We complex bash with $\Omega$ as the unit circle, so that $$|a|=|b|=|c|=1$$$$h=a+b+c$$$$o=0$$$$e=\frac12\left(a+b+c-\frac{ac}b\right)$$$$f=\frac12\left(a+b+c-\frac{ab}c\right)$$$$d=-\frac{bc}a$$$$m=\frac{b+c}2$$Let $T$ be the circumcenter of $\triangle DEF$. We have the vectors $$e' = e-d = \frac{(a+b)(ab-ac+2bc)}{2ab}$$$$f' = f-d = \frac{(a+c)(ac-ab+2bc)}{2ac}$$Then if $t'=t-d$, we have \begin{align*} t' &= \frac{e'f'(\overline{e'}-\overline{f'})}{\overline{e'}f'-e'\overline{f'}} \\ &= \frac{\frac{(a+b)(a+c)(ab-ac+2bc)(ac-ab+2bc)}{4a^2bc}\left[\frac{(a+b)(2a-b+c)}{2abc} - \frac{(a+c)(2a-c+b)}{2abc}\right]}{\frac{(a+b)(a+c)(2a-b+c)(ac-ab+2bc)}{4a^2bc^2} - \frac{(a+b)(a+c)(2a-c+b)(ab-ac+2bc)}{4a^2b^2c}} \\ &= \frac{(ab-ac+2bc)(ac-ab+2bc)\left[(a+b)(2a-b+c) - (a+c)(2a-c+b)\right]}{2ab(2a-b+c)(ac-ab+2bc) - 2ac(2a-c+b)(ab-ac+2bc)} \\ &= \frac{(b-c)(ab-ac+2bc)(ac-ab+2bc)(-b-c)}{2a(b-c)(-b-c)(2a^2-ab-ac+2bc)} \\ &= \frac{(ab-ac+2bc)(ac-ab+2bc)}{2a(2a^2-ab-ac+2bc)} \\ &= \frac{2a^2bc-a^2b^2-a^2c^2+4b^2c^2}{2a(2a^2-ab-ac+2bc)} \end{align*}and so $$t = t'+d = -\frac{(b+c)(ab+ac-2bc)}{2(2a^2-ab-ac+2bc)}$$Then $$x = \frac{t}{d\overline{t}} = \frac{ab+ac-2bc}{2a-b-c}$$and (renaming $I$ to $J$) we have $$j = \frac{ax(b+c) - bc(a+x)}{ax-bc} = \frac{a^2b^2+a^2c^2-2ab^2c-2abc^2+2b^2c^2}{a^2b+a^2c-4abc+b^2c+bc^2}$$Let line $MX$ meet $\Omega$ again at $P$, so that $$p = \frac{x-m}{x\overline{m}-1} = \frac{bc}a$$Let line $PO$ intersect line $BC$ at $Q$, so that $$q = \frac{bc(p-p) - p(-p)(b+c)}{bc - p(-p)} = \frac{bc(b+c)}{a^2+bc}$$It suffices to show that $E, F, J, Q$ are concyclic -- since from this it will follow that $Q = G$, and then point $P$ lies on $\Omega$, line $MX$, and line $OG$. Now we have the vectors $$e-j = \frac{(a-b)(b-c)(a^2b+a^2c-2abc-b^2c+bc^2)}{2b(a^2b+a^2c-4abc+b^2c+bc^2)}$$$$f-j = \frac{(a-c)(b-c)(a^2b+a^2c-2abc-bc^2+b^2c)}{2c(a^2b+a^2c-4abc+b^2c+bc^2)}$$$$e-q = \frac{(a+b)(a^2b-a^2c+2abc-b^2c-bc^2)}{2b(a^2+bc)}$$$$f-q = \frac{(a+c)(a^2c-a^2b+2abc-b^2c-bc^2)}{2c(a^2+bc)}$$Then $$\frac{(e-j)(f-q)}{(f-j)(e-q)} = \frac{(a-b)(a+c)(a^2b+a^2c-2abc-b^2c+bc^2)(a^2c-a^2b+2abc-b^2c-bc^2)}{(a-c)(a+b)(a^2b+a^2c-2abc-bc^2+b^2c)(a^2b-a^2c+2abc-b^2c-bc^2)}$$and we verify that this does indeed equal its own conjugate, so it is real, and $E,F,J,Q$ are concyclic. $\blacksquare$

Assume $FE\cap DX=J$.Easy to know $J\in BC$. Assume $AD\cap BC=K$.Since $E,F,M,K$ are concyclic,we have $D,X,K,M$ and $D,X,G,I$ are concyclic.Also easy to get $D,G,O$ are colinear and $\angle DXM=90^{\circ}$.