Opposite angles of a quadrilateral $ABCD$ are equal. If $AB = 4 \, cm$,determine $CD$ (in $cm$).

$E$ and $F$ are respectively the mid-points of the non-parallel sides $AD$ and $BC$ of a trapezium $ABCD$. Prove that $EF \parallel AB$ and $EF = \frac{1}{2}(AB + CD)$.

Prove that the line segment joining the midpoints of the diagonals of a trapezium is parallel to the parallel sides of the trapezium and is equal to half the difference of the lengths of the parallel sides.

Perimeter of rectangle $PQRS$ is $40\, cm$. If $PQ : QR = 3 : 5$,then $RS = \ldots \ldots \ldots \, cm$.

The quadrilateral formed by joining the mid-points of the sides of a quadrilateral $PQRS,$ taken in order,is a rhombus,if

$XYZW$ is a rhombus. If the diagonals $XZ$ and $YW$ intersect at $P$,then $\angle XPY = \ldots \ldots \ldots$ (in $^o$)