How I can show that the point P (Miquel point) is in the circle formed by the centers of the other 4 circles?

Question

This is the theorem

If the points $A', B', C'$ on the sides $BC, CA, AB$ of a triangle ABC are collinear, then the centers of the circumcircles of the triangles $\triangle AB'C'$ $\triangle A'BC$ $\triangle 'A'B'C$ and $\triangle ABC$ form a cyclic quadrilateral. Besides the circumcircle of the quadrilateral passing through the point of concurrency of the four circumcircles.

This is my picture

It remains for me to prove that the circumcircle formed by the centers of the other 4 is circuncirculos point P or point Miquel

Anyone can help me please?

Thanks for your help

Answer 1

Hint: We recognize that the circumcircles are hard to work with directly. Instead, let's consider the points $O^*, A^*, B^*, C^*$, which are obtained by expansion from point $P$ by a factor of 2.

We then want to show that these 5 points are concyclic.We focus on showing that $O^*, B^*, P, C^*$ are con cyclic. This follows easily from slight angle chasing. Hence we are done.

---

Judging from the phrasing of your question, I think you have shown that the 4 circles are concurrent. If not, here is an approach:

Let $\Gamma$ be the circumcircle of $ABC$ and $\Gamma_A$ be the circumcircle of $AB'C'$ and the rest defined similarly. Show that $\Gamma, \Gamma_A, \Gamma_B$ are concurrent by slight angle chasing. Hence, the 4 circles are concurrent.