证明：EF平分BD,DE=BE,∴∠EDB=∠EBD
BD平分∠ABC,∠ABD=∠CBD
∠EAB为△ABD外角,∴∠EAB=∠EDB+∠ABD
∠EBC=∠EBD+∠CBD.
∴∠EAB=∠EBC