Let OACB be paralelogram with O at the origin & OC a diagonal. Let D be the mid point of OA.We have to prove that BD & CO intersect in the same ratio.
I have proved ot using coordinate geometry by assigning coordinates .
But How can we prove it using vectors .
Let $\vec a $ and $\vec b $ be the position vectors of $A$ and $B$ respectively. Hence, $\overrightarrow{OA} =\vec a $ and $\overrightarrow{OB} =\vec b $. Then $\overrightarrow {OC} =\overrightarrow{OA} +\overrightarrow{OB} =\vec a+ \vec b $. Also, $\overrightarrow{OD} =0.5\vec a $.
Let $BD $ and $OC $ intersect at $E $ which divides $BD $ in $\alpha:1$ ratio, then $$ \overrightarrow {OE} = \frac {0.5 \alpha \vec a + \vec b}{\alpha +1} \tag {1} $$ Let $E $ divides $CO $ in the ratio $\beta :1$, then we have, $$\overrightarrow{OE} =\frac {\vec a +\vec b}{\beta +1} \tag {2} $$
Can you take it from it here?