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So my question is this: Points $A, B,$ and $C$ lie on one line, point $P$ lies outside of this line. Prove that the centers of the circumscribed circles of triangles $ABP, BCP,$ and $ACP$ and point $P$ lie on one circle.

I'm self taught and fairly new to proofs in general and geometric proofs in particular. I'm not really sure where to start with this problem, though I think the Simson line might be applicable. Any help would be seriously appreciated.

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Let $M_A, M_B, M_C$ be the midpoints of segments $AP, BP, CP$ respectively. They lie on a common line $s_P$ which is parallel to line $AB \equiv AC$ because $M_A, M_B, M_C$ determine midsegments in triangles $ABP, BCP, CAP$.

Now, draw the orthogonal bisectors $l_A, l_B, l_C$ of segments $AP, BP, CP$ respectively. Then $M_A \in l_A,\,\, M_B \in l_B, \,\, M_C \in l_C$. The intersection points $$O_{AB} = l_A \cap l_B, \,\,\, O_{BC} = l_B \cap l_C, \,\,\, O_{CA} = l_C \cap l_A$$ are the circumcircles of triangles $ABP, BCP, CAP$ respectively.

Finally, focus on the triangle $O_{AB}O_{BC}O_{CA}$. Observe that $M_A, M_B, M_C$ are the orthogonal projections of the point $P$ on the extended edges of triangle $O_{AB}O_{BC}O_{CA}$. However, as pointed out already, the tree points $M_A, M_B, M_C$ lie on a common line $s_P$, which is called the Simson line. This is possible if and only if $P$ lies on the circumcircle of triangle $O_{AB}O_{BC}O_{CA}$ by the theorem about the Simson line $s_P$.

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