Question

Ray AE‖ Ray BD, ray AF is the bisector of <EAB and ray BC is the bisector of <ABD, Prove that Ray AF is parallel to ray BC ,​

Answer 1

Answer:

Since, ray AF bisects ∠EAB and ray BC bisects ∠ABD, then

∠EAF = ∠FAB = ∠x = ½ ∠EAB and ∠CBA

=∠DBC = ∠y = ½ ∠ABD

∴∠x = ½ ∠EAB and ∠y = ½ ∠ABD ....(1)

Since, ray AE ∥ ray BD and segment AB is a transversal intersecting them at A and B, then

∠EAB =∠ ABD (Alternate interior angles)

On multiplying both sides by ½ , we get ½ ∠EAB = ½ ∠ABD

Now, using (1), we get

∠x = ∠y

But ∠x and ∠y are alternate interior angles formed by a transversal AB of ray AF and ray BC.

∴ ray AF ∥ ray BC (Alternate angles test)

Hope it will you mate..