Question

$D$ is a point on side BC of a $△ABC$ such that $BD:CD=2:1$, $O$ is the point mid point of AD. Find $ar\left(△OBD\right):ar\left(△ABC\right)$

Answer 1

Since $AD$ is the median of $△ABC$ and median divides a triangle into two triangles of equal area.

$\therefore$area of $△ABD=$area of $△ADC$

$\Rightarrow$area of $△ABD=\frac{1}{2}$area of $△ABC$ .......$\left(1\right)$

In $△ABD,BO$ is the median

$\therefore$ area of $△BOD=$area of $△BAO$

$\Rightarrow$ area of $△BOD=\frac{1}{2}$area of $△ABD$

$\Rightarrow$ area of $△BOD=\frac{1}{2}\times \frac{1}{2}$area of $△ABC$

$\Rightarrow$ area of $△BOD=\frac{1}{4}$area of $△ABC$

$\Rightarrow \frac{area\left(△BOD\right)}{area\left(△ABC\right)}=\frac{1}{4}$