In a given acute triangle $\triangle ABC$ with the values of angles given (known as $a$, $b$, and $c$), the inscribed circle has points of tangency $D,E,F$ where $D$ is on $BC$, $E$ is on $AB$, and $F$ is on $AC$. Circle $\gamma$ has diameter $BC$, and intersects $\overline{EF}$ at points $X$ and $Y$. Find $\tfrac{XY}{BC}$ in terms of the angles $a$, $b$, and $c$.
1. **Diagram and Initial Observations**:
   - Draw the acute triangle $\triangle ABC$ with inscribed circle touching $BC$ at $D$, $AB$ at $E$, and $AC$ at $F$.
   - Let $I$ be the incenter of $\triangle ABC$.
   - Circle $\gamma$ has diameter $BC$ and intersects $\overline{EF}$ at points $X$ and $Y$.

2. **Lemma**: $B$, $I$, and $X$ are collinear.
   - **Proof**:
     - Note that $\angle IDC = \angle IFC = 90^\circ$ because $D$ and $F$ are points of tangency.
     - Therefore, points $I$, $F$, $C$, and $D$ are concyclic.
     - Since $BX$ bisects $\angle EBD$ and $BD = BE$, $\triangle BDX \cong \triangle BEX$.
     - This implies $\angle XDC = 180^\circ - \angle XDB = 180^\circ - \angle XEB = \angle AEF = \angle AFE = \angle XFC$.
     - Hence, $X$, $D$, $F$, and $C$ are concyclic, implying $X$, $I$, $D$, and $C$ are concyclic.
     - Therefore, $\angle IXC = \angle IDB = 90^\circ$, so $BX \perp XC$.
     - Thus, $B$, $I$, and $X$ are collinear.

3. **Collinearity of $C$, $I$, and $Y$**:
   - By symmetry, $C$, $I$, and $Y$ are also collinear.

4. **Similarity of Triangles**:
   - $\triangle BIC \sim \triangle XIY$.
   - The ratio of their corresponding sides can be obtained by comparing their heights from $I$ to $XY$ and $I$ to $BC$.

5. **Height Calculation**:
   - Let $d_1$ be the distance from $I$ to $XY$.
   - Let $d_2$ be the distance from $I$ to $BC$.
   - Let $H$ be the intersection of $XY$ and $AI$.
   - Since $XY$ and $AI$ are perpendicular, $\triangle AEI \sim \triangle EIH$.
   - Therefore, $\angle HEI = \frac{a}{2}$, and $d_1 = r \sin \frac{a}{2}$.
   - The distance $d_2$ is simply the radius of the inscribed circle, $d_2 = r$.

6. **Ratio Calculation**:
   - Therefore, $\frac{XY}{BC} = \frac{d_1}{d_2} = \frac{r \sin \frac{a}{2}}{r} = \sin \frac{a}{2}$.

The final answer is $\boxed{\sin \frac{a}{2}}$