Lozada's cyclologic triangles

Let $ABC$ be a triangle, $P$ a point, $A'B'C'$ the antipedal triangle of $P$ and $A''$, $B''$, $C''$ the reflections of $P$ in $A$, $B$, $C$, respectively. Then, for any $P$ distinct from the Incenter of $ABC$, $A'B'C'$ and $A''B''C''$ are cyclologic. (César E. Lozada)

Proof. Notice that as $B''$ and $A''$ are reflections around $A'C'$ and $B'C'$, respectively, $\angle{BC'P}=\angle{BC'B''}$ and $\angle{AC'P}=\angle{AC'A''}$. Since $\angle{BC'P}+\angle{AC'P}=\angle{A'C'B'}$ it follows that $\angle{B''C'A''}=2\angle{A'C'B'}$. Analogously, $\angle{B''A'C''}=2\angle{C'A'B'}$ and $\angle{C''B'A''}=2\angle{A'B'C'}$. 

Supose $Q$ is the second intersection of circles $(A'C''B'')$ and $(B'A''C'')$, then, $\angle{A''QB''}=\angle{C''QB''}-\angle{C''QA''}$. But $$\angle{C''QB''}=\angle{C''A'B''}=2\angle{B'A'C'}$$ and $$\angle{C''QA''}=180^\circ-2\angle{A'B'C'},$$ so $$\angle{A''QB''}=2\angle{B'A'C'}+2\angle{A'B'C'}-180^\circ.$$ Now,  $\angle{A''QB''}+\angle{B''C'A''}=2\angle{B'A'C'}+2\angle{A'B'C'}-180^\circ+2\angle{A'C'B'}=180^\circ$. Which means that $QB''C'A''$ is cyclic. 

$\square$

The concurrency of circles $(A'B'C'')$, $(A'C'B'')$ and $(B'C'A'')$ is guaranteed by the fact that the cyclologic theorem is symmetric. 

Remark: $Q$ lies on the circumcircle of $A'B'C'$.

Proof: The facts that $ABC$ is homothetic with $A''B''C''$ and $A'CPB$ and $B'CPA$ are cyclic quadrilaterals lead us to infer that

$$\angle{A'QB''}=\angle{A'C''B''}=(90^\circ-\angle{CA'P})-\angle{BA'P}=90^\circ-\angle{B'A'C'}.$$ 

Similarly,

$$\angle{A''QB'}=\angle{A''C''B'}=(90^\circ-\angle{CB'P})-\angle{AB'P}=90^\circ-\angle{A'B'C'}.$$

We already know that $\angle{A''QB''}=2\angle{B'A'C'}+2\angle{A'B'C'}-180^\circ$. As $\angle{A'QB'}=\angle{A'QB''}+\angle{B''QA''}+\angle{A''QB'}$, it follows that

$$\angle{A'QB'}+\angle{A'C'B'}=\angle{B'A'C'}+\angle{A'B'C'}+\angle{A'C'B'}=180^\circ.$$

$\square$ 

Related material: Cyclologic triangles!