已知,△ABC中,角平分线BB'和CC'相等,求证:AB=AC
不妨设AB＜AC,则∠1>∠2,在AB'内作D使∠DBB'=∠2,在△BCD中,∠3+∠1＞∠C,则BD>DC.故可在CD上取CE=BD,过E作BD的平行线交CC'于F,因BD=CE,∠3=∠ECF,∠4=∠5,∴△BDB'≌△CEF,∴BB'=CF＜CC'矛盾.
同理AB>AC也不成立.
故AB=AC.