A Parallelogram whose Diagonals Intersect at Right Angles is a Rhombus
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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