Question

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

Answer 1

In a square $ABCD,$ $P,Q,R$ and $S$ are the mid-points of $AB,BC,CD$ and $DA$ respectively.

$\Rightarrow$ $AB=BC=CD=AD$ [ Sides of square are equal ]

In $△ADC,$

$SR\parallel AC$ and $SR=\frac{1}{2}AC$ [ By mid-point theorem ] ---- ( 1 )

In $△ABC,$

$PQ\parallel AC$ and $PQ=\frac{1}{2}AC$ [ By mid-point theorem ] ---- ( 2 )

From equation ( 1 ) and ( 2 ),

$SR\parallel PQ$ and $SR=PQ=\frac{1}{2}AC$ ---- ( 3 )

Similarly, $SP\parallel BD$ and $BD\parallel RQ$

$\therefore$ $SP\parallel RQ$ and $SP=\frac{1}{2}BD$

and $RQ=\frac{1}{2}BD$

$\therefore$ $SP=RQ=\frac{1}{2}BD$

Since, diagonals of a square bisect each other at right angle.

$\therefore$ $AC=BD$

$\Rightarrow$ $SP=RQ=\frac{1}{2}AC$ ----- ( 4 )

From ( 3 ) and ( 4 )

$SR=PQ=SP=RQ$

We know that the diagonals of a square bisect each other at right angles.

$\angle EOF={90}^o.$

Now, $RQ\parallel DB$

$RE\parallel FO$

Also, $SR\parallel AC$

$\Rightarrow$ $FR\parallel OE$

$\therefore$ $OERF$ is a parallelogram.

So, $\angle FRE=\angle EOF={90}^o$ (Opposite angles are equal)

Thus, $PQRS$ is a parallelogram with $\angle R={90}^o$ and $SR=PQ=SP=RQ$

$\therefore$ $PQRS$ is a square.