Show that the diagonals of a …

Given: A square ABCD.

To Prove : (i) AC = BD and (ii) Diagonals bisect each other at right angles.
Proof :

1. In {tex}\triangle{/tex}ADB and {tex}\triangle{/tex}BCA, we have
   AD = BC ...[As sides of a square are equal]
   {tex}\angle{/tex}BAD = {tex}\angle{/tex}ABC ...[All interior angles are of 90^o]
   AB = BA ...[Common]
   {tex}\triangle{/tex}ADB {tex}\cong{/tex} {tex}\triangle{/tex}BCA ...[By SAS rule]
   AC = BD ...[c.p.c.t.]
2. Now in {tex}\triangle{/tex}AOB and {tex}\triangle{/tex}COD, we have
   AB = CD ...[Sides of a square]
   {tex}\angle{/tex}AOB = {tex}\angle{/tex}COD ...[Vertically opp. angles]
   {tex}\angle{/tex}OBA = {tex}\angle{/tex}ODC ...[Alternate interior angles are equal]
   {tex}\triangle{/tex}AOB {tex}\cong{/tex} {tex}\triangle{/tex}COD ...[By ASA rule]
   OA = OC and OB = OD ...[c.p.c.t.] ...(1)
   Now consider {tex}\triangle{/tex}s AOD and COD.
   AD = CD ...[Sides of square]
   OA = OC ...[As proved above]
   OD = OD ...[Common]
   {tex}\triangle{/tex}AOD {tex}\cong{/tex} {tex}\triangle{/tex}COD ...[By SSS rule]
   {tex}\angle{/tex}AOD = {tex}\angle{/tex}COD ...[c.p.c.t.]
   But {tex}\angle{/tex}AOD + {tex}\angle{/tex}COD = 180° ...[linear pair]
   or {tex}\angle{/tex}AOD + {tex}\angle{/tex}AOD = 180° ...[As {tex}\angle{/tex}AOD = {tex}\angle{/tex}COD]
   or 2{tex}\angle{/tex}AOD = 180° {tex}\therefore{/tex} {tex}\angle{/tex}AOD = 90° ...(2)
   From equation (1) and (2) it is clear that diagonals of a square bisect each other at right angles.