Question

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

Answer 1

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

Join the diagonals $AC$and $BD.$

Let $BD$cut $SR$at $F$and $AC$cut $RQ$at $E.$

Let $O$be the intersection point of $AC$and $BD.$

In $△ABC,$we have

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

Again, in $△DAC$, the points $S$ and $R$are the midpoints of $AD$ and $DC,$respectively.

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

Now, $PQ\parallel AC$and $SR\parallel AC$

$\Rightarrow PQ\parallel SR$

Also, $PQ=SR$ [ Each equal to $\frac{1}{2}AC$ ]$\dots \dots \left(i\right)$
So, $PQRS$ is a parallelogram.

Now, in $△SAP$ and $△QBP,$ we have
$AS=BQ$
$\angle A=\angle B={90}^{\circ }$
$AP=BP$
i.e., $△SAP\cong △QBP$​ [by SAS congruence rule]
$\therefore PS=PQ$ [C.P.C.T]$\dots \dots \left(ii\right)$
Similarly, $△SDR\cong △RCQ$
$\therefore SR=RQ\dots \dots \left(iii\right)$
From $\left(i\right),\left(ii\right)$ and $\left(iii\right),$ we have
$PQ=PS=SR=RQ\dots \dots \left(iv\right)$
We know that the diagonals of a square bisect each other at right angles.
$\therefore \angle EOF={90}^{\circ }$
​Now, $RQ\parallel DB$
$\Rightarrow RE\parallel FO$
Also, $SR\parallel AC$
$\Rightarrow FR\parallel OE$
$\therefore OERF$ is a parallelogram.
So, $\angle FRE=\angle EOF={90}^{\circ }$​ (Opposite angles of parallelogram are equal)
Thus, $PQRS$ is a parallelogram with $\angle R={90}^{\circ }$ and $PQ=PS=SR=RQ.$
$\therefore PQRS$ is a square.