∵EF//BC∴∠EOB=∠OBC、∠FOC=∠OCB∵BO平分∠EBC、CO平分∠FCB∴∠EBO=∠OBC、∠FCO=∠OCB∴∠EBO=∠EOB、∠FCO=∠FOC∴EB=EO、FC=FO∵AB+BC+CA=24、BC=10∴AB+AC=14∴AE+EN+AF =AE+EO+OF+AF =AE+EB+FC+AF =AB+AC=14即ΔAEF的周长为18cm
