We have a quadrilateral ABCD such that angleO is the mid-point of AC and BD. Also AC ⊥ BD. Now, in ΔAOD and ΔAOB, we have AO = AO [Common] OD = OB [∵ O is the mid-point of BD] ∠AOD = ∠AOB [Each = 90°] ∴ ΔAOD ≌ ∠AOB [SAS criteria] ∴Their corresponding parts are equal. ⇒ AD = AB ...(1) Similarly, we have AB = BC ...(2) BC = CD ...(3) CD = DA ...(4) From (1), (2), (3) and (4) we have: AB = BC = CD = DA ∴Quadrilateral ABCD is having all sides equal. In ΔAOD and ΔCOB, we have AO = CO [Given] OD = OB [Given] ∠AOD = ∠COB [Vertically opposite angles] ∴ ΔAOD ≌ ΔCOB ⇒Their corresponding pacts are equal. ⇒ ∠1 = ∠2 But, they form a pair of interior alternate angles. ∴AD || BC Similarly, AB || DC ∴ ABCD is' a parallelogram. ∵ Parallelogram having all of its sides equal is a rhombus. ∴ ABCD is a rhombus. Now, in ΔABC and ΔBAD, we have AC = BD [Given] BC = AD [Proved] AB = BA [Common] ΔABC ≌ ΔBAD [SSS criteria] Their corresponding angles are equal. ∠ABC = ∠BAD Since, AD || BC and AB is a transversal. ∴∠ABC + ∠BAD = 180° [Interior opposite angles are supplementary] i.e. The rhombus ABCD is having one angle equal to 90°. Thus, ABCD is a square.