Excellent problem for practicing application of the following lemma: If $0^{\circ} <x,y < \alpha < 180^{\circ}$ such $\frac{\sin {x}}{\sin {(\alpha-x)}}=\frac{\sin y}{\sin {(\alpha-y)}}$ then $x=y$. Note that $\frac{\sin {\angle BIH}}{\sin {\angle BHI}}=\frac{BH}{BI}$, $\frac{\sin {\angle CIH}}{\sin {\angle CHI}}=\frac{CH}{CI}$ so $\frac{\sin {\angle BIH}}{\sin {\angle CIH}}=\frac{BH \cdot CI}{CH \cdot BI}$. We know that $\frac{BH}{AH}=\cot {\beta}$, $\frac{CH}{AH}=\cot {\gamma}$, $\frac{BI}{CI}=\frac {\sin {\frac{\beta}{2}}}{\sin {\frac{\gamma}{2}}}$. On the other side, we know that $\angle IBD=90^{\circ}-\frac{\beta}{2}$, $\angle ICD=90^{\circ}-\frac{\gamma}{2}$. Now we have $\frac{\sin {\angle CID}}{\sin {\angle ICD}}=\frac{CD}{ID}$, $\frac{\sin {\angle BID}}{\sin {\angle IBD}}=\frac{BD}{ID}$. This implies $\frac{\sin {\angle CID}}{\sin {\angle BID}}=\frac{CD \cdot \cos {\frac{\beta}{2}}}{BD \cdot \cos {\frac{\gamma}{2}}}$. Also, we know that $\frac{CD}{BD}=\frac{\cos {\beta}}{\cos {\gamma}}$. The rest is now a simple calculation.

Consider inverting in $A$ swapping $B, C$. Is known I goes to the $A$-excenter $J, D$ goes to $H$, and said $M$ the midpoint of arc $BC$, we have: $\angle AHI=\angle AJD=\angle JID$, the first is just inversion, the second is becouse $\angle AMD=\pi /2$, and is known $M$ is the midpoint of $IJ$, so now $$\angle BIH=-\beta /2 +90-\angle AHI, \angle DIC=\angle MIC-\angle JID=90-\beta /2-\angle AHI$$ Q. E. D.

Angle divider theorem From #2, we only need to show $\frac{\sin\angle BIH}{\sin\angle HIC}=\frac{\sin\angle CID}{\sin\angle BID}$. From the angle devider theorem, left side $$\begin{align}=&\frac{BH}{HC}\times\frac{CI}{BI}=\frac{AB\cos B}{AC\cos C}\times\frac{\sin\frac B2}{\sin\frac C2}\nonumber\\=&\frac{\sin C\cos B}{\sin B\cos C}\times\frac{\sin\frac B2}{\sin\frac C2}=\frac{\cos B\cos\frac C2}{\cos C\cos\frac B2}.\end{align}$$ Applying Ceva theorem to point $P$ and $\triangle BIC$, $$\begin{align}\frac{\sin\angle IBD}{\sin\angle CBD}\times\frac{\sin\angle BCD}{\sin\angle ICD}\times\frac{\sin\angle CID}{\sin\angle BID}=1.\tag{2}\end{align}$$ Notice that $\angle CBD=\angle CAD=\frac\pi2-\angle ADC=\frac\pi2-B$, so $$\angle IBD=\angle IBC+\angle CBD=\frac B2+\frac\pi2-B=\frac\pi2-\frac B2,$$ Similarly, $\angle ICD=\frac\pi2-\frac C2,\angle BCD=\frac\pi2-C$. Applying the induction formula, plugging into $(2)$, $$\frac{\cos\frac B2}{\cos B}\times\frac{\cos C}{\cos\frac C2}\times\frac{\sin\angle CID}{\sin\angle BID}=1.$$ Lastly, $\frac{\sin\angle BIH}{\sin\angle HIC}=\frac{\sin\angle CID}{\sin\angle BID}=\frac{\cos B\cos\frac C2}{\cos C\cos\frac B2}$.

I proposed this problem and here you could find three solutions (in Russian): https://youtu.be/6k4Sgb8fmJQ First one is angle chasing, second is trigonimetric and third one is with harmonic quadruples and isogonal conjugation with respect to the quad. Also, as it was mentioned by Michael Greenberg this problem was posted previously on aops: https://artofproblemsolving.com/community/c6h2249136p17322840

I have seen this problem 7 years ago...