Let $I$ be the incenter of $\triangle ABC$ and let $H_a$, $H_b$, and $H_c$ be the orthocenters of $\triangle BIC$ , $\triangle CIA$, and $\triangle AIB$, respectively. The lines $H_aH_b$ meets $AB$ at $X$ and the line $H_aH_c$ meets $AC$ at $Y$. If the midpoint $T$ of the median $AM$ of $\triangle ABC$ lies on $XY$, prove that the line $H_aT$ is perpendicular to $BC$