∵AE=EF+BFCE=BF∴AE=EF+CE=CF∵AC=BCCE=BF∴△ACE≌△BCF（SSS）∴∠ACE=∠CBF∵BF⊥CD即∠CBF+∠BCF=90°∴∠BCF+∠ACE=90°即∠BCA=90°∴AC⊥BC
