Emily from Wimbledon High School solved this by using vectors to find the cartesian co-ordinates of the points.

Let $B$ be the origin. Then align the triangles so $A$ is vertically above $B$, so the co-ordinates of $A$ are $(0,a)$.By the right angle $C$ is horizontally across from $B$, so $C$ is $(c,0)$. Then $Q$ is $(\frac{c}{2},\frac{a}{2})$.

By some basic trigonometry we can find the coordinates of $D$, as $BCD$ is an equilateral triangle we know all its angles are $60\circ$ and all its sides are equal. By considering the perpendicular from $BC$ to $D$, we find the horizontal component of $D$ is $d\times cos60$ and the length $BD=d$ is equal to $BC=c$ so $d\times cos60=\frac{c}{2}$. The vertical component is $d\times sin60=c\frac{3^(1/2)}{2}$.

But the vertical component does not matter, the point is that $D$ and $Q$ have the same horizontal component, so a straight line between them will cross through $BC$ and will be vertical, and hence parallel to $BA$