3.41: Vector geometry in a parallelogram

The diagram shows the parallelogram $OABC$ with $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OC} = \mathbf{c}$. Point $D$ lies on $CB$ such that $CD:DB = 3:1$. Extended, $AB$ and $OD$ meet at $E$, with $\overrightarrow{OE} = h\overrightarrow{OD}$ and $\overrightarrow{BE} = k\overrightarrow{AB}$.