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Given a triangle $\Delta ABC$ and a point $P$, we define $P_A, P_B, P_C$ as the reflections of $P$ around $BC, AC, AB$ respectively.
Now, $P_A, P_B, P_C$ are collinear if and only if $P\in(ABC)$. (1)

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Furthermore, define $c$ to be a circle concentric with $(ABC)$, i.e. $c$ has the circumcenter of $\Delta ABC$ as its center.
Then $A_{\Delta ABC}$ stays invariable as $P$ varies along $c$. (2)

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I managed to prove (1) (using a homothety of the Simson line), but how can you prove (2)?
I'm especially interested in a synthetic proof.

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  • $\begingroup$ See cut-the-knot.org/triangle/PedalTriangle.shtml $\endgroup$ – Anubhab Ghosal Jun 14 '19 at 16:20
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    $\begingroup$ The second part follows from taking a homothety centred at $P$ with ratio $\frac{1}{2}$ and Euler's Pedal Triangle Theorem. $\endgroup$ – Anubhab Ghosal Jun 14 '19 at 16:22
  • $\begingroup$ @Anubhab Ghosal Very clever ! (in the same spirit of what OP has done by homothetising Simson line). Please transform your two comments into a complete answer. I will be very happy to upvote it. $\endgroup$ – Jean Marie Jun 14 '19 at 16:58
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Let $PP_A\cap BC=\{A_1\}$, $PP_B\cap AC=\{B_1\}$ and $PP_C\cap AB=\{C_1\}.$

Thus, quadrilaterals $AC_1PB_1$, $CA_1B_1P$ and $BC_1PA_1$ are cyclic.

Now, let $P_A,$ $P_B$ and $P_C$ are collinear.

Thus, since $B_1C_1||P_BP_C$ and $B_1A_1||P_BP_A,$ we see that $A_1,$ $B_1$ and $C_1$ are collinear and we obtain: $$\measuredangle PAC=\measuredangle PC_1B_1=\measuredangle PBA_1=\measuredangle PBC,$$ which says that $PABC$ is cyclic.

Now, if $PABC$ is cyclic and $P$ is placed on the arc $AC$,

which without a point $B$, so by the same way we obtain: $$\measuredangle PC_1B_1=\measuredangle PBC$$ and $$\measuredangle PA_1B_1=\measuredangle PBA,$$ which says that $$\measuredangle PC_1B_1+\measuredangle PA_1B_1=\measuredangle ABC=180^{\circ}-\measuredangle A_1PC_1$$ and we got that $A_1,$ $B_1$ and $C_1$ are collinear,

which gives that $P_A,$ $P_B$ and $P_C$ are collinear and we are done!

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