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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 $\left(\frac{c}{2},\frac{a}{2}\right)$.
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 $d\times \cos{60}$ and the length $BD=d$ is equal to $BC=c$ so $d\times \cos{60}=\frac{c}{2}$. The vertical component is $d\times \sin{60}=c\frac{3^{1/2}}{2}$.
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 $BA$