设正方体棱长为a，则AE=AF=C'E'=C'F'=a/2EF=√(AE�0�5+AF�0�5)=a√2/2A'E=√(AA'�0�5+AE�0�5)=a√5/2同理可以求得E'F'=a√2/2A'F=a√5/2CE'=a√5/2CF'=a√5/2∴△A'EF≌△CE'F'∴∠EA'F=∠E'CF'
