In the figure (3) given below, AD is a median of △ABC and P is a point in AC such that area of △ADP : area of △ABD = 2 : 3. Find

(i) AP : PC

(ii) area of △PDC : area of △ABC

(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

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

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,

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

Since, AD is median of ∆ABC so,

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

Substituting above value in 1 we get,

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

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