Show that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a rhombus is a rectangle.

Consider $△ABC$

We know that $P$ and $Q$ are the midpoints of $AB$ and $BC$

By using the midpoint theorem

We know that $PQ\parallel AC$ and $PQ=1/2AC$

Consider $△ADC$

We know that $RS\parallel AC$ and$RS=1/2AC$

It can be written as $PQ\parallel RS$ and 

$PR=RS=1/2AC$....(1)

Consider $△BAD$

We know that $P$ and $S$ are the midpoints of $AB$ and $AD$

Based on the midpoint theorem

We know that $PS\parallel BD$ and $PS=1/2DB$

Consider $△BCD$

We know that $RQ\parallel BD$ and $RQ=1/2DB$

It can be written as $PS\parallel RQ$ and

$PS=RQ=1/2DB$.....(2)

By considering equations (1) and (2)

the diagonals intersect at right angles in a rhombus

so we get

We know that $RQ\parallel DB$

so we get $RE\parallel FO$

In the same way $SR\parallel AC$

so we get $FR\parallel OE$

We know that the opposite angles are equal in a parallelogram

so we get

So we know that $PQRS$ is a parallelogram having $\angle R={90}^o$

Therefore, it is proved that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a rhombus is a rectangle.