Question

如图,D在AB上,E在AC上,BE、CD交于F,∠ACD=35°,∠ABE=20°,求∠BFC

Answer 1

设∠BFC=X,∠FBC+∠FCB=a,则x+a=pi

∠DFE=x

∠ADC为ΔBDC外角,有

∠ADF=a+20°=pi-x-20°

同理

∠AEF=a+35°=pi-x-35°

四边形ADFE内角和为360°,得

∠A+∠ADF+∠DFE+∠AEF=360°

∠A+pi-x-20°+x+pi-x-35°=360°

得 x=∠A+20°+35°

故∠BFC=∠A+20°+35°

很显然∠BFC不固定