Prove that $GX$ is a symmedian of $\triangle GHI$. Problem:

Let $\omega$ be the incircle of $\triangle \rm ABC$. Suppose $\omega$ touches $\rm BC$, $\rm  CA$, $\rm AB$ at points
$\rm D$, $\rm E$, $\rm F$ respectively. $\rm AD$, $\rm BE$, $\rm CF$ intersect with $\omega$ at points $\rm G$, $\rm H$, $\rm I$
respectively. $\rm HI$ and $\rm AD$ intersect at point $\rm X$. If $\rm GH = 8$, $\rm HI = 9$, $\rm IG = 10%$, Find $\rm IX$.

(In Edit 2 is written my doubt).
My attempt:
After drawing a sketch of the problem I concluded that I could find $\rm IX$ just by looking at $\triangle \rm GHI$ since the rest of the information didn´t seem useful.
So I drew the following sketch:

Now we know,
$\begin{equation*} (GM_1)^2 + (M_1H)^2 = (GH)^2 = 8^2 \space \text{(I)} \\ (M_1I)^2 + (M_1H)^2 = (HI)^2 = 9^2 \space \text{(II)} \end{equation*}$
so by subtracting $\text{(II)}-\text{(I)}$, $17 = (M_1I)^2 - (GM_1)^2 = (GM_1 + M_1I)(M_1I - GM_1) = GI(M_1I - GM_1) = 10(M_1I - GM_1)$ $\implies M_1I - GM_1$$=\frac{17}{10}$ and since we know that $GM_1 + M_1I = 10$ then by adding both we get $2M_1T = \frac{117}{10} \implies M_1I = \frac{117}{20}$ and $GM_1 = \frac{83}{20}$. Furthermore, $(GH)^2 = (GM_1)^2+(M_1H)^2 \implies 8^2 = (\frac{83}{20})^2 +(M_1H)^2 \implies M_1H = \frac{9\sqrt{231}}{20}$. Now we know,
$\begin{equation*} (\frac{117}{20} - M_2I)^2 + (\frac{9\sqrt{231}}{20} - M_2X)^2 = (9 - IX)^2 \space \text{(I)} \\ (M_2I)^2 + (M_2X)^2 = (IX)^2 \space \text{(II)} \end{equation*}$
Since there are 3 unknowns I would need one more equation to find the solution but I am unable to come up with it, any hints would be appreciated.
Thanks in advance.
Edit 1:
Thanks to everyone that has helped, I have been pondering on the problem and this is what I have got:

I am trying to find some relation that helps me find $IX$ using all the information given in the problem. First of all, I want to point out that $EB$, $FC$ and $DA$ are the three symmedians of $\triangle FDE$ since there is a theorem that states that the line segment that connects a vertex to the intersection of the tangents from the other two vertices to the circumcircle is a symmedian of such triangle. Luckily, symmedians have many properties that we can use to get more information (I will try seeing what I get knowing that symmedians bisect the
antiparallels to the opposite side), however, I don´t find a way to use this new insight on $\triangle GHI$.
Any more help would be hugely appreciated, thanks to everyone.
Edit 2:
Finally, I wasn´t able to solve it. I don´t have access to the full solution but I read that the crux is realising that $GX$ is a symmedian of $\triangle GHI$. The rest just follows from a property that says that symmedians divide the opposite side in the ratio of
the square of the sides i.e $\frac{HX}{IX} = (\frac{GH}{IG})^2$ so $\frac{9 - IX}{IX} = (\frac{8}{10})^2 \implies 9 - IX = \frac{16}{25}IX \implies 9 = \frac{41}{25}IX \implies IX = \frac{225}{41}$.
Now my doubt is how to prove that $GX$ is a symmedian of $\triangle GHI$. I would appreciate any help.
 A: 
According to the opposite of the Ceva theorem, $AD, BE, CF$ are concurrent and this point is called $G_{e}$. Let the tangent drawn from the point $G$ to the circle $\omega$ intersect the lines $AC$ and $AB$ at the points $P$ and $Q$, respectively. It seems that the Nagel line of the triangle $APQ$ intersects the $\omega$ circle, which is the outer tangent circle, at the point $D$. Then the intersections of the tangent drawn from the point $D$ to the circle $\omega$ with the rays $[AQ$ and $[AP$ are called $B$ and $C$, respectively. Using this information:
\begin{equation}
    A, G, G_{e}, X, D \ \ \text{are linear.}\tag{1}
\end{equation}
Let's perform similar steps for the points $H$ and $I$ and name them as shown in the figure. If similar steps are applied for $A\rightarrow A', \ G\rightarrow D, \ D\rightarrow G$, then $\displaystyle A', D, G$ is linear. Using $(1)$; $A, G, G_{e}, X, D$ and $A', D, G$ if is linear then $A', D, X, G, G_{e}$ is also linear. Similarly, if the same thing is applied to the points $B'$ and $C'$, the triangle $GHI$ becomes the Gergonne triangle of the triangle $A'B'C'$. The problem can be solved by using: "The Gergonne point of the triangle $A'B'C'$ is the Symmedian point of the Gergonne triangle of the same triangle". Or;
let's draw $HD$ ve $DI$.
$$\angle{HGD}=\angle{DHA'} \Longrightarrow \triangle{GHA'}\sim \triangle{HDA'},$$
$$\frac{8}{HD}=\frac{A'H}{A'D},$$
$$\angle{IGA'}=\angle{A'ID} \Longrightarrow \triangle{GIA'}\sim \triangle{IDA'},$$
$$\frac{10}{DI}=\frac{A'I}{A'D},$$
$$A'D=\frac{A'H \ HD}{8}=\frac{A'I \ DI}{10}, \ \ A'H=A'I,$$
\begin{equation}
    8DI=10HD \Longrightarrow \frac{HD}{DI}=\frac{8}{10}.
\end{equation}
Now, using the similarities;
$$\triangle{HDX}\sim \triangle{GIX}: \frac{DX}{m}=\frac{HD}{10}\Longrightarrow DX=\frac{m \ HD}{10},$$
$$\triangle{IDX}\sim\triangle{GHX}: \frac{DX}{9-m}=\frac{DI}{8}\Longrightarrow DX= \frac{(9-m)DI}{8},$$
$$DX=\frac{m \ HD}{10}=\frac{(9-m)DI}{8},$$
$$\frac{HD}{DI}=\frac{10(9-m)}{8m}=\frac{8}{10},$$
$$m=IX=\frac{225}{41}.$$
