Question

Let $OACB$ be a parallelogram with $O$ at the origin and $OC$ a diagonal. Let $D$ be the mid-point of $OA$. Using vector methods prove that $BD$ and $CO$ intersect in the same ratio. Determine this ratio.

Answer 1

$OACB$ is a parallelogram with $O$ as origin. Let
$\overrightarrow{OA}=\overrightarrow{a},\overrightarrow{OB}=\overrightarrow{b},\overrightarrow{OC}=\overrightarrow{a}+\overrightarrow{b}$

and
$\overrightarrow{OD}=\frac{\overrightarrow{a}}{2}$
$\overrightarrow{CO}$ and $\overrightarrow{BD}$ meets at $P$
$\therefore \overrightarrow{OP}=\frac{\lambda \cdot 0+1(\overrightarrow{a}+\overrightarrow{b})}{\lambda +1}$ [along $\overrightarrow{OC}]$
$\Rightarrow \overrightarrow{OP}=\frac{\overrightarrow{a}+\overrightarrow{b}}{\lambda +1}.....$(i)
Again, $\overrightarrow{OP}=\frac{\mu \left(\frac{\overrightarrow{a}}{2}\right)+1(\overrightarrow{b})}{\mu +1}$ [along $\overrightarrow{BD}$ ] ......(ii)
From Eqs. (i) and (ii),
$\frac{\overrightarrow{a}+\overrightarrow{b}}{\lambda +1}=\frac{\mu \overrightarrow{a}+2\overrightarrow{a}}{2(\mu +1)}\Rightarrow \frac{1}{\lambda +1}=\frac{\mu }{2(\mu +1)}$
and $\frac{1}{\lambda +1}=\frac{1}{\mu +1}$
On solving, we get $\mu =\lambda =2$
Thus, required ratio is $2:1$.