Lema: In $\triangle ABC$ where $A’, B’, C’$ are the foot of the altitudes from $A,B,C$ respectively and $H$ is the orthocenter. If we draw a circle $\omega$ with center $H$ and radius $HA’$ that intersects the line $B’C’$ in $D$ and $E$. Then $D$ and $E$ are the second tangency points from $B$ and $C$ to $\omega$. Proof: Let $(BC’H)$ intersect for the second time $B’C’$ in $E’$. Then $\angle BC’H = \angle BE’H = 90º$ and $\angle E’C’H = \angle E’BH$. Because $BC’B’C$ is cyclic $\angle B’C’C = \angle B’BC$ which is equivalent to $\angle E’C’H = \angle HBA’$. Then $\triangle BE’H$ and $\triangle BA’H$ are congruent because they are similar by angles and share side $BH$, so $HA’ = HE’$ and $E = E’$, but as $\angle BE’H = 90º$ then $BE$ is tangent to $\omega$. Now if $BE$ and $DC$ intersects at $K$, $H$ is the incenter of $\triangle BKC$. As the line $DE$ (between two tangency points) intersects $CH$ at $C’$ and $\angle BC’H = 90º$ it belongs to the circle $(BEH)$, which is a circle with diameter $BH$.

Let $AD$ and $BC$ meet at $E$. Then $k$ is incircle of $\triangle EDC$ and $L$ and $K$ are touching point. From $In-touch\ chord$ Lemma feet of perpendicular from $C$ to $DO$ is on $LK$ and Let's say it $R$. Since $\angle ORC=90$ $R\in$ circle with diameter $OC$

Iran Lemma.