### In a triangle $ABC$, $AD$ is a median. $F$ is a point on $AC$ such that the line $BF$ bisects $AD$ at $E$. If $AD = 9 \space cm$ and $AF = 3 \space cm$, find the measure of $AC$.

9 cm

Step by Step Explanation:
1. We are given that $AD$ is the median of $\triangle ABC$ and $E$ is the midpoint of $AD.$

Let us draw a line $DG$ parallel to $BF$.

2. Now, in $\triangle ADG$, $E$ is the midpoint of $AD$ and $EF \parallel DG.$

By converse of the midpoint theorem we have $F$ as midpoint of $AG.$ \begin{aligned} \implies & AF = FG && \ldots \text(1) \end{aligned}

Similarly, in $\triangle BCF$, $D$ is the midpoint of $BC$ and $DG \parallel BF.$

By converse of midpoint theorem we have $G$ is midpoint of $CF.$ \begin{aligned} \implies FG = GC && \ldots \text(2) \end{aligned}
3. From equations (1) and (2), we get \begin{aligned} AF = FG = GC && \ldots \text(3) \end{aligned} Also, from the figure we see that \begin{aligned} AF + FG + GC = AC \\ AF + AF + AF = AC && \text { [from (3)] } \\ 3 AF = AC \end{aligned}
4. We are given that $AF$ = 3 cm.
Thus, $AC = 3 AF = 3 \times 3 \text{ cm} = 9 \text{ cm}$.