解法一：过点A作AG∥CF，∵BE∥CF，∴AG∥CF∥BE，∴∠GAB=∠ABE=70°，∠GAC=∠ACF=30°，∴∠BAC=∠GAB-∠GAC=40°．答：∠BAC的度数是40°．解法二：∵BE∥CF，∴∠EBC+∠BCF=180°，即∠EBA+∠ABC+∠BCF=180°；...