Consider a parallelogram with diagonals that are congruent and perpendicular.
In triangles Δ AOB and Δ AOD,
OA = OA (common)
OB = OD (Since the diagonals of a parallelogram bisect each other)
∠ AOB = ∠ AOD = 90° (diagonals are perpendicular)
Therefore, Δ AOB ≅ Δ AOD (By SAS postulate)
Since, corresponding parts of congruent triangles are equal,
AB = AD
Similarly, we can prove
BC = CD and CD = DA
So, AB = BC = CD = DA.
Also, it is given that AC = BD.
Hence, ABCD is a square.
Therefore, the correct statement is:
A parallelogram with diagonals that are congruent and perpendicular.
