答案:ΔAEF的周长=14cm
过程:解证:∵ EF∥ BC
∴∠EOB=∠OBC
又∵ BO平分∠ABC
∴∠EBO=∠OBC
∴∠EOB=∠EBO
∴ EB=EO
同理: FC=OF
所以,ΔAEF的周长=AE+EF+AF=AE+EO+OF+AF
=(AE+EB)+(AF+FC)=AB+AC
=ΔABC的周长-BC=24-10=14(cm)
答案:ΔAEF的周长=14cm
过程:解证:∵ EF∥ BC
∴∠EOB=∠OBC
又∵ BO平分∠ABC
∴∠EBO=∠OBC
∴∠EOB=∠EBO
∴ EB=EO
同理: FC=OF
所以,ΔAEF的周长=AE+EF+AF=AE+EO+OF+AF
=(AE+EB)+(AF+FC)=AB+AC
=ΔABC的周长-BC=24-10=14(cm)