(1) AM⊥AB,BN⊥ABA、B异点,所以,AM//BN（2）E为切点,B为切点,OE⊥CE,OB⊥BC,OE=OB,OC平分∠EOB,∠EOC=∠BOC同理,∠AOD=∠EOD∠DOE+∠EOC=∠COD=180°/2=90°即,OD⊥OC,∠AOD和∠BOC互余∴ RtΔAOD∽RtΔBOC∴ y:OA=...