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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