Given that ABCD is a square.
To prove : AC = BD and AC and BD bisect each other at right angles.
(i) In a Δ ABC and Δ BAD,
AB = AB ( common line)
BC = AD ( opppsite sides of a square)
∠ABC = ∠BAD ( = 90° )
Δ ABC ≅ Δ BAD ( By SAS property)
AC = BD ( by CPCT).
(ii) In a Δ OAD and Δ OCB,
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 a ΔOBA and ΔODA,
OB = OD ( from (ii) )
BA = DA
OA = OA ( common line )
ΔAOB = ΔAOD ----(iii) ( by CPCT
∠AOB + ∠AOD = 180° (linear pair)
2∠AOB = 180°
∠AOB = ∠AOD = 90°
∴AC and BD bisect each other at right angles.