∠BMC=120°
证明:
∵等边△ABD、等边△ACE
∴AB=AD,AC=AE,∠ABD=∠ADB=∠BAD=∠CAE=60
∵∠BAE=∠BAC+∠CAE,∠DAC=∠BAC+∠BAD
∴∠BAE=∠DAC
∴△ABE≌△ADC (SAS)
∴∠ABE=∠ADC
∴∠BMC=∠DBE+∠BDC
=∠ABD+∠ABE+∠BDC
=∠ABD+∠ADC+∠BDC
=∠ABD+∠ADB
=120°
当然也可以证明:∠BMD=∠CME=60°,∠DME=120°
∠BMC=120°
证明:
∵等边△ABD、等边△ACE
∴AB=AD,AC=AE,∠ABD=∠ADB=∠BAD=∠CAE=60
∵∠BAE=∠BAC+∠CAE,∠DAC=∠BAC+∠BAD
∴∠BAE=∠DAC
∴△ABE≌△ADC (SAS)
∴∠ABE=∠ADC
∴∠BMC=∠DBE+∠BDC
=∠ABD+∠ABE+∠BDC
=∠ABD+∠ADC+∠BDC
=∠ABD+∠ADB
=120°
当然也可以证明:∠BMD=∠CME=60°,∠DME=120°