In the given figure, AB || CD and a transversal t cuts them at E and F

1.	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» We know that AB \|\| CD and t is a transversal From the figure we know that ∠AEF and ∠EFD are alternate angles So we get ∠AEF = ∠EFD Dividing both the sides by 2 we get (1/2) ∠AEF = (1/2) ∠EFD So we get ∠PEF = ∠EFQ The alternate interior angles are formed only when the transversal EF cuts both FQ and EP. Therefore, it is proved that EP \|\| FQ.

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»

We know that AB || CD and t is a transversal

From the figure we know that ∠AEF and ∠EFD are alternate angles

So we get

∠AEF = ∠EFD

Dividing both the sides by 2 we get

(1/2) ∠AEF = (1/2) ∠EFD

So we get

∠PEF = ∠EFQ

The alternate interior angles are formed only when the transversal EF cuts both FQ and EP.

Therefore, it is proved that EP || FQ.