, 2023

Solution 1. Let $I$ denote the incenter of $\triangle ABC$, let $N$ be the midpoint of the minor arc $BC$, let $ND$ meet $\omega$ again at $U$, and let $AI \cap BC = L$. Consider the inversion centered at $A$ with radius $\sqrt{AB \cdot AC}$, composed with reflection about $AI$. This inversion sends $D$ to $T$, and $L$ to $N$, so
$$\angle ATL = \angle AND = \angle ANU = \angle ATU,$$
that is, $T, L, U$ are collinear. Let $N'$ be the midpoint of arc $BAC$; it is well known that $T, I, N'$ are collinear, so we have
$$-1 = (A, L; I, I_A) \stackrel{T}{=} (A, U; N', S) \stackrel{N}{=} (I_A, D; \infty_{I_A D}, NS \cap I_A D),$$
that is, $NS$ bisects $I_A D$, so $N, S, M$ are collinear.

Let the perpendicular bisector of $AD$ (i.e. $YZ$) meet $BC$ at $W$. Then $WD^2 = WA^2$, so $W$ lies on the radical axis $\ell$ of the $A$-excircle and the point $A$. Let the $A$-excircle touch $AC, AB$ at $E, F$ respectively; then the midpoint of $AE$ and the midpoint of $AF$ also lie on $\ell$. Let $EF \cap BC = P$ and $AN' \cap BC = Q$. Since $EF \parallel AN'$, we know that $\ell$ is equidistant from $EF$ and $AN'$, so $W = \ell \cap BC$ is the midpoint of $PQ$.

Consider harmonic conjugation with respect to $BC$, that is, inversion about the circle with diameter $BC$. Let $X'$ be the harmonic conjugate of $W$ with respect to $BC$, and let $M_{BC}$ be the midpoint of $BC$. Then
$$-1 = (P, Q; W, \infty_{BC}) = (D, L; X', M_{BC}) \stackrel{N}{=} (D, I_A; NX' \cap DI_A, \infty_{DI_A}),$$
that is, $NX' \cap DI_A = M$, so $N, X', M$ are collinear. Also, as shown above, $N, S, M, X$ are collinear, so $X = X'$. That is, $-1 = (B, C; X, W)$. Since $W = XY \cap BC$ (using the notation for the point $XY \cap BC$), it follows that $AX, BY, CZ$ are concurrent. $\square$

Solution 2. Let $I$ be the incenter of $\triangle ABC$, let $N$ be the midpoint of minor arc $BC$, and let $I_A D$ meet $TI$ at $P$.

We first prove again that $N, S, M$ are collinear: let $N'$ be the midpoint of arc $BAC$. Notice that
$$\angle IAD = \angle TAN = \angle TN'N = \angle IPD,$$
so $A, I, P, D$ are concyclic. Therefore
$$I_A T \cdot I_A S = I_A N \cdot I_A A = \frac{1}{2} I_A I \cdot I_A A = \frac{1}{2} I_A D \cdot I_A P = I_A M \cdot I_A P,$$
so $S, T, M, P$ are concyclic. Hence
$$\angle TSM = \angle TPM = \angle IPD = \angle IDA = \angle TAN,$$
which shows $N, S, M$ are collinear.

Let $YZ$ meet $BC$ at $X'$. Since the original problem is equivalent to $(B, C; X, X') = -1$, it suffices to prove that $\overline{M_A B}^2 = \overline{M_A X} \cdot \overline{M_A X'}$, where $M_A$ is the midpoint of segment $BC$.

Let $E$ be the point where the incircle of $\triangle ABC$ touches side $BC$, and let $K$ be the intersection of $NM_A$ and $AD$. It is well known that $\overline{M_A D} = \overline{M_A E}$ and that $AD$ passes through the reflection of $E$ about $I$; hence $\overline{KM_A} = \overline{IE}$ and $\overline{KE} = \overline{KD}$.

Notice that $MN \parallel DI$, so $\triangle NM_A X \sim \triangle IED$, and therefore
$$\overline{M_A X} = \overline{DE} \cdot \frac{\overline{NM_A}}{\overline{IE}}.$$
On the other hand, since
$$\angle KAX' = -\angle KDX' = -\angle KDE = \angle KED = \angle KEX',$$
we have $A, K, E, X'$ concyclic. And
$$\overline{M_A X'} = \frac{\overline{M_A K}}{\tan \angle KX'E} = \frac{\overline{IE}}{\tan \angle KX'E} = \frac{\overline{IE}}{\tan \angle KAE} = \frac{\overline{IE}}{\tan \angle DAE},$$
so the original claim only requires proving that
$$\overline{M_A B} \cdot \overline{M_A C} = \frac{\overline{DE} \cdot \overline{NM_A}}{\tan \angle DAE}.$$
If we let $O'$ be the circumcenter of $\triangle ADE$, we have $\frac{\overline{DE}}{\tan \angle DAE} = 2\overline{O'M_A}$. And by the power of a point, we have $\overline{M_A B} \cdot \overline{M_A C} = \overline{NM_A} \cdot \overline{M_A N'}$. So finally it is equivalent to prove that $\overline{M_A N'} = 2\overline{O' M_A}$, which follows from the following lemma.

Lemma. The circumcenter $O'$ of $\triangle ADE$ is the midpoint of $\overline{N' M_A}$.

Proof of the lemma. Let the perpendicular from $A$ to $BC$, and $AI$, meet the circumcircle $\Gamma$ of triangle $ADE$ again at $U, V$ respectively. It is well known that $A(U, V; D, E)_\Gamma = -1$, so the second intersection point $A^*$ of $VM_A$ with $\Gamma$ satisfies $A^*U \parallel DE$, that is, $A^*$ is the antipode of $A$ on $\Gamma$. Hence $\angle AVA^* = 90^\circ = \angle VAN'$, that is, $AN \parallel VA^*$. Also, since $M_A U$ and $M_A V$ are symmetric with respect to $DE$, we obtain that $AUM_A N'$ is an isosceles trapezoid, so the midpoint of $\overline{M_A N'}$ lies on the perpendicular bisector of $AU$, which is exactly the circumcenter $O'$ of $\triangle ADE$. $\square$