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Based on the diagram, the following relation holds true.
Parallelogram ABCD is a rhombus ⇔ AC⊥BD
This proof will be written in two parts.
A rhombus is a parallelogram with four congruent sides. By the Parallelogram Diagonals Theorem, it can be said that its diagonals bisect each other. Let Let ABCD be a rhombus with P at the midpoint of both diagonals.
Parallelogram ABCD is a rhombus ⇒ AC⊥BD
Conversely, let ABCD be a parallelogram whose diagonals are perpendicular.
By the Parallelogram Diagonals Theorem, the diagonals of the parallelogram bisect each other. If P is the midpoint of both diagonals, then AP and CP are congruent.
Furthermore, by the Parallelogram Opposite Sides Theorem, AB is congruent to DC and AD is congruent to BC. By the Transitive Property of Congruence, it follows that all sides of the parallelogram are congruent.
This means that if the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus.
AC⊥BD ⇒ parallelogram ABCD is a rhombus