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

1.	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.
Answer» 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.

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.

Answer»

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.