Denote $\omega$ as the incircle $X$ as their concurrency point. Let $T$ be the point on $\omega$ such that $IT \parallel XD$. Claim 01. $XTID$ is a rhombus. Proof. Notice that since $AX$ and $AO$ are isogonal, then we must have $IA$ bisects $\angle XAO$. Notice that $IA \perp EF \parallel XI$, and hence $\angle XIA = 90^{\circ}$. Easy angle chase yields us $\triangle XID$ being isosceles. Since $IT \parallel XD$, and $IT = ID = XD$, we then have $XT \parallel TD \perp BC$, which therefore gives us $A,X,T$ collinear. Therefore, $XTID$ is a rhombus. This implies that $DT \perp XI \parallel EF$. It is well known that $AT$ is the line joining $A$ and the tangency point of $A$ mixtillinear incircle with the circumcircle of $\triangle ABC$. Since $AX \equiv AT$ is isogonal to $AK$ where $K$ is the tangency point of the $A$-excircle with $BC$. Hence, we have $A,O,K$ being collinear. By the diameter of the incircle lemma, $A,Y,O,K$ is collinear where $Y$ is the top point of the incircle. Hence, $YD = 2 OM$ since $M$ is the midpoint of $DK$. Since $AY \equiv AO \parallel TI$, and $AT = TX = ID = IY$, this gives us $AXDY$ is a parallelogram as well. Hence, \[ AX = YD = 2 OM \]Since $X$ lies on the A-altitude, this immediately implies that $X = H$, the orthocenter as $AH = 2OM$.

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Claim 1) $T_a - D - O$ are collinear: Proof. since $AH'$ and $AO$ are isogonal we know that $\angle H'AI = \angle OAI= x$ .Now since $(AIYR)$ cyclic we have that $\angle H'RY = \angle H'AI$ .which means that $\angle H'RY = x = \angle IRD$ since $(IDT_aR)$ cyclic means that $\angle IRD = \angle IT_aD = x$ ...(1) since $OT_a = OM^{bc}$ we have that $\angle OM^{bc}T_a = \angle OT_aI = x $ ...(2) from (1) and (2) we have that $T_a - D - O$ are collinear. Claim 2) $A-H'-T_a$ collinear Proof: Now since $T_a-D-O$ collinear from desargues theorem we have that $A-H'-T_a$ collinear (we use desargues in the green triangles) now if we show that Y is the midpoint of $T_aH'$ we are done. $\angle AT_aM^{bc} = \angle M^{bc}T_aD$...(3) also since $H'D || AO$ we have that $\angle T_aH'D = \angle T_aAI$ ..(4) now from (3) and (4) we have that $\angle T_aH'D = \angle H'T_aD$ and since we know that $DB\perp H'T_a$ we have that $Y$ is the midpoint of $H'T_a$ $\blacksquare$ Note: Configurations like $R-T_a-M_{bc}$ being collinear, also $\angle M_{bc}AM^{bc} = 90$, aren;t proven above,but if you want its easy to prove them.

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Let $G$ be concurrency point and Let $AI$ meet $BC$ at $R$. Let $AI$ meet $GD$ at $K$ Note that $AI \perp EF || IG$. $\angle GAO = \angle C - \angle B$ and $\angle GAI = \angle 90 - \angle AGI = \angle 90 - \angle IRD = \angle 90 - \angle B - \frac{\angle A}{2} = \frac{\angle C-B}{2}$ so $AI$ is angle bisector of $\angle GAO$ so $\angle AGK = \angle 180 - \angle GAO \implies \angle AKG = \angle KAG \implies AI = IK$ Now Note that $ID = \frac{GK}{2} = \frac{GA}{2}$ so if $M$ is where $AG$ intersects incircle then $AM = MG$ so we have $ID = \frac{AG}{2}$ so for proving $G$ is orthocenter we can instead prove $OI || BC$. Let $G'$ be reflection of $G$ across $BC$ and Let $A'$ be antipode of $A$ in $ABC$ so we have $KA' || IO$ since $AO = OA'$ and $AI = IK$. Note that $G'A' || BC$ since $AG' \perp G'A'$ and $G'K || BC$ since $GIKG'$ is cyclic so $G',K,A'$ are collinear so $IO || KA' || BC$.

Let the altitude from $D$ to $EF$ intersect $(DEF)$ at $R$. Let $K$ be the concurrency point. \[\measuredangle KDI=\measuredangle (AO,DI)=\measuredangle (AO,AK)=\measuredangle B-\measuredangle C\]\[\measuredangle DIK=\measuredangle (DI,EF)=90-\frac{\measuredangle B}{2}+\frac{\measuredangle C}{2}\]Hence $\measuredangle IKD=90-\frac{\measuredangle B}{2}+\frac{\measuredangle C}{2}=\measuredangle DIK$ which implies $DI=DK$. Also $IK\perp DL$ thus, $K$ is the reflection of $I$ with respect to $DL$. Note that $IK$ is the perpendicular bisector, of $DR$ subsequently, $DIRK$ is parallelogram. We get that $RK\perp BC\perp AK$ so $A,R,K$ are collinear. Let's work on the complex plane with unit circle $(DEF)$. \[dr+ef=0\iff r=-\frac{ef}{d}\]\[k=\frac{\overline{d}r-d\overline{r}}{\overline{d}-\overline{r}}=\frac{-\frac{ef}{d^2}+\frac{d^2}{ef}}{\frac{1}{d}+\frac{d}{ef}}=\frac{d^4-e^2f^2}{def+d^3}=d-\frac{ef}{d}\]We have that \[-\frac{ef}{d}=r=a+d=\frac{2ef}{e+f}+d\implies d(de+ef+fd)+ef(d+e+f)=0\]To get $BK\parallel IE,$ \[(b-k)(\overline{e}-\overline{j})\overset{?}{=}(\overline{b}-\overline{k})(e-j)\iff (\frac{2df}{d+f}-d+\frac{ef}{d})\overset{?}{=}e^2(\frac{2}{d+f}-\frac{1}{d}+\frac{d}{ef})\]\[d^2f-d^3+def+ef^2\overset{?}{=}de^2-e^2f+\frac{d^2e(d+f)}{f}\]\[d^2f^2-d^3f+def^2+ef^3\overset{?}{=}de^2f-e^2f^2+d^3e+d^2ef\]\[-def(d+e+f)=d^2(de+ef+fd)\overset{?}{=}d^2f^2+e^2f^2+ef^3+def^2-de^2f\]\[d^2ef+2def^2+d^2f^2+e^2f^2+ef^3\overset{?}{=}0\]Which is true since $f(d^2e+2def+d^2f+e^2f+ef^2)=f(d(de+ef+fd)+ef(d+e+f))=0$ as desired.$\blacksquare$