Show that diagonals of a square …

Given: Let ABCD is a quadrilateral.
Let its diagonal AC and BD bisect each other at right angle at point O.

$\therefore$OA = OC, OB = OD
And $\angle$AOB = $\angle$BOC = $\angle$COD = $\angle$AOD = ${90^ \circ }$
To prove: ABCD is a rhombus.
Proof: In $\triangle$AOD and $\triangle$ BOC,
OA = OC [Given]
$\angle$AOD = $\angle$BOC [Given]
OB = OD [Given]
$\therefore$ $\triangle$AOD $\cong$ $\triangle$COB [By SAS congruency]
$\Rightarrow$ AD = CB [By C.P.C.T.]……….(i)
Again, In $\triangle$AOB and $\triangle$COD,
OA = OC [Given]
$\angle$ AOB = $\angle$COD [Given]
OB = OD [Given]
$\therefore$ $\triangle$AOB  $\cong$ $\triangle$COD [By SAS congruency]
$\Rightarrow$ AB = CD [By C.P.C.T.]……….(ii)
Now In $\triangle$AOB and $\triangle$BOC,
OA = OC [Given]
$\angle$ AOB = $\angle$BOC [Given]
OB = OB [Common]
$\therefore$ $\triangle$AOB $\cong$ $\triangle$COB [By SAS congruency]
$\Rightarrow$ AB = BC [By C.P.C.T.]……….(iii)
From eq. (i), (ii) and (iii),
AD = BC = CD = AB
And the diagonals of quadrilateral ABCD bisect each other at right angle.
Therefore, ABCD is a rhombus.