1. **Reflecting $R$ in $M$**: Let $R'$ be the reflection of $R$ in $M$. Since $M$ is the midpoint of $BC$, $R'$ lies on the circumcircle of $\triangle ABC$.

2. Projection and Reflection: Let $H$ be the projection of $D$ onto line $\tau$ (which is parallel to $BC$). Denote $V \equiv AT \cap MN$, $X \equiv AT \cap BC$, $Y \equiv AS \cap BC$, and let $D'$ be the reflection of $D$ in $M$.

3. Cross Ratio: Note that $-1 = A(U, T; R, \infty) = (U, X; M, D)$. This implies that the cross ratio of $(U, X; M, D)$ is harmonic.

4. Similarity of Triangles: From $\triangle MUR \sim \triangle MDA$, we infer:
$$\frac{XM}{XD} = \frac{UM}{UD} = \frac{RM}{RA}$$

5. Circumcenter and Right Angle: Note that $M$ is the circumcenter of $\triangle RR'S$, which implies that $\angle R'SR = 90^\circ$. Therefore, $\triangle RR'S \sim \triangle RAH$.

6. Ratio of Segments: From the similarity $\triangle RR'S \sim \triangle RAH$, we have:
$$\frac{RM}{RA} = \frac{RR'}{2RA} = \frac{RS}{2RH} = \frac{MY}{2MD}$$
Hence,
$$\frac{XM}{XD} = \frac{MY}{2MD}$$

7. Combining Ratios: It follows that:
$$\frac{XY}{XM} = \frac{YM}{XM} - 1 = 1 + 2\left(\frac{MD}{XD} - 1\right) = 1 + \frac{2XM}{XD} = 1 + \frac{MY}{MD} = \frac{D'Y}{D'M}$$

8. Parallel Lines and Harmonic Division: Because $AD' \parallel MN$, we obtain:
$$-1 = A(Y, M; X, D') = (K, M; V, \infty)$$
This shows that $AT$ bisects $MK$.

$\blacksquare$