仅提供思路:
在AB上找一点M,使
BM=BC
则易知,BCM为正三角形,
则可推出:
EM = CD
角EDA = 180 - EDB - ADC =120-ADC
角MAC = 180 - ABC - ACD =120-ACD
已知:
AD = AC
=>
ADC=ACD
=>
EDA=MAC
又
AED = AMC = 120
=>
AMC全等于AED(AAS)
AM = ED
=>
DE + DC = AE
仅提供思路:
在AB上找一点M,使
BM=BC
则易知,BCM为正三角形,
则可推出:
EM = CD
角EDA = 180 - EDB - ADC =120-ADC
角MAC = 180 - ABC - ACD =120-ACD
已知:
AD = AC
=>
ADC=ACD
=>
EDA=MAC
又
AED = AMC = 120
=>
AMC全等于AED(AAS)
AM = ED
=>
DE + DC = AE