Question

State true or false

P, Q, R, and S are the mid-point of the sides AB, BC, CD and DA of a quadrilateral ABCD such that AC $\perp$ BD. Then PQRS is a rectangle

• A. True
• B. False

Answer 1

The correct option is A True

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

Since, $AC\perp BD$

$\Rightarrow$ $\angle COD=\angle AOD=\angle AOB=\angle COB={90}^o.$

In $△ADC,$

$S$ and $R$ the mid-points of $AD$ and $DC$ respectively.

Then, by mid-point theorem

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

In $△ABC,$

$P$ and $Q$ are the mid-points of $AB$ and $BC$ respectively, then by mid-point theorem

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

From ( 1 ) and ( 2 ),

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

Similarly, $SP\parallel RQ$ and $SP=RQ=\frac{1}{2}BD$ ----- ( 4 )

Now, in quadrilateral $EOFR,$ $OE\parallel FR,$ $OF\parallel ER$

$\Rightarrow$ $\angle EOF=\angle ERF={90}^o$ [ Since $\angle COD={90}^o\Rightarrow \angle EOF={90}^o$ ] --- ( 5 )

From ( 3 ), ( 4 ) and ( 5 ) we can prove that,

$\therefore$ $PQRS$ is rectangle.

$\therefore$ The given statement is correct.