解：作AG⊥BC交EF于H,BC于G∵EF∥BC∴△AEF∽△ABC∴AE／AB＝EF／BC＝1／3∵EF∥BC∴AG⊥EF∵AD∥EF∥BC∴AE／BE＝AH／HG∵S△AEF／S△EBC＝（EF·AH／2）／（BC·HG／2）＝1／6∴S△AEF／S△EBC=1/6
