Given : ABCD is a square in which AC and BD are diagonals.
To Prove :
AC = BD and AC bisects BD at right angles, i.e. AC ⊥ BD. AO = OC, OB = OD
Proof : In ∆ABC and ∆BAD,
AB = AB [Common]
BC = AD [Sides of a square]
∠ABC = ∠BAD = 90° [Angles of a square]
∴ ∆ABC ≅ ∆BAD [SAS congruence]
⇒ AC = BD [CPCT]
Now in ∆AOB and ∆COD,
AB = DC [Sides of a square]
∠AOB = ∠COD [Vertically opposite angles]
∠OAB = ∠OCD [Alternate angles]
∴ ∆AOB ≅ ∆COD [AAS congruence]
∠AO = ∠OC [CPCT]
Similarly by taking ∆AOD and ∆BOC, we can show that OB = OD.
In ∆ABC, ∠BAC + ∠BCA = 90° [ ∠B = 90°]
⇒ 2∠BAC = 90° [∠BAC = ∠BCA, as BC = AD]
⇒ ∠BCA = 45° or ∠BCO = 45°
Similarly ∠CBO = 45°
In ∆BCO.
∠BCO + ∠CBO + ∠BOC = 180°
⇒ 90° + ∠BOC = 180°
⇒ ∠BOC = 90°
⇒ BO ⊥ OC ⇒ BO ⊥ AC
Hence, AC = BD, AC ⊥ BD, AO = OC and OB = OD. Proved.
