2016 AMC 10B Problem 19

We have $AE=AB-EB=4$ and $FC=DC-DF=3.$ Since $AE\parallel FC,$ the triangles $AEP$ and $CFP$ are similar. Therefore $$\frac{PF}{PE}=\frac{FC}{AE}=\frac34,$$ so $\frac{PF}{EF}=\frac37.$ Extend $AG$ to meet line $CD$ at $X.$ Because $BG=3,$ similarity gives $$\frac{AD}{DX}=\frac{BG}{AB},\qquad \frac4{DX}=\frac35,$$ and hence $DX=\frac{20}{3}$. Thus $FX=DX-DF=\frac{14}{3}.$ Since $AE\parallel FX,$ triangles $AEQ$ and $XFQ$ are similar, so $$\frac{QF}{QE}=\frac{FX}{AE}=\frac76.$$ Consequently $\frac{QF}{EF}=\frac7{13}.$ It follows that $$\begin{aligned} \frac{PQ}{EF}&=\frac{QF}{EF}-\frac{PF}{EF} \\ &=\frac7{13}-\frac37=\frac{10}{91}. \end{aligned}$$ Thus, the correct answer is D .