Consider the acute triangle ABC, having the orthocenter H and A ', B', C', the points where AH, BH, CH intersect the circle C (0, R), circumscribed to the triangle ABC. We note: OA'N BC = {A¡}, OBʼN AC = {B1}, OC'n AB = {C¡}. %3D Show that: a) HBC = \AʼBC. b) A¡H = A1 A'. c) m(«BOA1) = 180° – 2 m(
Consider the acute
$$
\mathrm{OA}^{\prime} \cap \mathrm{BC}=\left\{\mathrm{A}_{1}\right\}, \mathrm{OB'} \cap \mathrm{AC}=\left\{\mathrm{B}_{1}\right\}, \mathrm{OC'} \cap \mathrm{AB}=\left\{\mathrm{C}_{1}\right\} \text {. }
$$
Show that:
a) $\Varangle \mathrm{HBC} \equiv \Varangle \mathrm{A'}\mathrm{BC}$.
b) $\mathrm{A}_{1} \mathrm{H}=\mathrm{A}_{1} \mathrm{~A}^{\prime}$.
c) $m\left(\varangle \mathrm{BOA}_{1})=180^{\circ}-2 \mathrm{~m}(\varangle \mathrm{B})\right.$
d) $\frac{BA_{1}}{CA_{1}}=\frac{\sin 2 B}{\sin 2 C}$
e) lines $\mathrm{AA}_{1}, \mathrm{BB}_{1}, \mathrm{CC}_{1}$ are concurrent.
f) The set:
$$
\{M \in \triangle \mathrm{ABC} \mid \mathrm{MO}+\mathrm{MH}=\mathrm{R}\}
$$
has exactly three elements.
This problem with many subpoints all are for one problem.
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