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