In the given figure a transversal l cuts two lines AB and CD at E and F, respectively. EG is the bisector of ∠AEF and FH is the bisector of ∠EFD such that ∠a = ∠b. Show that EG\FH and AB \CD.

∴ ∠AEG = ∠GEF = a

Similarly, ∠EFH =∠HFD = b

∴∠GEF = ∠EFH[∵a=b]

[alternate angles]

∴EG || FH

Again ∠AEF = 2a

and ∠EFD = 2b[∵a=b]

∴ ∠AEF = ∠EFD

[∵alternate angles]

∴ AB || CD