(1)∵CF=CD∴∠CFD=∠D同理∠CEB=∠B又∠D=∠B(四边形ABCD为菱形)∴∠CFD=∠CEB∵△CFE为正三角形,∠CFD+∠CFE+∠AFE=∠CEB+∠CEF+∠AEF+180度∴∠AEF=∠AFE (2)连接AC,BD,AC交EF于G,BD交CE于H ∵∠AEF=∠AFE,∠GAE=...