Solution:

Denote by $k(O, R)$ the circumcircle of $\triangle ABC$ and by $Q$ the orthogonal projection of $H$ on $CL$. Let us further denote
$$\begin{gathered} K = HQ \cap LO, \quad S = k \cap LO, \quad P = CL \cap DS \\ M = AB \cap LO, \quad N = AB \cap CD \end{gathered}$$
Note that $N$ and $M$ are the midpoints of $HD$ and $KL$, respectively. Indeed, we have $AH = AD$ ($\angle AHD = \angle ABC = \angle ADH$) and analogously $BH = BD$. Hence $AB$ is the perpendicular bisector of the segment $HD$.

On the other hand, $DLSC$ is an isosceles trapezoid and $HK \parallel CS$. Therefore $DLKH$ is also an isosceles trapezoid and $AB$ is the perpendicular bisector of the segment $KL$. Then
$$\frac{LQ}{LC} = \frac{LK}{LS} = \frac{LM}{R}$$
whence
$$LQ = \frac{LC \cdot LM}{R}$$

We also have $LP = \frac{LO \cdot LS}{LC} = \frac{2R^2}{LC}$ since $\triangle LOP \sim \triangle LCS$ and $LB^2 = LS \cdot LM = 2R \cdot LM$ since $\triangle LBS \sim \triangle LMB$. Therefore
$$LP \cdot LQ = LB^2$$
On the other hand, using $\angle LBI = \frac{\angle B + \angle C}{2} = \angle LIB$ we obtain $LB = LI$. Then $LP \cdot LQ = LI^2$ and, in particular, $Q \equiv I \Longleftrightarrow P \equiv I$.

It remains to note that $\angle CIH = 90^\circ \Longleftrightarrow Q \equiv I$ (since $\angle CQH = 90^\circ$) and $\angle IDL = 90^\circ \Longleftrightarrow P \equiv I$ (since $\angle PDL = 90^\circ$). This completes the proof.

Remark. It can be proved that $\angle CIH = 90^\circ$ iff $\cos \angle A + \cos \angle B = 1$.