imc

First, we will define point $D$ as the origin. Then, we will find the equations of the following three lines: $AG$, $AC$, and $EF$. The slopes of these lines are $-\frac{3}{5}$, $-\frac{4}{5}$, and $2$, respectively. Next, we will find the equations of $AG$, $AC$, and $EF$. They are as follows: $$AG=f(x)=-\frac{3}{5}x+4$$ $$AC=g(x)=-\frac{4}{5}x+4$$ $$EF=h(x)=2x-4$$ After drawing in altitudes to $DC$ from $P$, $Q$, and $E$, we see that $\frac{PQ}{EF}=\frac{P^{\prime }{Q}^{\prime }}{E^{\prime }F}$ because of similar triangles, and so we only need to find the x-coordinates of $P$ and $Q$. Finding the intersections of $AC$ and $EF$, and $AG$ and $EF$ gives the x-coordinates of $P$ and $Q$ to be $\frac{20}{7}$ and $\frac{40}{13}$. This means that $P^{\prime }{Q}^{\prime }=D{Q}^{\prime }-D{P}^{\prime }=\frac{40}{13}-\frac{20}{7}=\frac{20}{91}$. Now we can find $\frac{PQ}{EF}=\frac{P^{\prime }{Q}^{\prime }}{E^{\prime }F}=\frac{\frac{20}{91}}{2}=\boxed{\frac{10}{91}}$