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

Given: 

$PQRS$ is a quadrilateral.

$[PA=AQ,QB=BR,RC=CS,SD=DP.]$    ....(1)

Let $ABCD$ be a rhombus.

To find out:

The kind of the quadrilateral $PQRS$

Construction:

We join $AC\&BD$, they intersect at $O$.

Solution:

$\because ABCD$ is a rhombus and its diagonals intersect at $O$

$\therefore [DO=BO,AO=CO,\angle DOC=\angle BOC=\angle DOA=\angle BOA={90}^o]$    ....(2)

Now, between $DOCS$ and $OCBR$, we have

$DO=BO$ (by 2) ,

$OC$ common side, $RC=CS$ (by 1 ), $\angle DOC=\angle BOC={90}^o$

$\therefore DOCS$ and $OCBR$ are identical rectangles.

Similarly, it can be proved that all the four sub figures of $PQRS$ are identical rectangles.

$\therefore PQRS$ is a rectangle.

So its diagonals are equal.

Hence, option D.