解：过A作AD⊥BC于D∵AB=AC，AB⊥AC，AD⊥BC∴AD=BD=CD∵BE=EF=FC∴AE=AF，DE=DF=EF/2，∠EAD=∠FAD=1/2∠EAF∴AD=3DE∴tan∠EAD=ED/AD=1/3∴tan∠EAF=2tan∠EAD/(1-tan²∠EAD)=(2/3)/[1-(1/3)²]=3/4
tan∠EAF=tan45度=1
