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# Need Some Help with a Geometry Question

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H is the orthocenter of acute triangle ABC and the extensions of AH, BH, and CH intersect the circumcircle of triangle ABC at $A', B' and C'.$ We know angle AHB:angle BHC:angle CHA=9:10:11. Find angle A'B'C' in degrees. Feb 23, 2020

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H is the orthocenter of acute triangle ABC and the extensions of AH, BH, and CH intersect the circumcircle
of triangle ABC at A', B' and C'.
We know angle AHB:angle BHC:angle CHA=9:10:11.
Find angle A'B'C' in degrees. $$\begin{array}{|rcll|} \hline \mathbf{\angle AHB:\angle BHC:\angle CHA} &=& \mathbf{9:10:11} \\\\ \frac{\angle AHB}{\angle CHA} &=& \frac{9}{11} \\ \angle AHB &=& \frac{9}{11} *( \angle CHA ) \\\\ \frac{\angle BHC}{\angle CHA} &=& \frac{10}{11} \\ \angle BHC &=& \frac{10}{11}*(\angle CHA ) \\\\ \angle AHB+\angle BHC+\angle CHA &=& 360^\circ \\ \frac {9}{11} *( \angle CHA )+\frac{10}{11}*(\angle CHA )+\angle CHA &=& 360^\circ \\ \frac {19}{11} *( \angle CHA )+\angle CHA &=& 360^\circ \\ \frac {30}{11} *( \angle CHA ) &=& 360^\circ \\ \angle CHA &=& \frac {11}{30} *360^\circ \\ \mathbf{\angle CHA} &=& \mathbf{132^\circ} \\ \hline \end{array}$$

$$\text{\angle A'B'C' in degrees is 2*42^\circ = 84^\circ}$$ Feb 24, 2020