reflections of a point around the sides of a triangle: the area stays constant

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)

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)

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.

• Commented Jun 14, 2019 at 16:20
• The second part follows from taking a homothety centred at $P$ with ratio $\frac{1}{2}$ and Euler's Pedal Triangle Theorem. Commented Jun 14, 2019 at 16:22
• @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. Commented Jun 14, 2019 at 16:58

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!