**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.**