Given: ABCD is a square.
To prove: AC = BD, AO = CO, BO = DO and ∠COD = 90°.
Solution: ΔBAD and ΔABC,
AD = BC [∵ Opposite sides of a square]
∠BAD = ∠ABC [∵ Each 90°]
AB = AB [∵ Common]
Hence, ΔBAD ≅ ΔABC [∵ SAS Congruency rule]
BD = AC [∵ CPCT]
In ΔAOB and ΔCOD,
∠OAB = ∠OCD [∵ Alternate angles]
AB = CD [∵ Opposite sides of a square]
∠OBA = ∠ODC [∵ Alternate angles]
Hence, ΔBAD ≅ ΔABC [∵ ASA Congruency rule]
AO = OC, BO = OD, [∵ CPCT]
In ΔAOB and ΔAOD.
OB = OD [∵ Proved above]
AB = AD [∵ Sides of a square]
OA = OA [∵ Common]
Hence, ΔBAD ≅ ΔABC [∵ SSS Congruency rule]
∠AOB = ∠AOD [∵ CPCT]
But. ∠AOB + ∠AOD = 180° [∵ Linear pair]
⇒ 2∠AOB = 180° [∵ ∠AOD = ∠AOB]
⇒ ∠AOB = (180/2) = 90°
Hence, the diagonals of a square are equal and bisect each at right angles.