8.证明：∵AC⊥CB,DB⊥CB∴∠ACB=∠DBC=90°在RT△ACB与RT△DBC中AB=DCCB=BC∴RT△ACB≌RT△DBC（HL）∴∠ABC=∠DCB∵∠ACB=∠DBC=90°∴∠ACB-∠ABC=∠DBC-∠DCB即∠ABD=∠ACD9.证明：∵BE=CF∴BE+EC=CF+EC即BC=EF...