In the quadrilateral given below, $AD=BC.P,Q,R$ and $S$ are mid-points of $AB,BD,CD$ and $AC$ respectively. Prove that $PQRS$ is a rhombus.

$AD=BC$ [ Given ]

The mid-point theorem states that the segment joining two sides of triangle at the mid-points of those sides is parallel to the third side and is half the length of the third side.

In $△BAD,$

$P$ and $S$ are the mid-points of sides $AB$ and $BD.$

So, By mid-point theorem,

$\Rightarrow$ $PS\parallel AD$ and $PS=\frac{1}{2}AD$ ------- ( 1 )

In $△CAD,$

$R$ and $Q$ are the mid-point of $CD$ and $AC.$

So, by mid-point theorem,

$\Rightarrow$ $OR\parallel AD$ and $QR=\frac{1}{2}AD$ ----- ( 2 )

Compare ( 1 ) and ( 2 ), we get

$\Rightarrow$ $PS\parallel QR$ and $PS=QR$

Since, one pair of opposite sides is equal as well as parallel then,

$\Rightarrow$ $PQRS$ is a parallelogram ---- ( 3 )

Now, In $△ABC,$ by mid-point theorem

$\Rightarrow$ $PQ\parallel BC$ and $PQ=\frac{1}{2}BC$ ----- ( 4 )

And $AD=BC$ ----- ( 5 )

Compare equations ( 1 ), ( 4 ) and (5) we get,

$\Rightarrow$ $PS=PQ$ ----- ( 6 )

Since, $PQRS$ is a parallelogram with $PS=PQ$ then $PQRS$ is a rhombus.