OABC is a trapezium. $\overrightarrow{OA}=a$, $\overrightarrow{OC}=2c$ and $\overrightarrow{CB}=2a$. $OP : PB = 1 : 3$.

1. Express $\overrightarrow{OB}$ by adding the vectors along a known route from $O$ to $B$.

Using the route $O \to C \to B$,
$\overrightarrow{OB}=\overrightarrow{OC}+\overrightarrow{CB}.$
Given $\overrightarrow{OC}=2c$ and $\overrightarrow{CB}=2a$,
$\overrightarrow{OB}=2c+2a=2a+2c.$

1. Find $\overrightarrow{OP}$ using the given ratio on $OB$.

Since $OP:PB=1:3$, $P$ divides $OB$ into $4$ equal parts, so
$\overrightarrow{OP}=\frac{1}{4}\overrightarrow{OB}.$
From part (i), $\overrightarrow{OB}=2a+2c$.
$\overrightarrow{OP}=\frac{1}{4}(2a+2c)=\frac{1}{2}a+\frac{1}{2}c.$

2. Express $\overrightarrow{CP}$ by adding vectors along the route $C \to O \to P$.

$\overrightarrow{CP}=\overrightarrow{CO}+\overrightarrow{OP}.$
Given $\overrightarrow{OC}=2c$, so $\overrightarrow{CO}=-2c$.
$\overrightarrow{CP}=-2c+\left(\frac{1}{2}a+\frac{1}{2}c\right)$
$\overrightarrow{CP}=\frac{1}{2}a-\frac{3}{2}c.$

1. Find a second vector from $C$ to another point on the supposed line, namely $\overrightarrow{CA}$.

$\overrightarrow{CA}=\overrightarrow{CO}+\overrightarrow{OA}$
Since $\overrightarrow{OC}=2c$, $\overrightarrow{CO}=-2c$.

$\overrightarrow{CA}=-2c+a=a-2c$

2. Compare $\overrightarrow{CP}$ with $\overrightarrow{CA}$ to decide whether the points are collinear.

From part (ii),
$\overrightarrow{CP}=\frac12a-\frac32c.$

If $C$, $P$ and $A$ were collinear, then $\overrightarrow{CP}$ would be a scalar multiple of $\overrightarrow{CA}$.

But
$\overrightarrow{CA}=a-2c.$
The coefficients are not in the same ratio:
$\frac{\frac12}{1}=\frac12 \quad \text{but} \quad \frac{-\frac32}{-2}=\frac34.$
So $\overrightarrow{CP}$ is not parallel to $\overrightarrow{CA}$. Therefore $C$, $P$ and $A$ do not lie on a straight line.

1. Compare the area of triangle $OBC$ with the area of trapezium $OABC$ using the common height.

Since $OA \parallel CB$, let the perpendicular height of the trapezium be $h$.

For triangle $OBC$, take $CB$ as the base:
$\text{area of }\triangle OBC=\frac12(2a)h=ah.$

For trapezium $OABC$:
$\text{area of trapezium }OABC=\frac12(a+2a)h=\frac32ah.$

Hence