$L'=DI\cap\Omega$, ($\Omega$=circuncirculo de $ABC$, $O$=circuncentro.) Entonces $LL'=$ diametro de $\Omega$. Como $CL$ es bisectriz de $\angle ACB$ es $\widehat{LA}=\widehat{LB}\rightarrow AB\perp LL'$,y como $CH\perp AB\rightarrow AB\perp DC$, entonces $DCL'L$ es un trapecio isosceles. $LA=LB=LI$ (un Lema muy conocido) $I=DL'\cap CL$, si esto no fuese asi entonces $\angle IDL\neq 90$. Vamos a demostrar que LM=QM y que H'=H. $\Omega_{1}=$ circunferencia de radio=$LI=R$ y centro $L$. $OL=OL'=r$. $P=LL'\cap \Omega_{1}$. $Q=$ interseccion de $LL'$ con la tangente a $\Omega_{1}$ en $I$. $S=$ proyeccion de $P$ sobre $IQ$, ($PS\parallel LI$) Como $IO\perp LL'$ y $IP$ es bisectriz de $\angle OIQ$ entonces $OP=PS=a$. $LM=m$, $MO=n$, $k=R/r$

Hint: Use the C-inmixtilinear circle.

I also have some key facts that may eventually lead to a synthetic solution:Let $X$ be the foot of perpendicular from $A$ and $Y$ be the foot of perpendicular from $B$.Let $Z=CI \cap AB$ and $W$ be the foot of the $A$-altitude.Then it is easy to see that $CI \perp IH \implies CXHIY$ is a cyclic pentagon.So since $CI$ is the $C$ bisector,we get $IX=IY$.Also note that $\angle{CIX}=\angle{CYX}=B=\angle{XBZ} \implies XIZB$ is cyclic.Thus $BI$ being the $B$ bisector yeilds $XI=IZ$.Thus $I$ is the circumcenter of $\triangle{XYZ}$.Now if $N$ is the nine-point center of $\triangle{ABC}$ then it is easy to get that $NI \perp XY$.We can then show that $OI \perp CD$(this is the main problem) which in turn implies that $\angle{IDL}=90^{\circ}$ by angle chasing.For the other part we can reverse the steps. I shall be highly excited to see a pure synthetic approach.@ricarlos can you please translate your post in english.

My solution: Let $ M $ be the midpoint of $ AB $ . Let $ O, N $ be the circumcenter, nine point center of $ \triangle ABC $ , respectively . Let $ Y=AH \cap CB , Z=BH \cap CA $ and $ T $ be the midpoint of $ CH $ . Let $ F $ be the tangent point of $ (I) $ with $ AB $ and $ E, F' $ be the reflection of $ F $ in $ I, M $, respectively . If $ \angle CIH=90^{\circ} $ : Since $ I $ is the midpoint of arc $ YZ $ of $ (CYZ) $ , so $ \triangle CYZ \cap T \cap I \sim \triangle CAB \cap O \cap L $ , hence we get $ \frac{OM}{OL}=\frac{CT}{CO}=\frac{CI}{CL} $ . ie. $ MI \parallel CO $ Since $ MN \parallel CO $ , so we get $ M, N, I $ are collinear . ... $ (*) $ Since $ I, N $ is the complement of Nagel point, circumcenter of $ \triangle ABC $, respectively , so from $ (*) $ we get $ CO $ is $ C- $ Nagel line of $ \triangle ABC $ , hence $ D $ is the tangent point of $ C- $ Mixtilinear incircle with $ (ABC) $ , so we get $ \angle LDI=90^{\circ} $ . If $ \angle LDI=90^{\circ} $ : Since $ D $ is the tangent point of $ C- $ Mixtilinear incircle with $ (ABC) $ , so we get $ CO $ is $ C- $ Nagel line of $ \triangle ABC $ . ie. $ C, E, O, F' $ are collinear . Since $ EF \parallel OM $ , so $ TC=TH=OM=IE=IF $ , hence combine with $ OM \parallel EI \parallel CT $ we get $ M, I, T $ are collinear . Since $ \angle CIT=\angle ICE=\angle TCI $ , so we get $ TH=TC=TI $ . ie. $ \angle CIH=90^{\circ} $ Q.E.D