Denote by $O_B$ the center of the excircle that touches $AC$, and by $E$ the center of the nine-point circle (fig. 23). Points $E$, $I$ and $F$, as well as the points $E$, $F_B$ and $O_B$ lie on the same line since $F$ and $F_B$ are the tangent points of the circles with the corresponding centers. Besides, $E \neq I$, since otherwise the circles either would not have any common points or would coincide. Taking this into account, if the point $E$ lies on the line $IO_B$, i.e. on a bisector of the angle $B$, then the points $F$ and $F_B$ lie on this line as well. Then the lines $IF$ and $BF_B$ both coincide with the bisector.

Now, suppose $E$ lies outside the line $IO_B$. Point $F_B$ is on the side $EO_B$ of triangle $EIO_B$ since the circles touch externally. In contrast, $B$ lies on the continuation of the side $O_B I$ of this triangle. Hence, the line $BF_B$ intersects the line $EI = IF$ at some point $X_B$, moreover, $X_B$ is the inner point of segment $EI$. Now we can apply Menelaus's theorem to the triangle $EIO_B$ and line passing through the points $B$, $X_B$ and $F_B$:

Fig. 23

$$\frac{EX_B}{X_B I} \cdot \frac{IB}{B O_B} \cdot \frac{O_B F_B}{F_B E} = 1.$$

Denote by $r$ the radius of the incircle of triangle $ABC$, by $r_B$ the radius of its excircle that touches side $AC$, and by $R_9$ the radius of the nine-point circle. If we drop the perpendiculars-radiuses $II'$ and $O_B O'_B$ from the points $I$ and $O_B$ to the line $AB$, it will form similar right triangles $BII'$ and $BO_B O'_B$ with the similarity coefficient $\frac{BO_B}{BI} = \frac{O_B O'_B}{II'} = \frac{r_B}{r}$. It means that

$$\frac{EX_B}{X_B I} = \frac{B O_B}{I B} \cdot \frac{F_B E}{O_B F_B} = \frac{r_B}{r} \cdot \frac{R_9}{r_B} = \frac{R_9}{r}.$$

Thus, line $BF_B$ passes through the unique point $X$ from the segment $EI$ which satisfies $\frac{EX}{XI} = \frac{R_9}{r}$. The same is also true in the case when $E$ lies on the line $IO_B$, because, as shown above, in this case all points of the segment $EI$ belong to the line $BF_B$.

By similar reasoning it follows that the lines $AF_A$ and $CF_C$ pass through the same point $X \in EI$, which satisfies $\frac{EX}{XI} = \frac{R_9}{r}$. To finish the proof, it remains to mention that point $X$, as well as the whole segment $EI$, also belongs to the line $IF$.