(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=∠GAF∴∠AGE=∠AGF∴AG垂直于EF即AC垂直于EF又BD垂直于AC∴EF//BD∴∠BHE=∠CEF=60度∴∠HEB=180度-60度-∠EHB又BD平分角ABC,角CEB=角ABC∴∠ABC=120度-1/2∠ABC可得∠ABC=80度即角B=80度