Given a trapezoid $ABDC$, and line segment $PQ$ where $P$ and $Q$ are points on $AC$ and $BD$, respectively, s.t. $AB||PQ||CD$. Suppose $PQ$ intersects $BC$ at $K$ and $AD$ at $J$, prove that $PK$ and $JQ$ are equal.
Below is an extreme case where $K$ and $J$ are at point $O$. Part of the reason must be because the lines are parallel. Also, I checked on GeoGebra and seen that they are numerically equal,i.e. $PK=JQ$.
As $\triangle ABC\sim\triangle PKC$, $AB:PK=BC:KC$.
As $\triangle ABD\sim\triangle JQD$, $AB:JQ=BD:QD$.
As $\triangle BCD\sim\triangle BKQ$, $BC:BK=BD:BQ$.
So, $BC:KC=BC:(BC-BK)=BD:(BD-BQ)=BD:QD$
Therefore, $AB:PK=AB:JQ$ and hence $PK=JQ$