A non-equilateral triangle $\triangle ABC$  of perimeter $12$ is inscribed in circle $\omega$ .Points $P$ and $Q$ are arc midpoints of arcs $ABC$ and $ACB$ , respectively. Tangent to $\omega$ at $A$ intersects line $PQ$ at $R$.
It turns out that the midpoint of segment $AR$ lies on line $BC$ . Find the length of the segment $BC$.

[i] (А. Кузнецов)[/i]
1. **Define the problem setup and key points:**
   - Let $\triangle ABC$ be a non-equilateral triangle inscribed in circle $\omega$ with perimeter 12.
   - Points $P$ and $Q$ are the midpoints of arcs $ABC$ and $ACB$, respectively.
   - The tangent to $\omega$ at $A$ intersects line $PQ$ at $R$.
   - The midpoint of segment $AR$ lies on line $BC$.

2. **Identify key properties and relationships:**
   - Since $P$ and $Q$ are arc midpoints, they are equidistant from $A$, $B$, and $C$.
   - The tangent at $A$ to $\omega$ is perpendicular to the radius $OA$.
   - The midpoint of $AR$ lying on $BC$ implies a specific geometric relationship.

3. **Use the incenter and excenter properties:**
   - Let $I$ be the incenter and $I_A$ be the $A$-excenter of $\triangle ABC$.
   - Let $D$ be the intersection of $AI$ with $BC$.
   - Since $K$ is the midpoint of $AR$, $\angle ADR = 90^\circ$.

4. **Analyze the intersection and angle properties:**
   - Since $PQ$ passes through $D$, and $\angle IBP = \angle IDP = \angle IDQ = \angle ICQ = 90^\circ$, $I$ is the Miquel point of $BPCQI_AD$.
   - By angle chasing, $W$ (the midpoint of arc $BC$) is the reflection of $I$ over $PQ$.

5. **Use the perpendicularity and reflection properties:**
   - Since $OI$ is perpendicular to $AW$, and $W$ is the midpoint of arc $BC$, this implies $AB + AC = 2BC$.

6. **Calculate the length of $BC$:**
   - Given the perimeter of $\triangle ABC$ is 12, we have $AB + BC + CA = 12$.
   - Using $AB + AC = 2BC$, we substitute into the perimeter equation:
     \[
     2BC + BC = 12 \implies 3BC = 12 \implies BC = 4
     \]

The final answer is $\boxed{4}$.