Question

Show that if the diagonals of quadrilateral are equal and bisects each other at Right angle then it is a square.

Answer 1

Let $ABCD$ be the quadrilateral and its diagonals are equal.

$\therefore$ $AC=BD$ ----- ( 1 )

Diagonals bisect each other at right angles.

$\Rightarrow$ $OA=OC$ and $OB=OD$ ----- ( 2 )

$\Rightarrow$ $\angle AOB=\angle BOC=\angle COD=\angle AOD={90}^o$ ----- ( 3 )

In $△AOB$ and $△COB,$

$\Rightarrow$ $OA=OC$ [ From ( 2 ) ]

$\Rightarrow$ $\angle AOB=\angle COB$ [ From ( 3 ) ]

$\Rightarrow$ $OB=OB$ [ Common side ]

$\therefore$ $△AOB\cong △COB$ [ $SAS$ congruence rule ]

$\therefore$ $AB=CB$ [ CPCT ]

Similarly we can prove

$△AOB\cong △DOA$, so $AB=AD$

and $△BOC\cong △COD,$ so $CB=DC$

So, $AB=AD=CB=DC$

Now we can say that

$\Rightarrow$ $AB=CD$ and $AD=BC$

In $ABCD$, both pairs of opposite sides are equal.

$\therefore$ $ABCD$ is a parallelogram.

In $△ABC$ and $△DCB,$

$\Rightarrow$ $AC=BD$ [ From ( 1 ) ]

$\Rightarrow$ $AB=DC$ [ Opposite sides of parallelogram are equal ]

$\Rightarrow$ $BC=CB$ [ Common side ]

$\therefore$ $△ABC\cong △DCB$ [ SSS Congruence rule ]

$\Rightarrow$ $\angle ABC=\angle DCB$ [ CPCT ] ---- ( 4 )

Now, $AB\parallel CD$ and $BC$ is transveral [ Opposite sides of parallelogram are equal ]

$\Rightarrow$ $\angle B+\angle C={180}^o$ [ Sum of adjacent angles in parallelogram is supplementary ]

$\Rightarrow$ $\angle B+\angle B={180}^o$ [ From ( 4 ) ]

$\Rightarrow$ $2\angle B={180}^o$

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

Thus, $ABCD$ is a parallelogram with all sides equal and one angle ${90}^o$.

So, $ABCD$ is a square.