(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.
∴
.
(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.
∴
.