I like it too! Solved with CyclicISLscelesTrapezoid. First, if $X = AC \cap A_1C_1$, then by Menelaus and Ceva, $(X, BO \cap AC; A, C) = -1$. Projecting from $B$, we get $(P, B'; A, C) = -1$ where $P = BX \cap (ABC)$ and $B'$ is the $B$-antipode. Note that $M, K \in (OX)$, and $-1 = (B, O; BO \cap XK, BO \cap AC) \overset X= (P', O; K, M)$ where $P'$ is the midpoint of $BP$. Also, if $M'$ is the reflection of $M$ over $O$ then $BHMM'$ is a parallelogram, so $\angle{OM'B} = \angle{BHM}$, which means it suffices to show that $OKM'B$ is cyclic. Now we invert about $(ABC)$. Because $P'KOM$ is harmonic, $K^*$ is the reflection of $M^*$ over $P'^*$. Also, $M'^*$ is the reflection of $M^*$ over $O$, and it suffices to show that $B, M'^*, K^*$. By midpoints, $OP'^* \parallel M'^*K^*$, and $M^*$ is the intersection of tangents at $A$ and $C$ so $M^*, B', P$ are collinear so $M^*B' \perp BP \implies M'^*B \parallel OP'^* \implies B, M'^*, K^*$ are collinear. $\square$

Cool result! Let $A_2$ and $C_2$ be the antipode of $A$ and $C$, respectively. By Pascal on the hexagon $AA_2C_2CBB$, the lines $A_1C_1$, $A_2C_2$, and the tangent line of $\odot(ABC)$ at $B$ are concurrent at a point $T$. Now let $M_2$ be the midpoint of $A_2C_2$. Since $\measuredangle OBT = \measuredangle OM_2T = \measuredangle OKT = 90^{\circ}$, then the points $O, K, M_2, B, T$ are concyclic. Therefore, $\measuredangle BKC_1 = \measuredangle BKT = \measuredangle BM_2T$. Finally, note that $A_2C_2 \parallel AC$, and also $MM_2 = 2OM = BH \implies BHMM_2$ is a parallelogram $\implies BM_2 \parallel HM$, which give us $\measuredangle BM_2T = \measuredangle(BM_2, M_2T) = \measuredangle(HM, AC) = \measuredangle HMA$. $\square$

VeRy InTeReStInG! Let $A_{2} = AA_{1}B \cap A_{1}C_{1}$ and $C_{2} = CC_{1}B \cap A_{1}C_{1}$. We have $\angle BA_{2}C_{2}=\angle BAA_{1}=90^\circ-\angle BCA=\angle HAC$ and $\angle BC_{2}A_{2}=\angle HCA$, so $\triangle BA_{2}C_{2} \sim \triangle HAC$. Let $A_{2}A \cap C_{2}C = T$, making $\triangle A_{2}C_{2}T$ an isosceles triangle with base angles equal to $\angle ABC$. So $\angle T=180^\circ-2\angle B=180^\circ-\angle AOC$, indicating that points $A$, $O$, $C$, $T$ is cyclic and $OT$ bisects $\angle ATC$, leading to $OA_{2}=OC_{2}$. Because $K$ is the foot of the perpendicular from $O$ to line segment $A_{2}C_{2}$, it is the midpoint of segment $A_{2}C_{2}$. Given $\triangle BA_{2}C_{2} \sim \triangle HAC$, we conclude that $\angle BKC_{1}=\angle HMA$.

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Can be done with complex bush We see that $A(a), M((a+c)/2), H(a+b+c), A_1((c+b)a^2/(a^2+bc))$, from which we easily compute $K$. Now we just need to prove $(a-m)/(h-m)$ and $(c_1-k)(b-k)$ have the same argument, i.e. that these two vectors are collinear, which is easy to see by dividing it and taking the conjugate