Question

In the given figure, AB || CD and a transversal t cuts them at E and F respectively. If EP and FQ are the bisectors of ∠AEF and ∠EFD respectively, prove that EP || FQ .

Answer 1

It is given that, AB || CD and t is a transversal.

∴ ∠AEF = ∠EFD .....(1) (Pair of alternate interior angles)

EP is the bisectors of ∠AEF. (Given)

∴ ∠AEP = ∠FEP = $\frac{1}{2}$∠AEF

⇒ ∠AEF = 2∠FEP .....(2)

Also, FQ is the bisectors of ∠EFD.

∴ ∠EFQ = ∠QFD = $\frac{1}{2}$∠EFD

⇒ ∠EFD = 2∠EFQ .....(3)

From (1), (2) and (3), we have

2∠FEP = 2∠EFQ

⇒ ∠FEP = ∠EFQ

Thus, the lines EP and FQ are intersected by a transversal EF such that the pair of alternate interior angles formed are equal.

∴ EP || FQ (If a transversal intersects two lines such that a pair of alternate interior angles are equal, then the two lines are parallel)