Question

The base BC of the $\Delta ABC$ is divided at D so that
$BD=\frac{1}{2}BC$. Prove that ar $\left(\Delta ABD\right)=\frac{1}{2}ar\left(\Delta ADC\right)$.

Answer 1

Draw $AX\perp BC$
$BD=\frac{1}{2}DC$
$\therefore DC=2BD$ ...(1)
BD+DC=BC
$\therefore BD+2BD=BC$ ...(From (1))
$\therefore 3BD=BC$
or $BD=\frac{1}{3}BC$ ....(2)
$ar\left(\Delta ABC\right)=\frac{1}{2}\times AX\times BC$ ...(3)
$ar\left(\Delta ABD\right)=\frac{1}{2}\times AX\times BD$
$\frac{1}{2}\times AX\times \frac{1}{3}BC...\left[From\right(2\left)\right]$
$\frac{1}{3}[\frac{1}{2}\times AX\times BC]$
$\frac{1}{3}ar\left(\Delta ABC\right)$ ....[From (3)]
$\Rightarrow ar\left(\Delta ABD\right)=\frac{1}{3}ar\left(\Delta ABC\right)$