Given: ABCD is a quadrilateral such that AC = BD, AO = CO, BO = DO and ∠COD = 90°.
To prove: ABCD is a square.
Solution: If the diagonals of a quadrilateral bisects each other at right angle, it is a rhombus .
Hence, AB = BC = CD = DA
In ΔBAD and ΔABC,
AD = BC [∵ Proved above]
BD = AC [∵ Given]
AB = AB [∵ Common]
Hence, ΔBAD ΔABC [∵ SSS Congruency rule]
∠BAD = ∠ABC [∵ CPCT]
But, ∠BAD + ∠ABC = 180° [∵ Co-interior angles]
⇒ 2∠ABC = 180° [∵ ∠BAD = ∠ABC]
⇒ ∠ABC = (180°/2) = 90°
Hence, if the diagonals of a quadrilateral are equal and bisect each other at right angles, than it is a square.