We know that
Area of △ ABD = ½ × BD × AL
Area of △ ADC = ½ × DC × AL
It is given that BD: DC = m: n
It can be written as
BD = DC × m/n
We know that
Area of △ ABD = ½ × BD × AL
By substituting BD
Area of △ ABD = ½ × (DC × m/n) × AL
So we get
Area of △ ABD = m/n × (1/2 × DC × AL)
It can be written as
Area of △ ABD = m/n × (Area of △ ADC))
We know that
Area of △ ABD/ Area of △ ADC = m/n
We can write it as
Area of △ ABD: Area of △ ADC = m: n
Therefore, it is proved that ar (△ ABD): ar (△ ADC) = m: n.