$PQRS$ is a trapezium with bases $PQ$ and $RS$. $T$ is the point where diagonals $PR$ and $QS$ intercept. $U$ is a point located in side $SP$. $V$ is the point where $QS$ and $RU$ intercept. $W$ is the point where $PV$ and $TU$ intercept. $X$ is the point where $SW$ and $PQ$ intercept.
Is $X$ the middle of $QP$?
$X$ middle of 
The Euclidean Geometry way:
Menelaus: ${TR\over PR}\cdot{PU\over SU}\cdot{SV\over TV} = 1$
Ceva: ${TY\over PY}\cdot{PU\over SU}\cdot{SV\over TV} = 1$
Therefore ${TR\over PR}={TY\over PY}$
Draw $PZ$ parallel to $QS$ intersecting $SX$ at $Z$.
Now ${TR\over PR}={TS\over QS}$, ${TY\over PY}={TS\over PZ}$
Therefore $QS=PZ$ and $PSQZ$ is a parallelogram. Therefore $QX=XP$.