Question

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

Answer 1

Consider $△ABC,$

It is given that $P$ and $Q$ are the midpoints of $AB$ and $BC.$

By using the midpoint theorem,

$PQ\parallel AC$ and $PQ=\frac{1}{2}AC\dots \left(i\right)$

Consider $△ADC,$

It is given that $S$ and $R$ are the midpoints of $AD$ and $DC.$

By using the midpoint theorem

$RS\parallel AC$ and $RS=\frac{1}{2}AC\dots \left(ii\right)$

So, from equation $\left(i\right)$ and $\left(ii\right),$ we get

$PQ\parallel RS$ and $PQ=RS=\frac{1}{2}AC\dots \left(iii\right)$

Consider $△BAD,$

It is given that $P$ and $S$ are the midpoints of $AB$ and $AD.$

By using the midpoint theorem,

$PS\parallel BD$ and $PS=\frac{1}{2}DB\dots \left(iv\right)$

Consider $△BCD,$

It is given that $Q$ and $R$ are the midpoints of $BC$ and $CD.$

By using the midpoint theorem,

$RQ\parallel BD$ and $RQ=\frac{1}{2}DB\dots \left(v\right)$

So, from equation $\left(iv\right)$ and $\left(v\right),$ we get

$PS\parallel RQ$ and $PS=RQ=\frac{1}{2}DB\dots \left(vi\right)$

We know that the diagonals of a rectangle are equal, so

$AC=BD\dots \left(vii\right)$

On comparing equations $\left(iii\right),\left(vi\right)$ and $\left(vii\right)$ we get,

$PQ\parallel RS$ and $PS\parallel RQ$

And,

$PQ=QR=RS=SP$

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