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.

{tex}\therefore {/tex}OA = OC, OB = OD
And {tex}\angle {/tex}AOB = {tex}\angle {/tex}BOC = {tex}\angle {/tex}COD = {tex}\angle {/tex}AOD = {tex}{90^ \circ }{/tex}
To prove: ABCD is a rhombus.
Proof: In {tex}\triangle{/tex}AOD and {tex}\triangle{/tex} BOC,
OA = OC [Given]
{tex}\angle {/tex}AOD = {tex}\angle {/tex}BOC [Given]
OB = OD [Given]
{tex}\therefore {/tex} {tex}\triangle{/tex}AOD {tex} \cong{/tex} {tex}\triangle{/tex}COB [By SAS congruency]
{tex} \Rightarrow {/tex} AD = CB [By C.P.C.T.]……….(i)
Again, In {tex}\triangle{/tex}AOB and {tex}\triangle{/tex}COD,
OA = OC [Given]
{tex}\angle {/tex} AOB = {tex}\angle {/tex}COD [Given]
OB = OD [Given]
{tex}\therefore {/tex} {tex}\triangle{/tex}AOB  {tex} \cong {/tex} {tex}\triangle{/tex}COD [By SAS congruency]
{tex} \Rightarrow {/tex} AB = CD [By C.P.C.T.]……….(ii)
Now In {tex}\triangle{/tex}AOB and {tex}\triangle{/tex}BOC,
OA = OC [Given]
{tex}\angle {/tex} AOB = {tex}\angle {/tex}BOC [Given]
OB = OB [Common]
{tex}\therefore {/tex} {tex}\triangle{/tex}AOB {tex} \cong{/tex} {tex}\triangle{/tex}COB [By SAS congruency]
{tex} \Rightarrow {/tex} 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.