Let the bisector of $\angle BAC$ intersect $BC$ at $S$. Assuming $X$ is the point between $A$ and $S$, we begin by showing $MX \parallel AC$: $\frac{SM}{XM}=\frac{\frac{ab}{b+c}-\frac{a}{2}}{\frac{b-c}{2}}=\frac{a}{b+c}$ (Here $a,b$ and $c$ denote the length of $BC,AC,AB$ respectively).$\frac{SC}{AC}=\frac{\frac{ab}{b+c}}{b}=\frac{a}{b+c}$, and so $\frac{SM}{XM}=\frac{SC}{AC}$, $MX \parallel AC$. Likewise $MY \parallel AB$. Since $\frac{AY}{SY}=\frac{AB+MY}{MY}=\frac{b+c}{b-c}=\frac{AX}{XS}$, $(A,S;X,Y)=-1$. Polarity principle shows that $TS$ is the polar of $A$, thus $TS \perp AM$. But we also have $TM\perp XY$, and so $S$ is the orthocenter of $\triangle AMT \implies AT\perp BC$.

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A remark for the above solution: Note that there's a synthetic solution for the first part ($MX$ is parallel to $AC$): Indeed, if $\omega$ meets $BC$ at $P$ and $Q$ ($P$ is closer to $B$), then $P$ and $Q$ are intouch and extouch points due to the length condition. Now using the config in USA TST 2015/1 , we obtain that $PX$ is an intouch chord, so Iran lemma gives $MX$ is midline. Otherwise, I really liked the finish with pole-polar+orthocenter in the above solution (which is the new thing in this configuration).