Question

A point D is taken on the side BC of a $△ABC$ such that BD = 2DC. Prove that ar $\left(△ABD\right)=2ar\left(△ADC\right)$

Answer 1

Given : In $△ABC,$ D is a point on BC such that BD = 2 DC
AD is joined
To prove : ar $\left(△ABD\right)=2ar\left(△ADC\right)$
Construction : Draw $AL\perp BC$

Proof : area $\left(△ABD\right)=\frac{1}{2}\times base\times height=\frac{1}{2}BD\times AL$
and area $\left(△ADC\right)=\frac{1}{2}DC\times AL$
But $BD=2DC\Rightarrow DC=\frac{1}{2}BD\therefore area\left(△ADC\right)=\frac{1}{2}\left(\frac{1}{2}BD\right)\times AL=\frac{1}{2}[\frac{1}{2}BD\times AL]=\frac{1}{2}\left[ar\right(△ABD\left)\right]\therefore ar\left(△ADC\right)=2area\left(△ABD\right)$