Here we will prove that if in a parallelogram the diagonals are equal in length and intersect at right angles, the parallelogram will be a square.
Given: PQRS is a parallelogram in which PQ ∥ SR, PS ∥ QR and diagonal PR ⊥diagonal QS.
To prove: PQRS is a square, i.e., PQ = QR = RS = SP and an angle, say ∠SPQ = 90°.
Proof:
In ∆PQR and ∆RSP,
∠QPR = ∠PRS (Since, PQ ∥ SR and QR is a transversal)
∠QRP = ∠SPR (Since, QR ∥ PS and PR is a transversal)
PR = PR (Common Side).
Therefore, ∆PQR ≅ ∆RSP (By AAS criterion of congruency).
Therefore, PQ = SR (CPCTC).
Similarly, ∆PQS ≅ ∆RSQ (By AAS criterion of congruency).
Therefore, PS = QR (CPCTC).
∆OPQ ≅ ∆ORS (By AAS criterion of congruency).
Therefore, OP = OR (CPCTC).
Similarly, ∆POQ ≅ ∆ROQ (By SAS criterion of congruency).
Therefore, PQ = QR (CPCTC).
Therefore, PQ = QR = RS = SP. (Proved)
∆SPQ ≅ ∆RQP (By SSS criterion of congruency).
Therefore, ∠SPQ = ∠RQP (CPCTC).
But ∠SPQ + ∠RQP = 180° (Since, PS ∥ QR).
Therefore, ∠SPQ = ∠RQP = \(\frac{180°}{2}\) = 90°. (Proved).
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