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

(N/A) Let $ABCD$ be a rhombus and $P, Q, R,$ and $S$ be the mid-points of sides $AB, BC, CD,$ and $DA,$ respectively. Join $AC$ and $BD.$
From triangle $ABD,$ by the mid-point theorem,we have:
$SP = \frac{1}{2} BD$ and $SP \parallel BD.$
Similarly,in triangle $BCD,$ we have:
$RQ = \frac{1}{2} BD$ and $RQ \parallel BD.$
Therefore,$SP = RQ$ and $SP \parallel RQ$ (Equation $1$).
Since one pair of opposite sides is equal and parallel,$PQRS$ is a parallelogram.
Also,the diagonals of a rhombus are perpendicular,so $AC \perp BD.$
In triangle $BAC,$ by the mid-point theorem,$PQ \parallel AC.$
Since $SP \parallel BD$ and $PQ \parallel AC,$ and $AC \perp BD,$ it follows that $SP \perp PQ.$
Thus,$\angle SPQ = 90^{\circ}$ (Equation $2$).
Since $PQRS$ is a parallelogram with one angle equal to $90^{\circ},$ $PQRS$ is a rectangle.