Question

△ABC是等边三角形，点D，E分别在BC,AC边上，且AE=DC，AD,BE交于点F。求∠BFD的度数

Answer 1

可证得△BAE ≌ACD
则有∠BEA=∠ADC
又∠ADC+∠CAD=120°
∴∠BEA+∠CAD=120°
则∠BFD=60°

Answer 2

如果是等边三角形,则
∵AE=CD,AC=AB,∠BAE=∠C
∴⊿BAE≌⊿ACD
∴∠DAC+∠AEB=∠DAC+∠ADC=180°-∠C=120°
∴∠AFE=180°-(∠DAC+∠AEB)=60°
∵∠BFD=∠AFE
∴∠BFD=60°