it into two triangles ABC and ADC.

In △ ABC and △ ADC,

⇒ AB = CD (Opposite sides of parallelogram are equal)

⇒ AD = BC (Opposite sides of parallelogram are equal)

⇒ AC = AC (Common side)

∴ △ ABC ≅ △ ADC (By S.S.S. axiom)

We know that,

Area of congruent triangle are equal.

∴ Area of △ ABC = Area of △ ADC.

Hence, proved that a diagonal divides a parallelogram into triangles of equal area.

(ii) From figure,

AQ and PQ are the altitude of triangle ABC and PRZ respectively wheres bases of both the triangles are equal i.e. BC = RZ = k (let).

We know that,

⇒ Area of triangle = $\dfrac{1}{2}$ × base × height

⇒ Area of △ ABC = $\dfrac{1}{2}$ × BC × AQ

⇒ Area of △ ABC = $\dfrac{1}{2} \times k \times AQ$ ……..(1)

⇒ Area of △ PRZ = $\dfrac{1}{2}$ × RZ × PQ

⇒ Area of △ PRZ = $\dfrac{1}{2} \times k \times PQ$ ……..(2)

Dividing equation (2) by (1), we get :

$\Rightarrow \dfrac{\text{Area of △ PRZ}}{\text{Area of △ ABC}} = \dfrac{\dfrac{1}{2} × k × PQ}{\dfrac{1}{2} × k × AQ} = \dfrac{PQ}{AQ}$.

Hence, proved that the ratio of the areas of two triangles of the same height is equal to the ratio of their bases.

(iii) We know that,

⇒ Area of triangle = $\dfrac{1}{2}$ × base × height

From figure,

⇒ Area of △ ABC = $\dfrac{1}{2}$ × AC × BM ……..(1)

⇒ Area of △ ADC = $\dfrac{1}{2}$ × AC × DN ……..(2)

Dividing equation (2) by (1), we get :

$\Rightarrow \dfrac{\text{Area of △ ADC}}{\text{Area of △ ABC}} = \dfrac{\dfrac{1}{2} × AC × DN}{\dfrac{1}{2} × AC × BM} = \dfrac{DN}{BM}$.

Hence, proved that the ratio of the areas of two triangles on the same base is equal to the ratio of their heights.