Let $H'$ be the reflection of $H$ about $BC$. Redefine $X$ as the point on $AB$ such that $BX=BH$ and $Y$ as the point on $AC$ such that $CY=CH$. Note that $\angle XHY=\angle BXH+\angle XYH-\angle A=(180^\circ-90^\circ+\angle A)-\angle A=90^\circ$. Since $\angle XBT=90^\circ=\angle XHY$ and $\frac{BX}{BT}=\frac{BH}{HC}=\frac{XH}{YH}$, so $X$ spirals $BT$ to $HY$. Thus, $\triangle BXH\sim\triangle TXY$, which means that $TX=TY$ and $\angle XTY=90^\circ-\angle XAY$.

Let $\triangle ABC$ be inscribed in the unit circle so that $$|a|=|b|=|c|=1$$$$h=a+b+c$$$$t = -a$$Now since $\angle XTY + \angle BAC = 90^{\circ}$ as directed angles, we have $$\left(\frac{t-x}{t-y}\right)\left(\frac{a-b}{a-c}\right) \in i\mathbb{R}$$$$\frac{(a-b)(t-x)}{(a-c)(t-y)} = -\frac{c(a-b)(\overline{t}-\overline{x})}{b(a-c)(\overline{t}-\overline{y})}$$So let $$\alpha = \frac{t-x}{t-y}$$Since $TX = TY$, we have $|\alpha| = 1$. Due to the angle condition above, we have $\alpha^2 = -\frac{c}b$. So we eliminate $c$, writing $$c = -\alpha^2b$$Then from the length condition, we have $$(t-x) = \alpha(t-y)$$$$(a+x) = \alpha(a+y)$$$$x-\alpha y = a(\alpha-1)$$Conjugating both sides gives $$\frac{a+c-x}{ac} - \frac1{\alpha}\frac{a+b-y}{ab} = \frac1a\left(\frac1{\alpha}-1\right)$$$$\alpha ab+\alpha bc-\alpha bx-ac-bc+cy = bc(1-\alpha)$$$$\alpha bx - cy = \alpha ab-ac+2bc(\alpha-1)$$$$\alpha bx + \alpha^2by = \alpha ab+\alpha^2ab-2\alpha^2b^2(\alpha-1)$$$$x + \alpha y = a(\alpha+1)-2\alpha b(\alpha-1)$$Adding the two linear equations in $x,y$ together gives $$x = \alpha a - \alpha b(\alpha-1)$$Subtracting gives $$y = \frac{a}{\alpha} - b(\alpha-1)$$Then since $h = a + b - \alpha^2b$, we have $$x-h = (\alpha-1)(a+b)$$$$y-h = (\alpha-1)\left(\alpha b-\frac{a}{\alpha}\right)$$(Note $\alpha\neq \pm1$ or else $b = -c$ and $\triangle ABC$ is not acute.) Then $$\frac{x-h}{y-h} = \frac{\alpha(a+b)}{\alpha^2b-a}$$which is pure imaginary. $\blacksquare$