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

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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


1 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.


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