Show that the quadrilateral formed by joining the mid points of the sides of a rhombus taken in order, form a rectangle .

Let ABCD be a rhombus and P, Q, R and S be the mid points of sides of AB, BC, CD and DA respectively.

In $\Delta$ABD and BDC we have

SP$||$BD and SP$=\frac{1}{2}BD\to \left(1\right)$

RQ$||$BD and RQ$=\frac{1}{2}BD\to \left(2\right)$

from $\left(1\right)$ and $\left(2\right)$ we get

PQRS is a parallelogram

As diagonals of rhombus bisect each other at right angle

$\therefore AC\perp BD$

Since SP$||$BD, PQ$||$AC and $AC\perp BD$

$\therefore SP\perp PQ$

LQPS$:{90}^o$ $\to \left(3\right)$

From above results we have parallelogram PQRS is rectangle.