证明连接AN,BM∵∠ACB＝90∴∠ACN＝90∵AC＝BC,CN＝CM∴△ACN全等于△BCM∴AN＝BM∵F是AM中点,O是AB中点∴OF∥BM,且OF＝BM/2∴∠AFO＝90∵E是BN中点,O是AB中点∴OE∥AN,且OE＝AN/2∴∠BEO＝90∴矩形OECF∴EO⊥FO∵OF...