Prove that any line parallel to the sides of a trapezium divides the oblique side proportionally.

$ABCD$ is a quadrilateral; $P, Q, R$ and $S$ are the points on the sides $AB, BC, CD$ and $DA$ respectively such that $AP:PB=AS:SD=CQ:QB=CR:RD$. Prove that $PQRS$ is a parallelogram

On three-line segments $OA, OB$ and $OC$ points $L, M, N$ respectively are so chosen that $LM\parallel AB$ and $MN\parallel BC$ but neither $L, M, N$ nor $A, B, C$ are collinear. Show that $LN\parallel AC$.

$ABCD$ is a parallelogram and $P$ is a point on the side $BC$. $DP$ when produced meets $AB$ produced at $L$, Prove that $\frac{DP}{LP}=\frac{DC}{BL}$

$ABCD$ is a parallelogram and $P$ is a point on the side $BC$. $DP$ when produced meets $AB$ produced at $L$, Prove that $\frac{DL}{DP}=\frac{AL}{DC}$