PLZ ANSWER THE ATTACHED QUESTION
Attachments:

Answers
Answered by
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
Similar questions