Show that the quadrilateral formed by joining the mid-points of the consecutive sides of a square is also a square.

(N/A) Given: $A$ square $ABCD$ in which $P, Q, R, S$ are the mid-points of sides $AB, BC, CD, DA$ respectively. $PQ, QR, RS$ and $SP$ are joined.
To prove: $PQRS$ is a square.
Construction: Join $AC$ and $BD$.
Proof: In $\triangle ABC$,$P$ and $Q$ are the mid-points of sides $AB$ and $BC$ respectively.
Therefore,$PQ \parallel AC$ and $PQ = \frac{1}{2} AC$ (Mid-point theorem) ... $(1)$
In $\triangle ADC$,$R$ and $S$ are the mid-points of $CD$ and $AD$ respectively.
Therefore,$RS \parallel AC$ and $RS = \frac{1}{2} AC$ (Mid-point theorem) ... $(2)$
From equations $(1)$ and $(2)$,we get $PQ \parallel RS$ and $PQ = RS$.
Thus,in quadrilateral $PQRS$,one pair of opposite sides is equal and parallel. Hence,$PQRS$ is a parallelogram.
Since $ABCD$ is a square,$AB = BC = CD = DA$.
Also,$PB = BQ = QC = CR = RD = DS = SA = AP = \frac{1}{2} AB$.
In $\triangle PBQ$ and $\triangle QCR$,$PB = QC$,$BQ = CR$,and $\angle PBQ = \angle QCR = 90^{\circ}$.
By $SAS$ congruence criterion,$\triangle PBQ \cong \triangle QCR$.
Therefore,$PQ = QR$ $(CPCT)$.
Since $PQRS$ is a parallelogram with adjacent sides equal,$PQRS$ is a rhombus.
Now,consider the diagonals of the square $ABCD$. $AC \perp BD$ and $AC = BD$.
Since $PQ \parallel AC$ and $QR \parallel BD$,and $AC \perp BD$,it follows that $PQ \perp QR$.
Thus,$PQRS$ is a rhombus with one angle equal to $90^{\circ}$.
Hence,$PQRS$ is a square.