I already posted on this before it was deleted anyway it's polarity: $$H_B,H_C,K \text{are collinear} \iff B,C-\text{midline} ,BC \text{are concurrent}$$

Just to clarify the above the poles and polars are considered WRT incircle of $\triangle ABC $.

Dear Mathlinkers, http://www.artofproblemsolving.com/community/c6t48f6h1103236_geometry sincerely Jean-Louis

Let $H_CI\cap AB = Y $ and $H_BI\cap AC = X $. Angle chasing shows that $BH_C || CH_B $. Also, $\angle IH_CB = \angle IAB = \angle IAC = \angle IH_BC $. So, $\Delta BH_CY\sim \Delta CH_BX $. This gives $\frac{BH_C}{CH_B} = \frac{BY}{CX} = \frac {BK}{CK} $. As, $BH_C || CH_B $, the result follows.

Solved with GrantStar. Construct $H_A$ the orthocenter of $BCI$ and note that $A,B,C,I,H_A,H_B,H_C$ lie on the Feuerbach hyperbola $\mathcal H$. It suffices to show $H_BH_C,BC,IH_A$ concur, or $(H_B,H_C),(B,C),(I,H_A)$ are pairs of an involution on $\mathcal H$. Project this involution through $I$. Letting $E,F$ be the tangency points of the incircle and lines $AC,AB$, and letting $M,N$ be the midpoints of $EK,FK$, we want to show that $(IE,IF),(IN,IM),(IO,IK)$ are involutive, where $O$ is the circumcenter of $ABC$ as it is well-known $IO$ is tangent to $\mathcal H$. But now this follows from DDIT from $I$ to quadrilateral $EMFN$, since it is well-known that $IO$ is the Euler line of triangle $EFK$.

In fact, the following problem is true. For isogonal conjugates $P$ and $Q$ in $\triangle{}ABC$, let $D$ be the foot of the altitude from $P$ to $\overline{BC}$, and let $H_B$ and $H_C$ be the orthocenters of $\triangle{}QCA$ and $\triangle{}QAB$, respectively. Then, we have that $\overline{H_BH_C}$ passes through $D$. To prove this, vary $P$ with degree $1$ on a line through $O$. Then, we see that $Q$ varies with degree $2$ on a rectangular hyperbola $\gamma$, so $H_B$ and $H_C$ also vary with degree $2$ on $\gamma$. When $Q$ is one of the intersections of $\gamma$ with the line at infinity we see that $H_B$ and $H_C$ are both the other intersection of $\gamma$ with the line at infinity, so $\overline{H_BH_C}$ has degree $2+2-2=2$. When $Q=H$ we see that $\overline{H_BH_C}=\overline{BC}$, so $\overline{H_BH_C}\cap\overline{BC}$ has degree $1$. We want that this is equal to $D$, so it suffices to check $3$ cases. When $Q$ is such that $\angle{}ABQ=90^\circ$ we see that $\angle{}PBC=90^\circ$, so $\overline{H_BH_C}\cap\overline{BC}=D=B$. Similarly, when $Q$ is such that $\angle{}ACQ=90^\circ$ we see that $\angle{}PCB=90^\circ$, so $\overline{H_BH_C}\cap\overline{BC}=D=C$. Then, when $Q$ is the intersection of $(ABC)$ with $\gamma$ other than $A,B,$ and $C$, we see that $\overline{H_BH_C}\cap\overline{BC}=D$ is the point at infinity on $\overline{BC}$ since $BCH_BH_C$ is a parallelogram. This gives $3$ cases, proving the claim.