Though E, draw a line parallel to p intersecting L at G and n at H respectively. Since l | | m ⇒ AG | | BE and AB | | GE [by construction] ∴ Opposite sides of quadrilateral AGEB are parallel . ∴ AGEB is a parallelogram . Similarly , we can prove that BEHC is a parallelogram . Now, AB = GE [opposite sides of | | gm AGEB] and BC = EH [opposite sides of | | gm BEHC] But, given that AB = BC . Thus, GE = EH Now, △DEG and △FEH, we have ∠DEG = ∠FEH [vertically opposite angles] GE = EH [proved above] and ∠DGE = ∠FHE [alternate interior angles] By ASA congruence axiom, we have △DEG ≅ △FEH Hence, DE = EF