Show $AC \perp BD$ first - this can be done by projecting $(A, B; L, \infty )$ at $E$, onto $FD$. By Newton's Theorem and right triangle similarity, we will see that $MC=MD=ME$. This will give $AC \perp BD$. Apparently this is Brahmagupta's Theorem! Now easy angle chase - it will give us $\triangle AEF \sim \triangle EPM$. We can calculate $\frac{EP^2}{EQ}$ easily now - set $Z$ so that $EP \perp PZ$ and $Z \in EM$. Now $\triangle EPZ \sim \triangle AEB$, so $\frac{EP^2}{EQ}= EZ = \frac{AB \cdot EP}{AE} = \frac{AB \cdot EM}{AF} = \frac{AB \cdot AC}{2 \cdot AF}$. Using Ptolemy to simplify the RHS will give us the desired equality. Apparently a straightforward trig bash works as well. I wish geo was harder.