Orthocenters and concyclic points

[ins-]

In the acute-angled triangle $ABC$ - $AA'$, $BB'$, $CC'$ are the heights. $H$ is the orthocenter. $O$ is the circumcenter. $H_A$, $H_B$, $H_C$ are the orthocenters of the triangles $AB'C'$, $BC'A'$, $CB'A'$. $H_1$, $H_2$, $H_3$ are the symmetric points of $H_A$, $H_B$, $H_C$ with respect to $B'C'$, $C'A'$, $A'B'$. Prove that $H_1$, $H_2$, $H_3$, $H$, $O$ lie on a circle.

[Sergic Primazon]

ins- в сообщении #1122950 писал(а):

In the acute-angled triangle $ABC$ - $AA'$, $BB'$, $CC'$ are the heights. $H$ is the orthocenter. $O$ is the circumcenter. $H_A$, $H_B$, $H_C$ are the orthocenters of the triangles $AB'C'$, $BC'A'$, $CB'A'$. $H_1$, $H_2$, $H_3$ are the symmetric points of $H_A$, $H_B$, $H_C$ with respect to $B'C'$, $C'A'$, $A'B'$. Prove that $H_1$, $H_2$, $H_3$, $H$, $O$ lie on a circle.

$C'H_b=HA'=B'H_c \Leftrightarrow (C'B' \parallel H_bH_c $ и $C'B'= H_bH_c)$

Аналогично: $(A'C' \parallel H_cH_a $ и $A'C'= H_cH_a)$, $(B'A' \parallel H_aH_b $ и $B'A'= H_aH_b)$

$H_1 \in \omega _a, H_2 \in \omega _b, H_3 \in \omega_c$

$HH_1 \parallel C'B'$, $HH_2 \parallel C'A'$ , $HH_3 \parallel A'B'$

$H_0$ - ортоцентр $\triangle H_aH_bH_c$

Получаем: $\angle H_0H_1H = \angle H_0H_3H = \angle H_0H_2H = \frac{\pi}{2}$ $\Rightarrow$ точки $H_0, H_1, H_2,H_3, H$ лежат на одной окружности.