证明:
∵AB‖DE
∴∠CAB=∠1 (同位角相等)
∵∠CAB=1/2∠BAD
∴∠BAD=2∠CAB
∴∠BAD=2∠1
∵∠BAD=∠CAB+∠DAC
∴∠BAD=∠1+∠DAC
∴∠DAC=∠1
∵∠1=∠ACB
∴∠ACB=∠DAC
∴AD‖BC (内错角相等,两直线平行)
证明:
∵AB‖DE
∴∠CAB=∠1 (同位角相等)
∵∠CAB=1/2∠BAD
∴∠BAD=2∠CAB
∴∠BAD=2∠1
∵∠BAD=∠CAB+∠DAC
∴∠BAD=∠1+∠DAC
∴∠DAC=∠1
∵∠1=∠ACB
∴∠ACB=∠DAC
∴AD‖BC (内错角相等,两直线平行)