Question

In the adjoining figure, the point $D$ divides the side $BC$ of $△ABC$ in the ratio $m:n$. Prove that $ar\left(△ABD\right):ar\left(△ADC\right)=m:n$.

Answer 1

We know that

Area of $△ABD=\frac{1}{2}\times BD\times AL$

Area of $△ADC=\frac{1}{2}\times DC\times AL$

It is given that $BD:DC=m:n$

It can be written as

$BD=DC\times \frac{m}{n}$

We know that

Area of $△ABD=\frac{1}{2}\times BD\times AL$

By substituting $BD$

Area of $△ABD=\frac{1}{2}\times (DC\times \frac{m}{n})\times AL$

so we get

Area of $△ABD=\frac{m}{n}\times (\frac{1}{2}\times DC\times AL)$

It can be written as

Area of $△ABD=\frac{m}{n}\times$ (Area of $△ADC\left)\right)$

We know that

Area of $△ABD$/ Area of $△ADC=\frac{m}{n}$

We can write it as

Area of $△ABD$: Area of $△ADC=m:n$

Therefore, it is proved that $ar\left(△ABD\right):ar\left(△ADC\right)=m:n$.