Question

Let AB, CD be two line segments such that AB || CD and AD || BC. Let E be the midpoint of BC and let DE extended meet AB in F. Prove that AB = BF.

Answer 1

Given: AB || CD, AD || BC, BE = CE

To Prove: AB = BF

Proof: Since AB || CD and AD || BC, ABCD is a parallelogram

⇒ AB = CD … (1)

In ΔEDC and ΔEFB:

∠DEC = ∠FEB (Vertically opposite angles)

CE = BE (Given)

∠DCE = ∠FBE (Interior alternate angles)

∴ΔEDC ≅ ΔEFB (By ASA congruency)

∴CD = BF … (2)

From (1) and (2), we get:

AB = BF