Suppose that $Y_3$ and $Z_3$ is the midpoint of $Y_1Y_2$ and $Z_1Z_2$ and $O$ is circumcenter. Observe that $ZYZ_2Y_2$ is cyclic because $\angle Y_2ZZ_2 = \angle Z_2YY_2 = \frac{\pi}{2}$ Thus we get that $$\angle OY_2Z_2 = \angle OZY = \frac{\pi}{2} - B$$ With same argument to cyclic quadrilateral $ZYZ_1Y_1$, we will also get $$\angle OY_1Z_1 = \angle OZY = \frac{\pi}{2} - B$$ Observe that $Y_1Z_1$ is parallel to $Y_2Z_2$, since $Y_3$ and $Z_3$ are midpoint, then we got $Y_3Z_3$ is parallel to $Y_2Z_2$ as well. Now, since parallel $Y_3Z_3$ is parallel to $Y_2Z_2$, then $\angle OY_3Z_3 = \frac{\pi}{2} - B$ as well, which means $\angle KY_3Z_3 = \frac{\pi}{2} - \angle OY_3Z_3 = B$ Also observe that quadrilateral $KY_3OZ_3$ is cyclic, thus we get $$\angle KOZ_3 = \angle KY_3Z_3 = B$$ But since $O$ is circumcenter, $OX \perp BC$, we also easily get $\angle XOZ_3 = B$. Thus $$\angle XOZ_3 = \angle KOZ_3 \rightarrow X \in OK$$ Since $K,X,O$ are collinear, that also means $K$ lies in the perpendicular bisector of $BC$, which obviously means $KB = KC$