过E点作EM‖AB,则EM‖CD,根据两直线平行,内错角相等,可知:
∠AEC=∠EAB+∠ECD=4(∠EAF+ECF)(1)
所以1/4∠AEC=∠EAF+ECF
连接EF,有外角性质可知,
∠AEC=∠EAF+∠EFA+∠ECF+∠EFC=(∠EAF+ECF)+(∠EFA+∠EFC)=(∠EAF+ECF)+∠AFC(2)
由(1)(2) 式得,∠AFC=3(∠EAF+ECF),而1/4∠AEC=(∠EAF+ECF)
所以∠AFC=3/4∠AEC
过E点作EM‖AB,则EM‖CD,根据两直线平行,内错角相等,可知:
∠AEC=∠EAB+∠ECD=4(∠EAF+ECF)(1)
所以1/4∠AEC=∠EAF+ECF
连接EF,有外角性质可知,
∠AEC=∠EAF+∠EFA+∠ECF+∠EFC=(∠EAF+ECF)+(∠EFA+∠EFC)=(∠EAF+ECF)+∠AFC(2)
由(1)(2) 式得,∠AFC=3(∠EAF+ECF),而1/4∠AEC=(∠EAF+ECF)
所以∠AFC=3/4∠AEC