Show that the quadrilateral formed by joining the mid-points of the consecutive sides of a square is also a square. Hint : Prove all the sides are equal and diagonals are equal

Show that the quadrilateral formed by joining the mid-points of the consecutive sides of a square is also a square.
Hint: Prove all the sides are equal and diagonals are equal

Answers (1)

Solution.
Given: In a square ABCD; P, Q, R and S are the mid-points of AB, BC, CD and DA.
To Prove: PQRS is a square
Construction: Join AC and BD

Proof : Here ABCD is a square
\ AB = BC = CD = AD
P, Q, R, S are the mid-points of AB, BC, CD and DA.
In $\triangle ADC$ ,
$SR\parallel AC$
$SR=\frac{1}{2}AC$   {using mid-point theorem} …..(1)
In $\triangle ABC$ ,
$PQ\parallel AC$
$PQ=\frac{1}{2}AC$ {using 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 $BD\parallel RQ$
\ $SP=\frac{1}{2}BD$ and $RQ=\frac{1}{2}BD$    {using mid-point theorem}
$SP=RQ=\frac{1}{2}BD$
Since diagonals of a square bisect each other at right angles.
AC = BD
$\Rightarrow SP=RQ=\frac{1}{2}AC$ …..(4)
From equation 3 and 4
SP = PQ = SP = RQ {All sides are equal}
In quadrilateral OERF
$OE\parallel FR$ and $OF\parallel ER$
$\angle EOF=\angle ERF=90^{\circ}$
In quadrilateral RSPQ
$\angle SRQ=\angle ERF=\angle SPQ=90^{\circ}$ {Opposite angles in a parallelogram are equal}
$\angle RSP + \angle SRQ + \angle RQP + \angle QRS = 360^{\circ}$
$90^{\circ} +90^{\circ}+ \angle RSP + \angle RQP = 360^{\circ}$
$\angle RSP + \angle RQP = 180^{\circ}$
$\angle RSP = \angle RQP = 90^{\circ}$
Since all the sides are equal and angles are also equal, so PQRS is a square.
Hence Proved