如图所示，OABC 是一个平行四边形，D 是 BC 的中点。BE 的延长线与 OC 的延长线交于点 F。 已知 BE : BF = 1 : 3，OC : OF = 1 : 2，OA = a，OC = c。

1. Express \(\overrightarrow{BF}\) using position vectors from \(O\).

Since $OABC$ is a parallelogram,
$\overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{OC}=\mathbf{a}+\mathbf{c}.$
Given $OC:OF=1:2$,
$\overrightarrow{OF}=2\overrightarrow{OC}=2\mathbf{c}.$
Hence
$\overrightarrow{BF}=\overrightarrow{OF}-\overrightarrow{OB}=2\mathbf{c}-(\mathbf{a}+\mathbf{c})=\mathbf{c}-\mathbf{a}.$

1. Express $\overrightarrow{BE}$ using the ratio $BE:BF=1:3$ and the known vector $\overrightarrow{BF}$.

Since $BE:BF=1:3$ and $B,E,F$ are collinear,
$\overrightarrow{BE}=\frac{1}{3}\overrightarrow{BF}.$
Using $\overrightarrow{BF}=\mathbf{c}-\mathbf{a}$,
$\overrightarrow{BE}=\frac{1}{3}(\mathbf{c}-\mathbf{a}).$

2. Add vectors along the path $A \to B \to E$ and simplify.

In parallelogram $OABC$,
$\overrightarrow{AB}=\overrightarrow{OC}=\mathbf{c}.$
Hence
$\overrightarrow{AE}=\overrightarrow{AB}+\overrightarrow{BE}$
$=\mathbf{c}+\frac{1}{3}(\mathbf{c}-\mathbf{a})$
$=\frac{4}{3}\mathbf{c}-\frac{1}{3}\mathbf{a}.$

1. Express the position vector of midpoint $D$ using the position vectors of $B$ and $C$.

In parallelogram $OABC$,
$\overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{AB}=\mathbf{a}+\mathbf{c}$ since $\overrightarrow{AB}=\overrightarrow{OC}=\mathbf{c}$.

Since $D$ is the midpoint of $BC$,
$\overrightarrow{OD}=\frac{1}{2}\left(\overrightarrow{OB}+\overrightarrow{OC}\right)$
$=\frac{1}{2}\left((\mathbf{a}+\mathbf{c})+\mathbf{c}\right)$
$=\frac{1}{2}\mathbf{a}+\mathbf{c}.$

1. Find $\overrightarrow{OE}$ using the earlier result for $\overrightarrow{AE}$.

From part (a)(ii),
$\overrightarrow{AE}=\dfrac{4}{3}\mathbf{c}-\dfrac{1}{3}\mathbf{a}$.

Hence
$\overrightarrow{OE}=\overrightarrow{OA}+\overrightarrow{AE}$
$=\mathbf{a}+\left(\dfrac{4}{3}\mathbf{c}-\dfrac{1}{3}\mathbf{a}\right)$
$=\dfrac{2}{3}\mathbf{a}+\dfrac{4}{3}\mathbf{c}.$

2. Compare $\overrightarrow{OE}$ with $\overrightarrow{OD}$.

From part (a)(iii),
$\overrightarrow{OD}=\dfrac{1}{2}\mathbf{a}+\mathbf{c}.$

Now
$\dfrac{4}{3}\overrightarrow{OD}=\dfrac{4}{3}\left(\dfrac{1}{2}\mathbf{a}+\mathbf{c}\right)$
$=\dfrac{2}{3}\mathbf{a}+\dfrac{4}{3}\mathbf{c}.$

Therefore
$\overrightarrow{OE}=\dfrac{4}{3}\overrightarrow{OD}.$

3. Conclude whether the points are collinear.

Since $\overrightarrow{OE}$ is a scalar multiple of $\overrightarrow{OD}$, the vectors are parallel and both start from $O$.

Therefore, $O$, $D$ and $E$ lie on a straight line.

1. Relate the base $CD$ of the triangle to the base $OA$ of the parallelogram.

Since $D$ is the midpoint of $BC$ and $BC=OA$ in the parallelogram,

$CD=\frac{1}{2}BC=\frac{1}{2}OA.$

2. Relate the height of triangle $CDE$ to the height of the parallelogram and calculate the area ratio.

From part (b),

$\overrightarrow{OE}=\frac{4}{3}\overrightarrow{OD}.$

Hence $E$ lies beyond $D$ on the straight line $OD$. Since $D$ is on $BC$, the perpendicular distance of $E$ from $BC$ is $\frac{1}{3}$ of the height of the parallelogram.

Therefore,

$=\frac{\frac{1}{2}\times \frac{1}{2}OA\times \frac{1}{3}h}{OA\times h}$

$=\frac{1}{12}.$