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.
∵ EG is the bisector of∠AEF.
∴ ∠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