In the diagram ABCD is a parallelogram. The diagonal AC and BD intersect at E. $\overrightarrow{AF} = 2p$ and $\overrightarrow{AB} = 6p + 4q$. F is a point on AD such that $\dfrac{AF}{AD} = \dfrac{1}{3}$. G is the mid-point of AE.

1. Find $\overrightarrow{AD}$ from the given ratio on side $AD$.

Since $F$ lies on $AD$ and $\dfrac{AF}{AD}=\dfrac13$,
$\overrightarrow{AF}=\frac13\overrightarrow{AD}.$
Given $\overrightarrow{AF}=2p$,
$\overrightarrow{AD}=3(2p)=6p.$

2. Use the diagonal property of a parallelogram to find $\overrightarrow{DE}$.

In parallelogram $ABCD$,
$\overrightarrow{DB}=\overrightarrow{DA}+\overrightarrow{AB}.$
Since $\overrightarrow{DA}=-\overrightarrow{AD}=-6p$ and $\overrightarrow{AB}=6p+4q$,
$\overrightarrow{DB}=-6p+(6p+4q)=4q.$
The diagonals of a parallelogram bisect each other, so $E$ is the midpoint of $DB$.
Hence
$\overrightarrow{DE}=\frac12\overrightarrow{DB}=\frac12(4q)=2q.$

1. Find $\overrightarrow{AD}$ using the given ratio $\dfrac{AF}{AD}=\dfrac{1}{3}$.

Since $\dfrac{AF}{AD}=\dfrac{1}{3}$ and $\overrightarrow{AF}=2p$,
$\overrightarrow{AD}=3\overrightarrow{AF}=3(2p)=6p.$

2. Find $\overrightarrow{AG}$ using the diagonal $AC$ and the midpoint properties of $E$ and $G$.

In parallelogram $ABCD$,
$\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{AD}.$
So
$\overrightarrow{AC}=(6p+4q)+6p=12p+4q.$
Since diagonals of a parallelogram bisect each other, $E$ is the midpoint of $AC$:
$\overrightarrow{AE}=\frac{1}{2}\overrightarrow{AC}=\frac{1}{2}(12p+4q)=6p+2q.$
Since $G$ is the midpoint of $AE$,
$\overrightarrow{AG}=\frac{1}{2}\overrightarrow{AE}=\frac{1}{2}(6p+2q)=3p+q.$

1. Express $\overrightarrow{BG}$ using a route through point $A$.

$\overrightarrow{BG}=\overrightarrow{BA}+\overrightarrow{AG}$
Since $\overrightarrow{AB}=6p+4q$,
$\overrightarrow{BA}=-(6p+4q).$

2. Substitute the known vector $\overrightarrow{AG}$ and simplify.

From part (b),
$\overrightarrow{AG}=3p+q.$
Therefore,
$\overrightarrow{BG}=-(6p+4q)+(3p+q)$
$=-6p-4q+3p+q$
$=-3p-3q.$

1. Find a second vector from $B$ to a point on the suspected straight line.

$\overrightarrow{BF}=\overrightarrow{BA}+\overrightarrow{AF}$

$=-(6p+4q)+2p$

$=-4p-4q$

2. Compare $\overrightarrow{BF}$ with the known vector $\overrightarrow{BG}$ to prove collinearity.

From part (c),

$\overrightarrow{BG}=-3p-3q$.

But

$\overrightarrow{BF}=-4p-4q$.

So

$\overrightarrow{BG}=\dfrac{3}{4}\overrightarrow{BF}$.

Since $\overrightarrow{BG}$ is a scalar multiple of $\overrightarrow{BF}$, $BG$ is parallel to $BF$. As both vectors start from $B$, the points $B$, $G$ and $F$ lie on a straight line.

1. Relate the height of triangle $ABF$ to the height of parallelogram $ABCD$.

Since $\dfrac{AF}{AD}=\dfrac{1}{3}$ and $F$ lies on $AD$, the perpendicular distance of $F$ from $AB$ is $\dfrac{1}{3}$ of the perpendicular height of parallelogram $ABCD$ from base $AB$.

2. Compare the area of triangle $ABF$ with the area of parallelogram $ABCD$.

Let the base $AB$ be $b$ and the height of parallelogram $ABCD$ be $h$.

$\text{Area of parallelogram }ABCD=bh$

$\text{Area of triangle }ABF=\frac{1}{2}\times b\times \frac{h}{3}=\frac{bh}{6}$

Hence,

$\frac{\text{area of triangle }ABF}{\text{area of parallelogram }ABCD}=\frac{\frac{bh}{6}}{bh}=\frac{1}{6}.$