Let $E$ be the tangency point of the A-excircle $(I_a)$ with $BC.$ $K,E$ are symmetrical about the midpoint $M$ of $BC$ and the antipode $K'$ of $K$ WRT $(I)$ lies on $AE,$ since $A$ is the exsimilicenter of $(I) \sim (I_a).$ Thus, $MI$ is the K-midline of $\triangle KEK',$ i.e. $MI \parallel AE.$ If $T \equiv AH \cap MI,$ then $ATIK'$ is a parallelogram $\Longrightarrow$ $AT=IK'=IK.$ But $IO \parallel BC$ implies then $\frac{_1}{^2}AH=OM=IK=AT$ $\Longrightarrow$ $O,T$ are the midpoints of $EK',AH$ and the result follows.

Sorry to revive, I'm a little confused with the above solution, specifically the last sentence. Luis González wrote: But $IO \parallel BC$ implies then $\frac{_1}{^2}AH=OM=IK=AT$ $\Longrightarrow$ $O,T$ are the midpoints of $EK',AH$ and the result follows. Can someone please explain this a little more in detail? Thanks.

I think there are several steps skipped. $IO \parallel BC$ and $OM \parallel IK$ implies that quadrilateral $IKMO$ is a rectangle, so $IK=OM$. $\frac{_1}{^2}AH=OM$ is a well known fact, so the statement follows. Then since $KM=MD$, we have $OM$ is the midline of $\Delta KEK' \Longrightarrow$ O is the midpoint of $EK'$. Since $ATIK'$ is a parallelogram, we have $AHKK'$ is also a parallelogram, also since $O$ is on line $AK'$, so $AO \parallel HK$.

Once we already know that AO goes to the tangent point of A's excircle (using collinearity), we can let P be the intersection of AH and the circumcircle. Then O, K, P are collinear, HKP is isosceles, AOP is isosceles, so angle PHK = angle PAO.

Invader_2011 wrote: I think there are several steps skipped. $IO \parallel BC$ and $OM \parallel IK$ implies that quadrilateral $IKMO$ is a rectangle, so $IK=OM$. $\frac{_1}{^2}AH=OM$ is a well known fact, so the statement follows. Then since $KM=MD$, we have $OM$ is the midline of $\Delta KEK' \Longrightarrow$ O is the midpoint of $EK'$. Since $ATIK'$ is a parallelogram, we have $AHKK'$ is also a parallelogram, also since $O$ is on line $AK'$, so $AO \parallel HK$. D??

Let $M$ be the midpoint of $BC$, let $N$ be the midpoint of arc $BAC$. Note that $IOKM$ is a rectangle. As $OI\parallel BC\perp ON$ and $AI\perp AN$, we get that $ANOI$ is cyclic. It's well-known that $\angle INA=\angle IMB$ (for the proof, see here). Therefore, $\angle IOK=\angle IMB=\angle INA=\angle IOA$. Let $S=IK\cap AO$, thus as $OI\perp SK$ and $OI$ bisects $\angle SOK$, we get that $IK=IS$. As $AH=2OM=2IK=SK$ and $AH\parallel SK$, we obtain that $AHSK$ is a parallelogram, and hence $AO\parallel HK$.

Luis González wrote: Let $E$ be the tangency point of the A-excircle $(I_a)$ with $BC.$ $K,E$ are symmetrical about the midpoint $M$ of $BC$ and the antipode $K'$ of $K$ WRT $(I)$ lies on $AE,$ since $A$ is the exsimilicenter of $(I) \sim (I_a).$ Thus, $MI$ is the K-midline of $\triangle KEK',$ i.e. $MI \parallel AE.$ If $T \equiv AH \cap MI,$ then $ATIK'$ is a parallelogram $\Longrightarrow$ $AT=IK'=IK.$ But $IO \parallel BC$ implies then $\frac{_1}{^2}AH=OM=IK=AT$ $\Longrightarrow$ $O,T$ are the midpoints of $EK',AH$ and the result follows. This is a nice solution but it took me much too long to understand, so here is an easier to understand version that has essentially the same things. Let $M$ be the midpoint of side $BC$, let $K'$ be the reflection of $K$ over $I$, let $E$ be the intersection of $\overline{AK'}$ with side $\overline{BC}$, and let $T$ be the intersection of line $\overline{AH}$ with line $\overline{MI}$. We claim that $E$ is the tangency point of the $A$-excircle with $BC$. Indeed, consider $BC$ as horizontal, $K'$ is the top point of the incircle of $\triangle ABC$. By homothety centered at $A$, we get that $K'$, the top point of the incircle, will be sent to $E$, the top point of the $A$-excircle. Then, it is well-known that $M$ is the midpoint of $K$ and $E$. Next, we claim that $AK'IT$ is a parallelogram. Indeed, consider the homothety $\mathcal{H}(K, \frac{1}{2})$ which sends $\overline{KE}$ to $\overline{IM}$. Thus, $AK'$ and $TI$ are parallel. Since $\overline{AT}$ and $\overline{K'I}$ are both perpendicular to $BC$, $AK'IT$ is a parallelogram. Since $KI = IK'$ and the two are parallel, $AIKT$ is also a parallelogram. Next, we claim that $IOMK$ is a rectangle. Indeed, the problem statement tells us that $\overline{IO} || \overline{KM}$, and $\overline{IK}$ and $\overline{OM}$ are both perpendicular to $\overline{BC}$. Since $AIKT$ and $IOMK$ are parallelograms, we can join these two to form a new parallelogram: $AOMT$. This means that $\overline{AO}$ is parallel to $\overline{IM}$. however, we already proved that $\overline{IM}$ is parallel to $\overline{AE}$. Thus, $O$ lies on $\overline{AE}$. Next, we claim that $\frac{1}{2}AH = OM$. This is a lemma that holds for any triangle. By constructing the orthic triangle of $ABC$ and noting that $AH$ is the diameter of the top triangle which is inversely similar to $ABC$, we find that $AH = 2R\cos A$. This is a nice general result, so we leave the details for the reader. Furthermore, $OM = OB \cos(\angle BOM) = R\cos(\angle BOC / 2) = R\cos A$. Thus, the claim is true. Using the claim, note that $\frac{1}{2} AH = OM = IK = AT$, using the fact that $IOMK$ is a rectangle and $ATIK$ is a parallelogram proven earlier. Thus, $T$ is the midpoint of $AH$. Let $N$ be the intersection of $AH$ with side $BC$. By homothety centered at $N$, since $T$ is the midpoint of $AH$, $M$ is the midpoint of $KE$, and $\overline{TM}$ is parallel to $\overline{AE}$, we get that $\overline{HK}$ is also parallel to $\overline{AE}$. Since $O$ lies on $\overline{AE}$, we get that $\overline{HK}$ is parallell to $\overline{AO}$, as desired.