证明：连接AC∵ABCD是菱形∴AB=BC=CD=DA∵∠B= 60°∴△ABC,△ADC是等边三角形∵∠BAE= 20°,∠EAF=60°∴∠CAF=20°在△ABE和△ACF中∵∠BAE=∠CAF=20°AB=AC∠B=∠ACF=60°∴△ABE≌△ACF∴AE=AF∴△AEF是等边三角...