Let $ABC$ an isosceles triangle, $P$ a point belonging to its interior. Denote $M$, $N$ the intersection points of the circle $\mathcal{C}(A, AP)$ with the sides $AB$ and $AC$, respectively. 

Find the position of $P$ if $MN+BP+CP$ is minimum.
1. **Setup and Definitions:**
   - Let $ABC$ be an isosceles triangle with $AB = AC$.
   - Let $P$ be a point inside the triangle.
   - Let $M$ and $N$ be the intersection points of the circle $\mathcal{C}(A, AP)$ with sides $AB$ and $AC$, respectively.
   - We need to find the position of $P$ such that $MN + BP + CP$ is minimized.

2. **Lemma Application:**
   - We use the lemma: For any $x \in \mathbb{R}$, $a \sin x + b \cos x \leq \sqrt{a^2 + b^2}$.
   - Particularly, for any $x > 0$, $x \sin \frac{A}{2} + \cos \frac{A}{2} \leq \sqrt{1 + x^2}$, with equality if and only if $x = \tan \frac{A}{2}$.

3. **Coordinate System and Simplification:**
   - Assume $D$ is the midpoint of $BC$, so $DB = DC = 1$ and $AD = a > 0$.
   - Suppose $P$ lies on $AD$ with $PD = x \in (0, a)$.
   - Then $AP = a - x$.

4. **Calculations:**
   - The length $MN$ is given by $MN = 2AP \sin \frac{A}{2} = 2(a - x) \sin \frac{A}{2}$.
   - The distances $PB$ and $PC$ are equal due to symmetry, and $PB = PC = \sqrt{1 + x^2}$.
   - The angle $\widehat{PBC}$ has $\tan \widehat{PBC} = x$.

5. **Objective Function:**
   - We need to minimize $MN + BP + CP$.
   - This translates to minimizing $2(a - x) \sin \frac{A}{2} + 2 \sqrt{1 + x^2}$.
   - Simplifying, we need to minimize $(a - x) \sin \frac{A}{2} + \sqrt{1 + x^2}$.

6. **Using the Lemma:**
   - From the lemma, $(a - x) \sin \frac{A}{2} + \sqrt{1 + x^2}$ is minimized when $x = \tan \frac{A}{2}$.
   - Therefore, $x = \tan \frac{A}{2}$ implies $\tan \widehat{PBC} = \tan \frac{A}{2}$.
   - This means $\widehat{PBC} = \frac{A}{2}$, implying $PB \perp AC$.

7. **Conclusion:**
   - The point $P$ is the orthocenter of $\triangle ABC$ when $MN + BP + CP$ is minimized.

\[
\boxed{\text{The point } P \text{ is the orthocenter of } \triangle ABC.}
\]