Let there be a scalene triangle $ABC$, and denote $M$ by the midpoint of $BC$. The perpendicular bisector of $BC$ meets the circumcircle of $ABC$ at point $P$, on the same side with $A$ with respect to $BC$. Let the incenters of $ABM$ and $AMC$ be $I,J$, respectively. Let $\angle BAC=\alpha$, $\angle ABC=\beta$, $\angle BCA=\gamma$. Find $\angle IPJ$.
1. **Define the problem and setup:**
   - Let $ABC$ be a scalene triangle.
   - $M$ is the midpoint of $BC$.
   - The perpendicular bisector of $BC$ meets the circumcircle of $ABC$ at point $P$, on the same side as $A$ with respect to $BC$.
   - The incenters of $\triangle ABM$ and $\triangle AMC$ are $I$ and $J$, respectively.
   - We need to find $\angle IPJ$.

2. **Reflect $I$ across $PM$ to get $I'$:**
   - Let $I'$ be the reflection of $I$ with respect to $PM$.

3. **Claim: $I'$ and $J$ are isogonal conjugates with respect to $\triangle PMC$:**
   - To prove this, we need to show that $\angle PMI' = \angle JMC$ and $\angle PCJ = \angle I'CM$.

4. **Calculate $\angle PMI'$:**
   - $\angle PMI' = \angle PMI = 90^\circ - \angle IMB$.
   - $\angle IMB = \frac{\angle AMB}{2}$ because $I$ is the incenter of $\triangle ABM$.
   - $\angle AMB = 180^\circ - \angle BMC = 180^\circ - \gamma$ (since $M$ is the midpoint of $BC$ and $P$ lies on the circumcircle).
   - Therefore, $\angle IMB = \frac{180^\circ - \gamma}{2}$.
   - Thus, $\angle PMI' = 90^\circ - \frac{180^\circ - \gamma}{2} = 90^\circ - 90^\circ + \frac{\gamma}{2} = \frac{\gamma}{2}$.

5. **Calculate $\angle JMC$:**
   - $\angle JMC = \frac{\angle AMC}{2}$ because $J$ is the incenter of $\triangle AMC$.
   - $\angle AMC = 180^\circ - \angle BMA = 180^\circ - \beta$ (since $M$ is the midpoint of $BC$ and $P$ lies on the circumcircle).
   - Therefore, $\angle JMC = \frac{180^\circ - \beta}{2}$.

6. **Verify $\angle PMI' = \angle JMC$:**
   - From the previous steps, $\angle PMI' = \frac{\gamma}{2}$ and $\angle JMC = \frac{180^\circ - \beta}{2}$.
   - Since $\gamma = 180^\circ - \alpha - \beta$, we have $\frac{\gamma}{2} = \frac{180^\circ - \alpha - \beta}{2}$.
   - Therefore, $\angle PMI' = \angle JMC$.

7. **Calculate $\angle PCJ$:**
   - $\angle PCJ = \angle PCM - \angle JCM$.
   - $\angle PCM = 90^\circ - \frac{\angle BPC}{2}$ (since $P$ lies on the perpendicular bisector of $BC$).
   - $\angle JCM = \frac{\angle ACM}{2}$ because $J$ is the incenter of $\triangle AMC$.
   - Therefore, $\angle PCJ = 90^\circ - \frac{\alpha}{2} - \frac{\gamma}{2}$.

8. **Calculate $\angle I'CM$:**
   - $\angle I'CM = \angle IBM$ because $I'$ is the reflection of $I$.
   - $\angle IBM = \frac{\angle ABM}{2}$ because $I$ is the incenter of $\triangle ABM$.
   - Therefore, $\angle I'CM = \frac{\beta}{2}$.

9. **Verify $\angle PCJ = \angle I'CM$:**
   - From the previous steps, $\angle PCJ = 90^\circ - \frac{\alpha}{2} - \frac{\gamma}{2}$ and $\angle I'CM = \frac{\beta}{2}$.
   - Since $\alpha + \beta + \gamma = 180^\circ$, we have $\frac{\beta}{2} = 90^\circ - \frac{\alpha}{2} - \frac{\gamma}{2}$.
   - Therefore, $\angle PCJ = \angle I'CM$.

10. **Conclude that $I'$ and $J$ are isogonal conjugates with respect to $\triangle PMC$:**
    - Since $\angle PMI' = \angle JMC$ and $\angle PCJ = \angle I'CM$, $I'$ and $J$ are isogonal conjugates with respect to $\triangle PMC$.

11. **Calculate $\angle IPJ$:**
    - By the claim, $\angle MPI = \angle MPI' = \angle JPC$.
    - Therefore, $\angle IPJ = \angle IPM + \angle MPJ = \angle JPC + \angle MPJ = \angle MPC$.
    - $\angle MPC = \frac{\angle BPC}{2}$ because $P$ lies on the circumcircle of $\triangle ABC$.
    - $\angle BPC = \angle BAC = \alpha$.
    - Therefore, $\angle MPC = \frac{\alpha}{2}$.

The final answer is $\boxed{\frac{\alpha}{2}}$