Let $H'$ be the reflection of $H$ across $BC$. $H'$ lies on $(ABC)$. Note that $\angle BH'O=90^\circ-\angle BAH'=\angle AEO$, and $BO=CO=EY$, so $EBH'YO$ is concyclic. Similarly, $H'XODC$ is concyclic. Thus, $\angle XHY=\angle XH'Y=\angle XH'O+\angle YH'O=\angle XCO+\angle YBO=\angle COD+\angle BOE=180^\circ-\angle BOC=180^\circ-\angle BAC$, as desired.

Complex bash with $\triangle ABC$ inscribed in the unit circle, so that \begin{align*} |a|=|b|=|c|&=1 \\ o &= 0 \\ h &= a+b+c \\ d = \frac{ac(i\sqrt{bc}-i\sqrt{bc}) - i\sqrt{bc}(-i\sqrt{bc})(a+c)}{ac - i\sqrt{bc}(-i\sqrt{bc})} &= \frac{b(a+c)}{b-a} \\ e = \frac{ab(i\sqrt{bc}-i\sqrt{bc}) - i\sqrt{bc}(-i\sqrt{bc})(a+b)}{ab - i\sqrt{bc}(-i\sqrt{bc})} &= \frac{c(a+b)}{c-a} \\ x = b + d - o &= \frac{b(b+c)}{b-a} \\ y = c + e - o &= \frac{c(b+c)}{c-a} \end{align*}Then we find the vectors \begin{align*} h-x &= \frac{a(a+c)}{a-b} \\ h-y &= \frac{a(a+b)}{a-c} \end{align*}and so $$\left(\frac{h-x}{h-y}\right)\left(\frac{a-b}{a-c}\right)^2 = \frac{(a+c)(a-c)}{(a+b)(a-b)}\cdot\frac{(a-b)^2}{(a-c)^2} = \frac{(a-b)(a+c)}{(a+b)(a-c)}$$which is real. Then since $\angle XHY + 2\angle BAC$ is a multiple of $180^{\circ}$, and since $\angle BAC$ is acute, it must equal $180^{\circ}$. $\blacksquare$