如图，AD为△ABC的角平分线，M为BC的中点，ME‖AD交BA的延长线于E，交AC于F。求证BE=CF=½（AB+AC）

AD为△ABC的角平分线， AC/AB= CD/BD
M为BC的中点, AC/AB= (BM+DM)/(BM-DM)
(AC+AB)/AB = 2BM / (BM-DM)

ME‖AD交BA的延长线于E, △EBM ∽ △ABD
BF/AB = BM/BD = BM / (BM-DM) = ½(AC+AB)/AB
BF =½（AB+AC）

ME‖AD交BA的延长线于E,△CFM ∽ △CAD
CF/AC = CM/CD = BM/(BM+MD)
BF/AC = (BF/AB) * (AB/AC)
=[ BM / (BM-DM)] * [(BM-DM)/(BM+DM)]
= BM/(BM+MD)
= CF/AC

BF = CF