Let $\triangle ABC$ and $\omega$ the circumscribed circle and $P$ inside the triangle and $\omega_1$, $\omega_2$, $\omega_3$ the circumscribed circles of the triangles $\triangle PAB, \triangle PBC, \triangle PAC$ and $O_1, O_2, O_3$ the centres of $\omega_1$, $\omega_2$, $\omega_3$.
If $O_1, O_2, O_3$ lie on $\omega$ show that $P$ is the incenter of $\triangle ABC$.
I tried to make an inversion around $P$ of radius $1$. I denote $A^*$ the inverse of $A$ and so on. I have to prove that $\angle PAB=\angle PAC$ and this is equivalent to prove that $\angle PB^*A^*=\angle PC^*A^*$. Now I am stuck. I am not sure that is ok my picture after inversion.