Question

In the figure, ray AE || ray BD, ray AF is the bisector of angle EAB & ray BC is the bisector of angle ABD. Prove that line AF || line BC. ​

Answer 1

Step-by-step explanation:

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

∠EAF=∠FAB=∠x=21∠EAB and ∠CBA

=∠DBC=∠y=21∠ABD

∴∠x=21∠EAB and ∠y=21∠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 21, we get

21∠EAB=21∠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)

Answer 2

Step-by-step explanation:

refer to the above attachments

hope it helps