IMC 2002 — Keystage 3 (IWYMIC) — individual Question 15

In triangle $ABC$, $D$ lies on $BC$ and $AD$ bisects $\angle CAB$. Let $M$ be the midpoint of $BC$. Point $E$ lies on the extension of $BA$ with $ME\parallel AD$, and $ME$ meets $AC$ at $F$. Prove $BE=CF=\frac{AB+AC}{2}$.