已知:如图AB∥EF。说明:∠BCF=∠B+∠F解:经过C画CD∥AB∴∠B=∠1 ( )∵AB∥EF而CD∥AB(画图)∴CD∥EF ( )∴∠F=_______( )∴∠1+∠2=∠B+∠F( )即∠BCF=∠B+∠F