如图,BF⊥AC,CE⊥AB,BF与CE交与点D,且BD=CD,连结AD.(1)说明△BDE≌△CDF

连接BC
(1)
∵BF⊥AC,CE⊥AB
∴∠BED=∠DFC=90°
∵BD=DC
∠FDC=∠EDB
∴△BDE≌△CDF
(2)
∵△BDE≌△CDF
∴∠EBD=∠FCD
∵BD=CD
∴∠DBC=∠DCB
∴∠EBD+∠DBC=∠FCD+∠DCB
即：∠ABC=∠ACB
∴AB=AC
∵∠ABD=∠ACD
BD=DC
∴△ABD≌△ACD
∠BAD=∠CAD
∴D在∠BAC的平分线上