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About six months ago I came up with a nice property related to Ferma points and circular quadrilaterals, but I couldn't prove it: Let $ABCD$ be a cyclic quadrilateral. For each vertex, consider the triangle formed by the other three vertices, and define: $F_A$ as $X(13)$ of $\triangle BCD$, $F_B$ as $X(13)$ of $\triangle ACD$, $F_C$ as $X(13)$ of $\triangle ABD$, $F_D$ as $X(13)$ of $\triangle ABC$. Prove that the four points $F_A, F_B, F_C, F_D$ are concyclic. enter image description here

Is this feature known in advance? If so, please attach a source mentioning it.

Also how can it be proved?

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  • $\begingroup$ Do all the vertex angles have to measure $\le120°$? I sense a problem if one or two of them doesn't. $\endgroup$ Commented Apr 24 at 18:38
  • $\begingroup$ Okay, I've modified the question now, I used $X(13)$ This does not require conditions on angle measurements because the point is defined using paracentric coordinates and this does not require conditions on angle measurements @OscarLanzi $\endgroup$ Commented Apr 25 at 3:26
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    $\begingroup$ In Geogebra, your idea works. You move the blue points: geogebra.org/m/znrf8gcw $\endgroup$ Commented Apr 25 at 7:57

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Note that $ABF_DF_C$ is cyclic. Indeed, the (oriented) angles $AF_CB$ and $AF_DB$ are equal to $\dfrac{2\pi}{3}$. Similarly, $BCF_AF_D$, $CDF_BF_A$ and $DAF_CF_B$ are cyclic.

Now, \begin{align*} \angle F_AF_BF_C + \angle F_CF_DF_A & = 2\pi - \angle DF_BF_A - \angle F_CF_BD + 2\pi - \angle F_AF_DB - \angle BF_DF_C = \\ & = \angle F_ACD + \angle DAF_C + \angle BCF_A + \angle F_CAB = \\ & = \angle BCD + \angle DAB = \\ & = \pi. \end{align*} It follows that $F_AF_BF_CF_D$ is cyclic.

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  • $\begingroup$ This really reminds me a similar problem, but with incenters instead of Ferma points. And the proofs are basically the same! $\endgroup$ Commented 2 days ago
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    $\begingroup$ @richrow Yes. There is a general phenomenon known as Miquel's six circle theorem en.wikipedia.org/wiki/… $\endgroup$ Commented 2 days ago

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