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Let $ABC$ be an acute angled triangle with $\angle{BAC}$ = 60◦ and $AB$ > $AC$. Let $I$ be the incenter, and $H$ the orthocenter of the triangle $ABC$. Prove that $2\angle{AHI}$ = $3\angle{ABC}$.

I let $D,E,F$ be altitude on $AB,BC,AC$ respectively and $HI$ meet $AB$ at $X$. I observed that $\angle{AHD}$=$\angle{ABC}$. So I assume that the equality hold if $\angle{DHX}$=1/2$\angle{AHD}$ and to prove that is true I draw $HZ$ meet $AB$ at $Z$ so that $\triangle{AHZ}$ is isosceles triangle. But the problem is I don't know how to prove that $\angle{DHX}=\angle{XHZ}$. Please help

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Let $C'$ be on $AB$ so that $AB\bot CC'$ and let $\angle CBI = x ={1\over 2}\angle ABC$.

Then since $BCHI$ is cyclic ($\angle BIC = \angle BHC = 120^{\circ}$) we have $\angle IHC = \pi-x$ so $\color{red}{\angle IHC' = x}$.

Since $\angle BAH = {\pi\over 2}-2x $ we have $\color{red}{\angle AHC' = 2x} $ (observe the triangle $AHC'$.) We are done.

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