Question

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

Answer 1

Let $ABCD$ be the rectangle and $P,Q,R,S$ be the midpoints of $AB,BC,CD,DA$ respectively.

Join $AC$, a diagonal of the rectangle.
In $\Delta ABC$, we have:

$PQ||AC$ and $PQ=\frac{1}{2}AC$ [By midpoint theorem]

Again, in $\Delta DAC$, the points $S$ and $R$ are the mid points of $AD$ and $DC$, respectively.

$SR||AC$ and $SR=\frac{1}{2}AC$ [By midpoint theorem]

Now, $PQ||AC$ and $SR||AC$ and $PQ||SR$

Also, $PQ=SR$ [Each equal to $\frac{1}{2}$AC] . . . . . . . (i)

So, $PQRS$ is a parallelogram.

Now, in $\Delta SAP$ and $\Delta QBP$, we have:

$AS=BQ,\angle A=\angle B={90}^{\circ }andAP=BP$

i.e.,$\Delta SAP\sim \Delta QBP$

$PS=PQ$ . . . . . . . . . (ii)

Similarly, $\Delta SDR\sim \Delta QCR$

$SR=RQ$ . . . . . . . . (iii)

From (i), (ii) and (iii), we have:

$PQ=PS=SR=RQ$

Hence, $PQRS$ is a rhombus.