Question

In the figure, ray AE || ray BD, ray AF is the bisector of ∠EAB and ray BC is the bisector of ∠ABD. Prove that line AF || line BC.

Answer 1

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

∠EAF = ∠FAB = ∠x = `1/2`∠EAB and

∠CBA = ∠DBC = ∠y = `1/2`∠ABD

∴ ∠x = `1/2`∠EAB and ∠y = `1/2`∠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 `1/2`, we get

`1/2"∠EAB" = 1/2"∠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)