The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is

In figure,

ABCD is a rhombus and P, Q, R, S are mid-point of AB, BC, CD and DA.

In $△$ABD,

Since, P and S are mid-point of AB and AD

$\therefore$ By Mid-point theorem,

P $\left|\right|$BD;   PS = $\frac{1}{2}$BD    ----- (1)

In $△$BCD,

Since, R and Q are mid-point of sides CD & CB

$\therefore$ By Mid-point theorem,

RQ$\left|\right|$BD;  RQ = $\frac{1}{2}$BD   ------(2)

From (1) & (2),

PS$\left|\right|$QR and PS = QR    -------(3)

Similarly, 

PQ$\left|\right|$RS and PQ = RS     -------(4)

From (3) and (4) we get,

Quadrilateral PQRS is a parallelogram.

The diagonals of a rhombus bisect each other at right angles.

In quadrilateral OMQL,

since, QR$\left|\right|$PS

$\therefore$ OM$\left|\right|$QL;  OM = QL

As PQ$\left|\right|$RS

$\therefore$ QM$\left|\right|$OL; QM = OL

$\therefore$ Quadrilateral  OMQL is a parallelogram.

As, $\angle$O = 90${}^{\circ }$

$\angle$MQL = $\angle$MOL = 90${}^{\circ }$  ---(opposite angles of parallelogram are equal)

$\therefore$ PQRS is a rectangle.