Balti Tee 1997 · Ülesanne 15

Solution:

Let $I$ be the incenter of triangle $ABC$ (the intersection point of the angle bisectors $AA_{1}$, $BB_{1}$ and $CC_{1}$), and let $B_{1}C_{1}$ intersect the side $AC$ and the angle bisector $AA_{1}$ at $P$ and $Q$, respectively (see Figure 9).

Then

$$\angle AQ C_{1} = \frac{1}{2}\left(\widehat{AC_{1}} + \widehat{A_{1}B_{1}}\right) = \frac{1}{2} \cdot \left(\frac{1}{2} \widehat{AB} + \frac{1}{2} \widehat{BC} + \frac{1}{2} \widehat{CA}\right) = 90^{\circ}$$

Since $\angle AC_{1}B_{1} = \angle B_{1}C_{1}C$ (as their supporting arcs are of equal size), then $C_{1}B_{1}$ is the bisector of angle $AC_{1}I$. Moreover, since $AI$ and $C_{1}B_{1}$ are perpendicular, then $C_{1}B_{1}$ is also the bisector of angle $AMI$.

Similarly we can show that $B_{1}C_{1}$ bisects the angles $AB_{1}I$ and $API$. Hence the diagonals of the quadrangle $AMIP$ are perpendicular and bisect its angles, i.e. $AMIP$ is a rhombus and $MI$ is parallel to $AC$.

Similarly we can prove that $NI$ is parallel to $AC$, i.e. points $M$, $I$ and $N$ are collinear, q.e.d.

Figure 9