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

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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.

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