Just use pole polar wrt to the incircle

Here is my solution. It's well known that $I$,$O$, and orthocenter of $DEF$ are collinear. So the original problems can be rewritten as: Let $ABC$ be a triangle with $O$ be its circumcenter, $H$ its orthocenter. Let the tangent to circle $(AOB)$ at $B$ intersects the tangent to circle $(AOC)$ at $C$ at point $X$. Define $Y$ as pole of $BC$ WRT $(ABC)$. Then prove that $XY$,$OH$, and tangent to $(ABC)$ at $A$ are concurrent. Through out the solution we use pole and polar WRT to circle $(ABC)$. Let define $L_{1}$ to be the pole of $H$, and $L_{2}$ be the pole of $X$. By La Hire, we can get pole of $OH$ be point of infinity at $L_{1}$, and pole of $XY$ is point of intersection of $BC$ and $L_{2}$. Note that $XY$, $OH$ and tangent to $(ABC)$ from $A$ are concurrent iff point $A$, pole $XY$ and pole $OH$ are collinear. Define $K$ to be the pole of $XY$ so we only need to prove that $AK$ is parallel to $L_{1}$ iff $AK$ is perpendicular to OH. Now define point $K'$ as intersection of $BC$ with line through $A$ perpendicular to $OH$. So we only need to prove that $K'=K$ Let $BX$ and $CX$ intersects $(ABC)$ at $P$ and $Q$ respectively. By brokard we have intersection of $BC$ and $PQ$ is $K$. The rest is easy by trigonometry and using the fact that trilinear polar of $H$ is perpendicular to $OH$. Just prove that $\frac{BK}{CK}=\frac{BK'}{CK'}$

We present an easier solution. Let $K^*$ be the isogonal of $K$ in $\triangle{AEF}$ and let $D^*$ be the reflection of $K^*$ in the perpendicular bisector of $\overline{EF}.$ Since $AK$ and $AK^*$ are isogonal wrt $\angle{EAF}$, we have $D^*\in AK.$ Also, $$\angle{D^*FE}=\angle{K^*EF}=\angle{KEA}=\angle{EDC}=\angle{DFE}$$and in the same way $\angle{D^*EF}=\angle{DEF}$. It follows that $D^*$ is the reflection of $D$ in $\overline{EF}.$ Now, by this we know that $AD^*,OI$ and $BC$ are concurrent, so we are done.

Let $OI$ and $AK$ intersect line $BC$ on the points $X$ and $X'$ and compute $\frac {BX}{CX}$ and $\frac{BX'}{CX'}$. and for that , just use Cevian theorem for points $I$ and $K$ in triangles $\triangle BOC$ and $\triangle AEF$ respectively.

Let $\angle A = 2\alpha$,$\angle B = 2\beta$, $\angle C = 2 \theta$ By Sine Theorem on $\triangle FKE \rightarrow \frac{FK}{KE}=\frac{\sin{\theta -\alpha}}{\sin{\beta - \alpha}}$ Let $AK$ and $OI$ intersect $BC$ at $T_1$ and $T_2$. We will show $\frac{BT_1}{CT_1}=\frac{BT_2}{CT_2}$. $\frac{FK}{AK}=\frac{\sin{\angle BAK}}{\alpha + \theta}$ and similarly $\frac{EK}{AK}=\frac{\sin{\angle CAK}}{\alpha + \beta}$ $\frac{BT_1}{CT_1} = \frac{AB \sin \angle BAK}{AC \sin \angle CAK} = \frac{\sin{2\theta} \cdot \sin{\theta - \alpha} \cdot \sin{\theta + \alpha}}{\sin{2\beta} \cdot \sin{\beta - \alpha} \cdot \sin{\beta + \alpha}} $ Now lets calculate $\frac{BT_2}{CT_2}$ $\frac{\sin \angle BOI}{\sin {\theta - \alpha}}=\frac{OB}{OI} = \frac{OC}{OI} =\frac{\sin \angle COI}{\sin {\beta - \alpha}}$ $\frac{BT_2}{CT_2} = \frac{BI \cdot \sin \angle BOI}{CI \cdot \sin \angle COI}= \frac{\sin \theta \cdot \sin \theta - \alpha}{\sin \beta \cdot \sin \beta - \alpha}= \frac{\sin{2\theta} \cdot \sin{\theta - \alpha} \cdot \sin{\theta + \alpha}}{\sin{2\beta} \cdot \sin{\beta - \alpha} \cdot \sin{\beta + \alpha}}=\frac{BT_1}{CT_1}$ and we are done.

Take the isogonal conjugate $K'$ of $K$ in $\triangle AEF$. Clearly, we check that $\angle K'FE=\angle AFK=\angle BDF=\angle DEF$ thus $K'F\parallel DE$. Analogously, $K'E\parallel DF$, thus $K'EDF$ is a parallelogram. We $\sqrt{bc}$ invert at $A$. Now, the line $AK$ is sent to $AK'$, and $IO\cap BC$ is sent to $(AA'I_A)\cap (ABC)$ where $A'$ is the reflection of $A$ over $BC$ and $I_A$ the $A$-excenter. We wish to show that $AK'$ passes through this point. Yet this is clearly equivalent to showing that $K'$ lies on the radical axis of $(AA'I_A)$ and $(ABC)$. Note that by parallelogram properties $K'$ is the reflection of $D$ over the midpoint of $EF$, call it $N$. Further, let $M$ be the midpoint of minor arc $BC$. Define $f(\bullet)=\textrm{Pow}_{(AA'I_A)}(\bullet)-\textrm{Pow}_{(ABC)}(\bullet)$. By Linearity of Power of a Point, \[f(D')=\frac{D'D}{DN}f(N)+\frac{D'N}{DN}f(D)=2f(N)+f(D)\]Note that \[f(N)=\textrm{Pow}_{(AA'I_A)}(N)-\textrm{Pow}_{(ABC)}(N)=-(NA\cdot NI_A-NA\cdot NM)\]\[f(D)=\textrm{Pow}_{(AA'I_A)}(D)-\textrm{Pow}_{(ABC)}(D)=\textrm{Pow}_{(AA'I_A)}(D)+DB\cdot DC)\]We simply wish to show thus that, due to $NA\cdot NI_A-NA\cdot NM=NA\cdot MI_A$, and definition of radical axis, \[2NA\cdot MI_A-\textrm{Pow}_{(AA'I_A)}(D)-DB\cdot DC=0\]By flipping the signs. Now, we define $g(\bullet)=\textrm{Pow}_{(AA'I_A)}(\bullet)-\textrm{Pow}_{(D)}(\bullet)$. Let $I'$ be the reflection of $I$ over $D$. Note that by symmetry the center of $(AA'I_A)$ lies on $BC$, thus $I'$ and $I$ have equal power with respect to $(AA'I_A)$, as well as to $(D)$ by reflection. Thus, by linearity of Power of a Point, \[g(D)=\frac{1}{2}\cdot 2(\textrm{Pow}_{(AA'I_A)}(I)-\textrm{Pow}_{(D)}(I))=-IA\cdot II_A -r^2\]Where $r$ is the inradius. Note by Fact $5$, we have $2MI_A=II_A$. Now, reducing our expression, we wish to show \[II_A\cdot NA -II_A\cdot AI+r^2=DB\cdot DC\]Or \[II_A\cdot NI+r^2=DB\cdot DC\]Note that as $(AFN)$ is tangent to $FI$, we have $NI\cdot AI=r^2$, thus \[NI\cdot II_A+NI\cdot AI=AI_A\cdot NI=DB\cdot DC\]Note that by inversion, $AII_A\cdot AI=bc$, thus \[\frac{NI}{AI}=\frac{(s-b)(s-c)}{bc}\]Yet \[NI\sin\left(\frac{\angle A}{2}\right)=FI=\frac{NI}{\sin\left(\frac{\angle A}{2}\right)}\]Thus $\frac{NI}{AI}=\sin^2\left(\frac{\angle A}{2}\right)$ Thus, we only need to show that \[\sin^2\left(\frac{\angle A}{2}\right)=\frac{(s-b)(s-c)}{bc}\]This is well-known.

Let $ D' $ be the reflection of $ D $ WRT $ EF. $ From here we know that $ BC, OI, AD' $ are concurrent. On the other hand, since $ E(A, D'; K, F) = F(A, D'; K, E) $ we get $ A, D', K $ are collinear. Thus, we conclude that $ BC, OI, AK $ are concurrent. $ \qquad \blacksquare $