角ABC、角ACB的平分线相交于点F,过过点F作DE‖BC交AB于点D,交AC于点E,若AB=9cm,AC=8cm,则△ADE的周长是

来源：学生作业学帮网编辑：学帮网时间：2024/05/08 02:58:07

∵∠ABC、∠ACB的平分线相交于点F,DE‖BC
∴∠ABF=∠FBC=∠BFD DB = DF
∠ACF=∠FCB=∠CFE EC = FE
△ADE的周长 = AD + DF + FE + AE
= AD + DB + EC + AE
= AB + AC
= 9+ 8
= 17