Question

Question 5
P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and $AC\perp BD$. Prove that PQRS is a square.

Answer 1

Given in quadrilateral ABCD, P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively.

Also, $AC=BDandAC\perp BD$.

To prove PQRS is a square.

Proof Now, in $\Delta ADC$, S are R are the mid-points of the sides AD and DC respectively, then by mid-point theorem,

$SR\parallel ACandPQ=\frac{1}{2}AC$ ……(i)

In $\Delta ABC,P$ and Q are the mid-points of AB and BC , then by mid- point theorem,
$PQ||ACandPQ=\frac{1}{2}AC$ …(ii)
From Eqs. (i) and (ii) , $PQ||SRandPQ=\frac{1}{2}AC$ …(iii)
Similarly, in $\Delta ABD$, by mid-point theorem,
$SP||BDandSP=\frac{1}{2}BD=\frac{1}{2}AC$ [given, AC = BD] .....(iv)
And $\Delta BCD$, by mid-point theorem,
$RQ||BDandRQ=\frac{1}{2}BD=\frac{1}{2}AC$ [given, BD = AC] ...(v)

From Eqs. (iv) and (v),
$SP=PQ=\frac{1}{2}AC$
From Eqs. (iii) and (iv)
PQ = SR = SP= RQ
Thus, all four sides are equal,
Now, in quadrilateral OERF, OE || FR and OF || ER
$\therefore \angle EOF\angle ERF={90}^{\circ }$
$[\because AC\perp DB\Rightarrow =\angle DOC=\angle EOF={90}^{\circ }$ as opposite angles of a parallelogram]

$\therefore \angle QRS={90}^{\circ }$
Similarly $\angle RQS={90}^{\circ }$
So, PQRS is a square.
Hence proved.