Let ABC be any triangle and M, N and P be the points where the internal bisectors of ABC, relative respectively to the vertices A, B and C, intersect the circle circumscribed around the triangle (M ≠ A, N ≠ B and P ≠ C). Prove that the incenter of ABC is the orthocenter of MNP.

1. Consider the triangle ABC and let the internal bisectors of angles at vertices A, B, and C intersect the circumcircle of triangle ABC at points M, N, and P respectively.

2. Let I be the incenter of triangle ABC. This means I is the point of intersection of the internal bisectors of angles at A, B, and C.

3. Observe that the internal bisectors that intersect the circumcircle at M, N, and P also meet at the incenter I.

4. Since M, N, and P lie on the circumcircle and correspond to equidistant angles bisected by the internal angle bisectors, we can infer that angles at M, N, and P are subtended by the arcs opposite to the respective angles they bisect in triangle ABC.

5. To prove that the incenter I is the orthocenter of triangle MNP, examine the relationships through cyclic properties and angle chasing.

6. The bisectors divided angle ABC in such a way that the angles at points M, N and P in triangle MNP are exactly the external counterclockwise angles formed due to intersection by the internal bisectors at the circumcircle.

7. Since the incenter I of ABC lies at the concurrence of these cleaned segment of internally bisected angles, it implies the orthogonal topography constraint based angles converging at I.

8. Using properties of symmetry and cyclic quadrilaterals within triangle MNP surrounding I, it shows that the perpendiculars from the vertices of MNP to the opposite sides must meet at I, thus making I the orthocenter of MNP.

Thus, the incenter of triangle ABC is the orthocenter of triangle MNP.