Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Given: ABCD is a quadrilateral in which AO = CO, BO = DO and ∠COD = 90°.
To prove: ABCD is a rhombus.
Sulution: In ΔAOB and ΔAOD,
BO = DO [∵ Given]
∠AOB = ∠AOD [∵ Each 90°]
AO = AO [∵ Common]
Hence, ΔAOB ≅ ΔAOD [∵ SAS congruency rule]
AB = AD [∵ CPCT]
Similarly, AB = BC and BC = CD
Now, all the four sides of Quadrilateral ABCD are equal.
Hence, ABCD is a rhombus.