Q:

Proving two lines are parallel with intersections and midpoints

To prove : 

Fix points $A,C$ and set point $B$ to be the midpoint of segment $AC$. Fix 
  point $Y$ (anywhere) and consider an arbitrary point $X$ on line $YB$. If $P$ and $Q$ are the intersection points of two lines $AY \cap CX$ , $CY \cap AX$ respectively, show that line $PQ$ is parallel to line $AC$.

I first proved converse, to see what geometry this figure has in structure.And that if trapezoid $APQC$ is given and If $Y$ and $X$ are the intersection points of the sides of the trapezoid $AP\cap CQ$ and $AQ\cap PC$ respectively, then line $YX$ passes through midpoints of segment $AC$ and $PQ$. However this didn't help much.
Im trying to solve this by looking for some formula related with a point $B$.
So $PQ$ and $AC$ will be parallel when $B$ is midpoint of $AC$.
I took hours with this problem and I couldn't find a clue.
What would be the proof look like? 

A:

Hint: Recall Ceva's theorem:

If, in triangle $ABC$, points $D,E,F$ are on segments $BC,CA,AB$ respectively, then $AD,BE,CF$ concur if and only if
$$\frac{AF}{FB}\cdot\frac{BD}{DC}\cdot\frac{CE}{EA}=1.$$

Now, you know that, in triangle $ACY$, cevians $YB$, $AQ$, and $CP$ concur. What length ratios do you get when you apply Ceva's theorem? What length ratios would result in $PQ$ being parallel to $AC$?

