Question

$P,Q,R$ and $S$ are midpoints of $AB,BC,CD$ and $DA$ of quadrilateral $ABCD$ in which $AC=BD$, $AC\perp BD$. Prove that $PQRS$ is a square.

Answer 1

Now, in $△ADC,$

$S$ and $R$ are the mid-points of the sides $AD$ and $DC$

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 $BC$.

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, in $△ABD,$ by mid-point theorem,

$SP\parallel BD$ and $SP=\frac{1}{2}BD=\frac{1}{2}AC$ [ Since, $AC=BD$ ] ----- ( 4 )

And $△BCD,$ by mid-point theorem,

$\Rightarrow$ $RQ\parallel BD$ and $RQ=\frac{1}{2}BD=\frac{1}{2}AC$ [ Sice, $BD=AC$ ] ----- ( 5 )

From equations ( 4 ) and ( 5 ),

$\Rightarrow$ $SP=RQ=\frac{1}{2}AC$ ----- ( 6 )

From equation ( 3 ) and ( 6 ),

$\Rightarrow$ $PQ=SR=SP=RQ$

Thus, all four sides are equal.

Now, in quadrilateral $OERF,OE\parallel FR$ and $OF\parallel ER$

$\therefore$ $\angle EOF=\angle ERF={90}^o$ [ Since, $AC\perp BD$ ]

$\therefore$ $\angle QRS={90}^o$

Similarly, $\angle RQS={90}^o$

$\therefore$ $PQRS$ is a square.