Show that the only way these ratios can be equal is if both ratios equal one, and show that in that case, $\overline{XY}$ is parallel to $\overline{BC}$.
I have $\overline{AB}$$\overline{XP}$$\overline{CY}$=$\overline{BX}$$\overline{PY}$$\overline{AC}$ but I'm not sure what to do with it.
On
From $$\frac{XP}{PY} = \frac{\triangle AXP}{\triangle AYP} = \frac{\frac{1}{2} AX \cdot AP \cdot \sin \angle XAP}{\frac{1}{2} AY \cdot AP \cdot \sin \angle YAP} = \frac{AX \cdot \sin \angle XAP}{AY \cdot \sin \angle YAP}$$ and $$\frac{BQ}{CQ} = \frac{\triangle ABQ}{\triangle ACQ} = \frac{\frac{1}{2} AB \cdot AQ \cdot \sin \angle BAQ}{\frac{1}{2} AC \cdot AQ \cdot \sin \angle CAQ} = \frac{AB \cdot \sin \angle XAP}{AC \cdot \sin \angle YAP}$$
We have $$\frac{XP}{YP}=\frac{BQ}{CQ} \iff \frac{AX}{AY}=\frac{AB}{AC} \iff XY \parallel BC \iff \frac{AX}{XB} = \frac{AY}{YC}$$
By Ceva's Theorem, we have $$\frac{XP}{PY}=\frac{BQ}{CQ}=\frac{AX}{XB} \cdot \frac{BQ}{CQ} \cdot \frac{CY}{AY}=1$$
Let the intersection of $PQ$ and $XC$ be $O$. By Menelaus theorem,
$${PX\over XY}\cdot {u\over t+u}\cdot {AO\over PO}=1$$
$${PY\over XY}\cdot {s\over r+s}\cdot {AO\over PO}=1 $$
Divide the two equations,
$${PX\over PY}={s(t+u)\over u(r+s)}$$
By Ceva theorem,
$${BQ\over QC}\cdot {u\over t}\cdot {r\over s}=1$$
Hence
$${BQ\over QC}={st\over ur}$$
If they are equal, $${st+su\over ur+us}={st\over ur}={su\over us}=1$$
Since $st=ur$ we have ${s\over u}={r\over t}$ hence parallel.