⑴∠DFB不变,始终有∠DFB=60°.
证明:∵ΔABC是等边三角形,
∴AC=BC,∠DAC=∠ECB=60°,
∵AD=CE,∴ΔACD≌ΔCBE,
∴∠ACD=∠CBE,
∴∠DFB=∠BCF+∠CBE=∠BCF+∠ACD=∠ACB=60°.
⑵依然有∠DFB=60°.
证明:∠DAC=∠BCE=120°,AC=BC,AD=CE,
∴ΔACD≌ΔCBE,∴∠CBE=∠ACD,
∵∠DFB=∠ECF+∠E=∠ACD+∠E,
∴∠DFB=∠CBE+∠E=180°-∠BCE=60°.
⑴∠DFB不变,始终有∠DFB=60°.
证明:∵ΔABC是等边三角形,
∴AC=BC,∠DAC=∠ECB=60°,
∵AD=CE,∴ΔACD≌ΔCBE,
∴∠ACD=∠CBE,
∴∠DFB=∠BCF+∠CBE=∠BCF+∠ACD=∠ACB=60°.
⑵依然有∠DFB=60°.
证明:∠DAC=∠BCE=120°,AC=BC,AD=CE,
∴ΔACD≌ΔCBE,∴∠CBE=∠ACD,
∵∠DFB=∠ECF+∠E=∠ACD+∠E,
∴∠DFB=∠CBE+∠E=180°-∠BCE=60°.