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-∠B)。

如图

Answer 1

∠ACB=∠G+∠CFG
=∠G+∠AFE
=∠G+∠AEF
=∠G+(∠B+∠G)
2∠G=∠ACB-∠B
∠G=½（∠ACB-∠B).