Problems from Ukrainian Authors

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, $E\ne I$, 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 $I{O}_B$, 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 $B{F}_B$ both coincide with the bisector.

Now, suppose $E$ lies outside the line $I{O}_B$. Point $F_B$ is on the side $E{O}_B$ of triangle $EI{O}_B$ since the circles touch externally. In contrast, $B$ lies on the continuation of the side $O_{B}I$ of this triangle. Hence, the line $B{F}_B$ 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 $EI{O}_B$ and line passing through the points $B$, $X_B$ and $F_B$:

Fig. 23

$$\frac{E{X}_B}{X_{B}I}\cdot \frac{IB}{B{O}_B}\cdot \frac{O_B{F}_B}{F_{B}E}=1.$$

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 $I{I}^{\prime }$ and $O_B{O}_B^{\prime }$ from the points $I$ and $O_B$ to the line $AB$, it will form similar right triangles $BI{I}^{\prime }$ and $B{O}_B{O}_B^{\prime }$ with the similarity coefficient $\frac{B{O}_B}{BI}=\frac{O_B{O}_B^{\prime }}{I{I}^{\prime }}=\frac{r_B}{r}$. It means that

$$\frac{E{X}_B}{X_{B}I}=\frac{B{O}_B}{IB}\cdot \frac{F_{B}E}{O_B{F}_B}=\frac{r_B}{r}\cdot \frac{R_9}{r_B}=\frac{R_9}{r}.$$

Thus, line $B{F}_B$ 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 $I{O}_B$, because, as shown above, in this case all points of the segment $EI$ belong to the line $B{F}_B$.

By similar reasoning it follows that the lines $A{F}_A$ and $C{F}_C$ 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 $IF$.