Firoz has drawn a quadrilateral PQRS whose side PQ = SR and PQ || SR; I prove with reason that PQRS is a parallelogram.

Given: In a quadrilateral PQRS, PQ = SR and PQ || SR

To Prove: PQRS is a parallelogram

Proof:

Join QS.

In ∆PQS and ∆QRS,

PQ = SR {Given}

QS = QS {Common}

∠PQS = ∠RSQ {Alternate interior angles ∵ PQ||SR}

⇒ ∆PQS ≅ ∆QRS {By SAS criterion of congruency}

∴ ∠PSQ = ∠RQS {Corresponding angles of congruent triangles are equal}

But as the transversal QS intersects the straight lines PS and QR and the two alternate angles become equal.

∴ PS||QR

Now, ∵ PQ||SR and PS||QR in the quadrilateral PQRS

∴ PQRS is a parallelogram.

Hence, proved.