Let the lines through $B$ and $C$ parallel to $XY$ meet $AM$ at $P$ and $Q$, respectively. Since $BM = MC$, we have $PM = MQ$ and
$$\frac{BX}{XA} + \frac{CY}{YA} = \frac{PT}{TA} + \frac{QT}{TA} = \frac{PT + QT}{TA} = \frac{2MT}{TA} = 2.$$

Let $I_a$, $I_b$, and $I_c$ be the excenters of $\triangle ABC$ opposite $A$, $B$, and $C$, respectively. Since the figures $BH_aCI_a$, $CH_bAI_b$, and $AH_cBI_c$ are parallelograms, we have
$$\begin{aligned} 2 &= \frac{BX}{XA} + \frac{CY}{YA} = \frac{BH_a}{H_bA} + \frac{CH_a}{H_cA} = \frac{I_aC}{CI_b} + \frac{I_aB}{BI_c} \\ &= \frac{r_a}{r_b} + \frac{r_a}{r_c} = \frac{S/(p-a)}{S/(p-b)} + \frac{S/(p-a)}{S/(p-c)} = \frac{p-b}{p-a} + \frac{p-c}{p-a}, \end{aligned}$$
which is equivalent to $2a = b + c$.

On the other hand, $H_aT \perp BC \Leftrightarrow BH_a^2 - H_aC^2 = BT^2 - TC^2$. Let $U$ be the tangency point of the ex-circle opposite $A$ and $BC$. Then
$$BH_a^2 - H_aC^2 = CI_a^2 - I_aB^2 = CU_a^2 - UB^2 = (p-b)^2 - (p-c)^2 = a(c-b).$$

Also, since $BT$ and $CT$ are medians in $\triangle ABM$ and $\triangle ACM$, respectively, we have
$$\begin{aligned} BT^2 - TC^2 &= \frac{1}{2}(BA^2 + BM^2) - \frac{1}{4}AM^2 - \frac{1}{2}(CA^2 + CM^2) + \frac{1}{4}AM^2 \\ &= \frac{1}{2}(BA^2 - CA^2) = \frac{1}{2}(c-b)(c+b). \end{aligned}$$
If $b = c$, then $H_aT \perp BC$ by symmetry. If $b \neq c$, then the above implies that
$$H_aT \perp BC \Leftrightarrow BH_a^2 - H_aC^2 = BT^2 - TC^2 \Leftrightarrow a = \frac{1}{2}(b+c),$$
as needed.