OBC, ADC, AEB and OED are straight lines.
$\overrightarrow{OA}=a$ and $\overrightarrow{OB}=b$.
$OB=\dfrac{1}{3}OC$, $AE:EB=3:2$ and $OE:ED=1:1$.

1. Express $\overrightarrow{AB}$ in terms of $a$ and $b$.

$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=b-a$

2. Use the ratio $AE:EB=3:2$ to find $\overrightarrow{OE}$.

Since $AE:EB=3:2$, $\overrightarrow{AE}=\frac{3}{5}\overrightarrow{AB}=\frac{3}{5}(b-a).$ Hence $\overrightarrow{OE}=\overrightarrow{OA}+\overrightarrow{AE}=a+\frac{3}{5}(b-a)=\frac{2}{5}a+\frac{3}{5}b.$

1. Compare the area of $\triangle ODC$ with the area of $\triangle OBE$.

Since $OB=\dfrac13OC$, we have $OC=3OB$.

Since $OE:ED=1:1$, $E$ is the midpoint of $OD$, so $OD=2OE$.

Taking bases on the same straight line $OBC$ and corresponding heights from $E$ and $D$,
$\text{area of }\triangle ODC=3\times 2\times \text{area of }\triangle OBE=6\times \text{area of }\triangle OBE.$

2. Find the required ratio by subtracting $\triangle OBE$ from $\triangle ODC$.

The quadrilateral $BEDC$ is the part of $\triangle ODC$ outside $\triangle OBE$.

Let $\text{area of }\triangle OBE=x$.

Then
$\text{area of }BEDC=6x-x=5x.$

Therefore
$\frac{\text{area of }\triangle OBE}{\text{area of }BEDC}=\frac{x}{5x}=\frac15.$