Sabba has drawn two straight line segments AD and BC such that, AD || BC and AD = BC; I prove with reason that AB = DC and AB || DC.

Given: AD || BC and AD = BC

To Prove: AB = DC and AB || DC

Proof:

Join AB, CD and BD.

In ∆ABD and ∆CDB,

AD = BC {Given}

BD = BD {Common}

∠ADB = ∠DBC {Alternate interior angles ∵ AD || BC}

⇒ ∆ABD ≅ ∆CDB {By SAS criterion of congruency}

∴ ∠ABD = ∠CDB {Corresponding angles of congruent triangles are equal}

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

∴ AB || DC

Now, ∵ AB || DC and AD || BC in the quadrilateral ABCD

∴ ABCD is a parallelogram.

⇒ AB = DC {Opposite sides of a parallelogram are equal}

Hence, proved.