The figure formed by joining the mid-points of the adjacent sides of a square is a

The correct option is A. square

Let $ABCD$ be a square. Let $P,Q,R$ and $S$ be the mid points of the side $AB,BC,CD$ and $AD$ respectively.

Consider $\Delta ADC$, since $S$ and $R$ are the mis points of $AD$ and $DC$, therefore, by Mid Point Theorem,

$SR\parallel AC$ and $SR=\frac{AC}{2}$...(i)

Similarly, in $\Delta ABC$, since $P$ and $Q$ are the mis points of $AB$ and $BC$, therefore, by Mid Point Theorem,

$PQ\parallel AC$ and $PQ=\frac{AC}{2}$...(ii)

From (i) and (ii)

$SR\parallel PQ$ and $SR=PQ$ ...(iii)

Therefore, $PQRS$ is a parallelogram.

Now, consider $\Delta APS$ and $\Delta PBQ$

$AS=BQ$ [As $AS$ and $BQ$ are the sides of the square]

$AP=BP$ [$P$ is the mid point of $AB$]

$\angle SAP=\angle QBP={90}^{\circ }$ [Each angle is ${90}^{\circ }$ of a square]

Therefore, $\Delta SAP$ and $\Delta BQP$ by SAS congruence rule.

Hence, $SP=PQ$ (By CPCT) ...(iv)

From (iii) and (iv),

$\Rightarrow PQ=QR=RS=PS$...(v)

Now consider, quadrilateral $ERFO$,

Since, $EO\parallel RF$ and $ER\parallel OF$, therefore,

$\Rightarrow$ Quadrilateral $ERFO$ is a parallelogram.

Now, we know that diagonals of a rhombus bisect each other at ${90}^{\circ }$.

$\therefore \angle EOF={90}^{\circ }$

Hence, $\angle SRQ=\angle ERF=\angle EOF={90}^{\circ }$ [Opposite angles of parallelogram are equal]...(vi)

From (v) and (vi), $PQRS$ is a square.