Quadrilateral $PQRS$ is a trapezium. Quadrilateral $PQRS$ is a trapezium ($PQ||SR$ and $PQ>SR$). 
Points $X$ and $Y$ are mid points of diagonal $PR$ and $QS$ respectively. 
Prove that segment $XY$ parallel to side $PQ$  and $XY= \frac{1}{2}(PQ-SR)$.
I tried this question whole day but couldn't find answer please help me in doing it
 A: Draw the line passing through point $Q$ and parallel to edge $PS$ and let that line intersect the line $SR$ at point $N$ (see picture). Then $PQNS$ is a parallelogram because edge $SN$ is parallel to edge $PQ$ and edge $PS$ is parallel to edge $QN$. Consequently, the intersection point of the diagonals $PN$ and $QS$ is the midpoint of $QS$ (and $PN$), which is point $Y$. Furthermore,  $XY$ is a mid-segment in triangle $NPR$ parallel to $RN$ and half of its size. This means that $XY$ is parallel to $PQ$, because $PQ$ is parallel to $RN$, and $XY = \frac{1}{2} RN$. However, $RN = SN - SR$ and since $SN = PQ$ we get
$$XY = \frac{1}{2} (PQ - SR) \,\,\, \text{ as well as } \,\,\, XY \, || \, PQ$$     

A: Let $a||PQ$ and $X\in a$.
Let $a\cap QR=\{M\}$ and $a\cap SP=\{N\}$.
Since $X$ a middle point of $PR$, we get that $M$ is a middle point of $QR$ 
and
since $PQRS$ is trapezium, we get that  $N$ is a middle point of $PS$ and $Y\in a$.
Thus, $XY||PQ$ and $$XY=NY-NX=\frac{PQ}{2}-\frac{SR}{2}=\frac{PQ-SR}{2}$$ 
