(1)证明:连接OD,∵EF是⊙O的切线,∴OD⊥EF,又∵BH⊥EF,∴OD∥BH,∴∠ODB=∠DBH,∵OD=OB,∴∠ODB=∠OBD∴∠OBD=∠DBH,∴BD平分∠ABH;(2)解:过点O作OG⊥BC于点G,则BG=CG=4,在Rt△OBG中,OG===.
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