设AD、CE交于点F，作∠AFC角平分线FG交AC于G点∵∠B=60°，AD、CE为∠A、∠C角平分线∴∠AFE=∠CFD=∠FAC+∠FCA=(180°-60°)/2=60°∴∠AFG=∠CFG=60°在△AFE与△AFG中AF公用，∠EAF=∠GAF，∠AFE=∠AFG∴△AFE≌△AFG∴AE=AG同理可证CD=CG∴AE+CG=AG+CG=AC
