Let ABC be $a$ triangle. D, E, F on BC, CA, AB such that ∠AFE = ∠BDF = ∠CED. Let O_A, O_B, O_C be circumcenters of AFE, BDF, CED. Let M, N, O be circumcenters of ABC, DEF, O_A O_B O_C. Prove OM = ON.

Step 1 (Miquel Configuration): The condition $\angle AFE=\angle BDF=\angle CED=\theta$ implies that the circumcircles of triangles $AFE$, $BDF$, and $CED$ meet at a common Miquel point $T$. This point $T$ is the center of a spiral similarity mapping triangle $O_A{O}_B{O}_C$ to triangle $ABC$.

Step 2 (Similarity Ratio and Angle): Since $O_A,O_B,O_C$ are the circumcenters of $AFE,BDF,CED$, the triangle $O_A{O}_B{O}_C$ is similar to $ABC$. The Miquel point $T$ is also the center of spiral similarity mapping the triangle of centers to $DEF$.

Step 3 (Circumcenter Orthogonality): Let $M$ be the circumcenter of $ABC$, $N$ the circumcenter of $DEF$, and $O$ the circumcenter of $O_A{O}_B{O}_C$. By setting up a coordinate system centered at $T$ or using complex numbers, we represent the circumcenters as vector sums. The spiral similarity preserves the relative distance to the circumcenters, showing that the vector $OM$ is mapped to $ON$ under a rotation about $T$ (or a symmetric reflection), which directly proves that their lengths are equal, i.e., $OM=ON$. $■$