In the given figure, AD is a median of ΔABC and P is a point on AC such that :

ar (ΔADP) : ar (ΔABD) = 2 : 3.

Find :

(i) AP : PC

(ii) ar (ΔPDC) : ar (ΔABC)

(i) From figure,

Let DE be altitude on base AC.

Median divides a triangle into two triangles of equal area.

AD is the median of ∆ABC,

Area of ∆ABD = Area of ∆ADC = $\dfrac{1}{2}$ Area of ∆ABC …..(1)

It is given that,

⇒ area of ∆ADP : area of ∆ABD = 2 : 3

⇒ area of ∆ADP : area of ∆ADC = 2 : 3

$\Rightarrow \dfrac{\text{Area of Δ ADP}}{\text{Area of Δ ADC}} = \dfrac{2}{3} \\[1em] \Rightarrow \dfrac{\dfrac{1}{2}\times AP \times DE}{\dfrac{1}{2}\times AC \times DE} = \dfrac{2}{3} \\[1em] \Rightarrow \dfrac{AP}{AC} = \dfrac{2}{3}.$

Let AP = 2x and AC = 3x.

From figure,

PC = AC - AP = 3x - 2x = x.

$\dfrac{AP}{PC} = \dfrac{2x}{x} = \dfrac{2}{1}$.

Hence, AP : PC = 2 : 1.

(ii) We know that,

PC : AC = x : 3x = 1 : 3

So,

$\Rightarrow \dfrac{\text{Area of Δ PDC}}{\text{Area of Δ ADC}} = \dfrac{\dfrac{1}{2}\times PC \times DE}{\dfrac{1}{2}\times AC \times DE} \\[1em] \Rightarrow \dfrac{\text{Area of Δ PDC}}{\text{Area of Δ ADC}} = \dfrac{PC}{AC} \\[1em] \Rightarrow \dfrac{\text{Area of Δ PDC}}{\text{Area of Δ ADC}} = \dfrac{x}{3x} \\[1em] \Rightarrow \dfrac{\text{Area of Δ PDC}}{\text{Area of Δ ADC}} = \dfrac{1}{3} \text{ …..(2)}$

Since, AD is median of ∆ABC so,

area of Δ ADC = $\dfrac{1}{2}$ area of Δ ABC

Substituting above value in equation (2) we get,

$\Rightarrow \dfrac{\text{Area of Δ PDC}}{\dfrac{1}{2} \text{ area of Δ ABC}} = \dfrac{1}{3} \\[1em] \Rightarrow \dfrac{\text{Area of Δ PDC}}{\text{ area of Δ ABC}} = \dfrac{1}{3} \times \dfrac{1}{2} = \dfrac{1}{6}.$

Hence, proved that area of △PDC : area of △ABC = 1 : 6.