Let $B'$ be the reflection of $A$ through $B$. Since the perimeter of $\triangle ABC$ equals the length of the segment $AX$, $AB$ is less than half of $AX$ and, therefore, $B'$ lies on the segment $AX$.
Let the point $C'$ lie on $AY$ such that $B'C'$ touches $\omega$ in the point $Z$. Let $C''$ be the midpoint of $AC'$. Since $B$ and $C''$ are midpoints of the sides $AB'$ and $AC'$, respectively, we have that the perimeter of $\triangle AB'C'$ is double that of $\triangle ABC''$.
We also have that the perimeter of $\triangle AB'C'$ equals
$$\begin{aligned} |AB'| + |B'C'| + |C'A| &= |AB'| + |B'Z| + |ZC'| + |C'A| \\ &= |AB'| + |B'X| + |YC'| + |C'A| \\ &= |AX| + |AY| \\ &= 2|AX|. \end{aligned}$$
Therefore, the perimeter of triangles $\triangle ABC$ and $\triangle ABC''$ is the same, namely $|AX|$.

If we assume that $C''$ lies between $A$ and $C$ we have that $|BC''| + |AC''| = |BC| + |AC|$, so
$$|BC''| = |BC| + |CC''|.$$
Which contradicts the triangle inequality, so $C''$ does not lie between $A$ and $C$. Similarly, $C''$ cannot lie between $C$ and $Y$, and must therefore lie on $C$. Hence, $C$ and $C''$ are the same point so $C$ is the midpoint of $AC'$.
Now, $\omega$ is tangent to the extensions of the sides $AB'$ and $AC'$ of $\triangle AB'C'$ as well as being tangent to the side $B'C'$ and is therefore an excircle of the triangle.
Also, $B$ and $C$ are the midpoints of sides $AB'$ and $AC'$, respectively.
We have that the point $D$ lies on the line $B'C'$, for $D$, $B'$ and $C'$ are reflections of $A$ through points on $BC$. Also, since $BC\parallel B'C'$, and $AD \perp BC$, we have that $AD \perp B'C'$. So $D$ is the foot of the altitude from $A$ in $\triangle AB'C'$. Thus, the circle through $B, D, C$ is the nine-point circle of $\triangle AB'C'$. According to Feuerbach's theorem, it touches the excircle $\omega$, as required.