Question

Question 11
Show that the quadrilateral formed by joining the mid-points of the consecutive sides of a square is also a square.

Answer 1

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

To show PQRS is a square.

Construction Join AC and BD.

Proof Since, ABCD is a square.

AB = BC = CD = AD

Also, P, Q,R and S are the mid-points of AB, BC, CD and DA, respectively.

Then, in $\Delta ADC,SR||AC$

and $SR=\frac{1}{2}AC$ [by mid-point theorem]…(i)

In $\Delta ABC$, PQ||AC

and $PQ=\frac{1}{2}AC$ AC …(ii)

From Eqs. (i) and (ii),

SR||PQ and $SR=PQ=\frac{1}{2}AC$ AC …(iii)

Similarly, SP||BD and BD||RQ

$\therefore$ SP||RQ and $SP=\frac{1}{2}AC$ 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 …(iv)

From Eqs. (iii) and (iv), SR = PQ = SP = RQ [all side are equal]

Now, in quadrilateral OERF,

OE||FR and OF||ER

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

Hence, PQRS is a square. Hence proved.