Solution:

Let the tangent line to the incircle at $E$ (which is of course parallel to $BC$) meet $AB$ and $AC$ at $X$ and $Y$, respectively.

Note that
$$\angle IXE = \frac{\angle BXE}{2} = \frac{180^\circ - \angle DBX}{2} = 90^\circ - \angle DBI = BID.$$
Therefore the right triangles $IXE$ and $BID$ are similar. We get
$$\frac{XE}{EI} = \frac{ID}{BD}$$
that is, $XE \cdot BD = r^2$ where $r$ is the inradius. Likewise, $EY \cdot DC = r^2$. So
$$\frac{XE}{EY} = \frac{DC}{BD} = \frac{BF}{FC}$$
Now $XE / BF = XY / BC = AX / AB$, so $\triangle AXE \sim \triangle ABF$ and $A$, $E$, and $F$ are collinear.