Question:
In the adjoining figure, the point D divides the side BC of ∆ABC in the ratio m : n. Prove that ar(∆ABD) : ar(∆ADC) = m : n.
Solution:
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