Given : A ∆ABC in which AD is the median and P is a point on AD

To prove: (i) ar(∆BDP) = ar(∆CDP),

(ii) ar(∆ABP) = ar(∆ACP).

(i)

In ∆BPC, PD is the median. Since median of a triangle divides the triangles into two equal areas

So, area(∆BDP) = area(∆CDP)----1

Hence proved

(ii)

In ∆ABC AD is the median

So, area(∆ABD) = area(∆ADC) ----2 and

area(∆BDP) = area(∆CDP) [from 1]

Now subtracting area(∆BDP) from ---2 , we have

area(∆ABD) - area(∆BDP) = area(∆ADC) - area(∆BDP)

area(∆ABD) - area(∆BDP) = area(∆ADC) - area(∆CDP) [since area(∆BDP) = area(∆CDP) from –1]

area(∆ABP) = area(∆ACP)

Hence proved.