In the adjoining figure, the point D divides the side BC of ∆ABC in the

Given: D is a point on BC of ∆ ABC, such that BD : DC =  m : n
To prove:  ar(∆ABD) : ar(∆ADC) = m : n
Construction: Draw AL ⊥ BC.
Proof: 

$\operatorname{ar}(\triangle A B D)=\frac{1}{2} \times B D \times A L$       .........(i)

$\operatorname{ar}(\triangle A D C)=\frac{1}{2} \times D C \times A L$      ..........(ii)

Dividing (i) by (ii), we get:

$\frac{\operatorname{ar}(\triangle A B D)}{\operatorname{ar}(\Delta A D C}=\frac{\frac{1}{2} \times B D \times A L}{\frac{1}{2} \times D C \times A L}$

$=\frac{B D}{D C}$

$=\frac{m}{n}$

∴ ar(∆ABD) : ar(∆ADC) = m : n