$E$ is the mid-point of a median $AD$ of $\triangle ABC$ and $BE$ is produced to meet $AC$ at $F$. Show that $AF = \frac{1}{3} AC$.

(N/A) Given: $A$ $\triangle ABC$ in which $AD$ is a median,$E$ is the mid-point of $AD$,and $BE$ is produced to meet $AC$ at $F$.
To prove: $AF = \frac{1}{3} AC$.
Construction: Draw $DG \parallel BF$ intersecting $AC$ at $G$.
Proof: In $\triangle ADG$,$E$ is the mid-point of $AD$ and $EF \parallel DG$.
By the converse of the mid-point theorem,$AF = FG$ .....$(1)$.
In $\triangle FBC$,$D$ is the mid-point of $BC$ and $DG \parallel BF$.
By the converse of the mid-point theorem,$FG = GC$ .....$(2)$.
From equations $(1)$ and $(2)$,we get:
$AF = FG = GC$ .....$(3)$.
Since $AC = AF + FG + GC$,substituting from $(3)$ gives:
$AC = AF + AF + AF = 3AF$.
Therefore,$AF = \frac{1}{3} AC$.
Hence,proved.