Drop the perpendicular from $A,B,C$ to $BC, CA, AB$. Name them $A', B', C'$ respectively. Notice that $X_a$ is the Miquel Point of quadrilateral $BCB'C'$. Therefore, $AX_a, B'C', BC$ must concur, which means we could redefine $Y_a$ as the intersections of $BC$ and $B'C'$. Similarly, $Y_b = AC \cap A'C'$, $Y_c = AB \cap A'B'$. Then by Desargues Theorem, it suffices to prove that $AA', BB', CC'$ are concurrent, which is true since they all concur at the orthocenter $H$.

Observe that each of the $Y$'s lie on the radical axis of the nine point circle and the circumcircle of $ABC$ which gives the collinearity, the proof is simple using power of a point, one can look at math_pi_rate's blog post.

a solution by riadok from here riadok wrote: For G4: Use polar duality centered at $H$ such, that image of $A$ is line $BC$. Then this is self duality of $ABC$. Constuction of point $Y_a$ is construction of pole of $AM_a$ under this duality as $AX_a\perp X_aM_a$ and $Y_a$ lies on pole of $A$. As $AM_a$, $BM_b$, $CM_c$ are concurrent, it follows, that $Y_a$, $Y_b$, $Y_c$ are collinear. We even proved, that this line is perpendicular to line $HR$, where $R$ is centroid of $ABC$. $\square$

Isnt $\overline{Y_AY_bY_c}$ the Orthic Axis?