Question

如图所示在△ABC中,点F为EA上一点且FD⊥BC于D,AE平分∠BAC,且∠C大于∠B,求证：∠EFD=1/2（∠C-

Answer 1

∵AE平分∠BAC∴∠BAE=1/2∠BAC∵∠BAC+∠C+∠B=180°∴∠BAE=90°-1/2 ∠C-1/2∠B∵∠DEF=∠BAE+∠B∴∠DEF=90°-1/2 ∠C+1/2∠B ∵FD⊥BC∴∠DEF+∠EFD=90° ∴∠EFD=90°- ∠GEF=1/2（∠C-∠B）