如图,四边形ABCD是⊙O的内接四边形,延长BC,AD交于点E,且CE=AB=AC,连接BD,交AC于点F. (I)证明:BD平分∠ABC; (II)若AD=6,BD=8,求DF的长.
解:(Ⅰ)∵CE=AC,∴∠E=∠CAE, ∵AB=AC,∴∠ABC=∠ACB.∵∠DBC=∠CAE,∴∠DBC=∠E=∠CAE. ∵∠ABC=∠ABD+∠DBC,∠ACB=∠E+∠CAE,∴∠ABD=∠CAE, ∴∠ABD=∠DBC,即BD平分∠ABC. (Ⅱ)由(Ⅰ)知∠CAE=∠DBC=∠ABD.又∵∠ADF=∠ADB,∴△ADF∽△BDA,∴ , ∵AD=6,BD=8.∴ .