Problem source

Show that the line through the points with position
vectors $\bf x$ and $\bf y$ has equation 
\[{\bf r} = (1-\alpha){\bf x} +\alpha {\bf y}\,,
\]
where $\alpha$ is a scalar parameter.
The sides $OA$ and $CB$ of a trapezium $OABC$ are parallel, and $OA>CB$.
The point $E$ on $OA$ is such that $OE : EA = 1:2$, and $F$ is the midpoint of 
$CB$. The point $D$ is the intersection of $OC$ produced and $AB$ produced;
the point $G$ is the intersection of $OB$ and $EF$; and the point $H$ 
is the intersection of $DG$ produced and $OA$. Let $\bf a$ and $\bf c$ be the 
position vectors of the points $A$ and $C$, respectively, with respect to
the origin $O$.
\begin{questionparts}
\item Show that $B$ has position vector $\lambda {\bf a} + {\bf c}$ for
some scalar parameter $\lambda$.
\item Find, in terms of $\bf a$, $\bf c$ and $\lambda$
only, the position vectors of $D$, $E$, $F$, $G$ and $H$.
Determine the ratio  $OH:HA$.
\end{questionparts}