Question

如图,已知在△ABC中,AB>AC,∠AEF=∠AFE,延长EF与BC的延长线交于点G,求证∠G=½（∠ACB

Answer 1

∠ACB=∠G+∠CFG

=∠G+∠AFE

=∠G+∠AEF

=∠G+(∠B+∠G)

2∠G=∠ACB-∠B

∠G=½（∠ACB-∠B).