Show that the diagonals of a square are equal and bisect each other at right angles.

Given that ABCD is a square.
To prove : AC=BD and AC and BD bisect each other at right angles.
Proof:
(i) In ΔABC and ΔBAD, AB=AB (common line)
BC=AD (opposite sides of a square)
∠ABC=∠BAD ( = 90⁰ )
ΔABC≅ΔBAD (By SAS property)
AC=BD (by CPCT)
(ii) In ΔOAD and ΔOCB, AD=CB (opposite sides of a square)
∠OAD=∠OCB (transversal AC )
∠ODA=∠OBC (transversal BD )
ΔOAD≅ΔOCB (ASA property)
OA=OC ———(i)
Similarly OB=OD ———(ii)
From (i) and (ii) AC and BD bisect each other.
Now in ΔOBA and ΔODA, OB=OD (from (ii) )
BA=DA
OA=OA (common line )
ΔAOB=ΔAOD —-(iii) (by CPCT
∠AOB+∠AOD=180⁰ (linear pair)
2∠AOB=180⁰
∠AOB=∠AOD=90⁰
∴ AC and BD bisect each other at right angles.