Question

In the given figure, ray AE || ray BD, ray AF is the bisector of $\angle$EAB and ray BC is the bisector of $\angle$ABD.

Prove that line AF || line BC.

Answer 1

Since, ray AF bisects ∠EAB and ray BC bisects ∠ABD, then
∠EAF = ∠FAB = ∠x = $\frac{1}{2}$∠EAB and ∠CBA = ∠DBC = ∠y = $\frac{1}{2}$∠ABD
∴ ∠x = $\frac{1}{2}$∠EAB and ∠y = $\frac{1}{2}$∠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 $\frac{1}{2}$, we get
$\frac{1}{2}$∠EAB = $\frac{1}{2}$∠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)