Consider the isosceles triangle $ABC$ with $AB = AC$, and $M$ the midpoint of $BC$. Find the locus of the points $P$ interior to the triangle, for which $\angle BPM+\angle CPA = \pi$.
1. **Given**: An isosceles triangle \(ABC\) with \(AB = AC\) and \(M\) is the midpoint of \(BC\). We need to find the locus of points \(P\) inside the triangle such that \(\angle BPM + \angle CPA = \pi\).

2. **Step 1**: Denote \(S\) as the intersection of \(AP\) and \(BC\). Let \(\angle CBP = u\) and \(\angle BCP = v\). By Steiner's theorem, we have \(\angle BPM \equiv \angle CPS\).

3. **Step 2**: From Steiner's theorem, we know that \(\frac{SB}{SC} = \left(\frac{PB}{PC}\right)^2\). Using the Law of Sines in \(\triangle BPS\) and \(\triangle CPS\), we get:
   \[
   \frac{SB}{SC} = \left(\frac{\sin v}{\sin u}\right)^2
   \]

4. **Step 3**: Consider the triangles \(\triangle ABS\) and \(\triangle ACS\):
   \[
   \frac{PA}{PS} = \frac{BA}{BS} \cdot \frac{\sin (B - u)}{\sin u}
   \]
   \[
   \frac{PA}{PS} = \frac{CA}{CS} \cdot \frac{\sin (B - v)}{\sin v}
   \]

5. **Step 4**: Equating the two expressions for \(\frac{PA}{PS}\), we get:
   \[
   \frac{SB}{SC} = \frac{\sin (B - u)}{\sin u} \cdot \frac{\sin v}{\sin (B - v)}
   \]
   Substituting \(\frac{SB}{SC} = \left(\frac{\sin v}{\sin u}\right)^2\), we have:
   \[
   \left(\frac{\sin v}{\sin u}\right)^2 = \frac{\sin (B - u)}{\sin u} \cdot \frac{\sin v}{\sin (B - v)}
   \]

6. **Step 5**: Simplifying, we get:
   \[
   \sin (B - u) \sin u = \sin (B - v) \sin v
   \]
   This implies:
   \[
   \cos (B - 2u) = \cos (B - 2v)
   \]

7. **Step 6**: Solving \(\cos (B - 2u) = \cos (B - 2v)\), we get:
   \[
   u = v \quad \text{or} \quad u + v = B
   \]

8. **Step 7**: If \(u = v\), then \(PB = PC\), meaning \(P\) lies on the angle bisector of \(\angle BAC\), which is line \(AM\).

9. **Step 8**: If \(u + v = B\), then \(\angle ABP \equiv \angle BCP\) and \(\angle ACP \equiv \angle CBP\). This implies that the lines \(AB\) and \(AC\) are tangent to the circumcircle of \(\triangle BPC\).

10. **Conclusion**: The locus of points \(P\) is either on the line \(AM\) or on an arc (interior to \(\triangle ABC\)) of the circumcircle of \(\triangle BIC\), where \(I\) is the incenter of \(\triangle ABC\).

The final answer is the locus of points \( \boxed{ P } \) is either on the line \(AM\) or on an arc of the circumcircle of \(\triangle BIC\) interior to \(\triangle ABC\).