Let $PRUS$ be a trapezium such that $\angle PSR = 2\angle QSU$ and $\angle SPU = 2 \angle UPR$. Let $Q$ and $T$ be on $PR$ and $SU$ respectively such that $SQ$ and $PU$ bisect $\angle PSR$ and $\angle SPU$ respectively. Let $PT$ meet $SQ$ at $E$. The line through $E$ parallel to $SR$ meets $PU$ in $F$ and the line through $E$ parallel to $PU$ meets $SR$ in $G$. Let $FG$ meet $PR$ and $SU$ in $K$ and $L$ respectively. Prove that $KF$ = $FG$ = $GL$.