Here we will prove that a parallelogram, whose diagonals intersect at right angles, is a rhombus.

**Given:** PQRS is a parallelogram in which PQ ∥ SR, PS ∥ QR and ∠QOR
= ∠POQ = ∠ROS = ∠POS = 90°.

**To prove:** PQRS is a rhombus, i.e., PQ = QR = RS = SP.

**Proof:**

In ∆PQR and ∆RSP,

∠QPR = ∠SRP (Since, SR ∥ PQ)

∠QRP = ∠SPR (Since, PS ∥ QR),

PR = PR

Therefore, ∆PQR ≅ ∆RSP, (By AAS criterion of congruency)

Therefore, PQ = RS, QR = PS (CPCTC)

Therefore, ∆OPQ ≅ ∆ORS (By AAS criterion of congruency)

Therefore, OP = OR (CPCTC)

In ∆POQ and ∆QOR,

OP = OR,

OQ = OQ,

∠POQ = ∠QOR = 90°.

Therefore, ∆POQ ≅ ∆QOR (By SAS criterion of congruency)

Therefore, PQ= QR (CPCTC)

Therefore, PQ = RS = QR = PS.

So, PQRS is a rhombus.

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