Question

PLZ ANSWER THE ATTACHED QUESTION

Answer 1

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