<s>[INST] Geometry problem to prove Euler line of a triangle $I$ is the incenter of $\triangle{ABC}$ and $D$ is the tangent point of the incircle with side $BC$. Points $E$ and $F$ are on side $BC$ so that $\angle{BAE}=\angle{ACB}, \angle{CAF}=\angle{ABC}$. $G$ and $H$ are incenters of $\triangle{ABE}$ and $\triangle{ACF}$ respectively. Prove that $AD$ is the Euler Line of $\triangle{GHI}$.
I tried to find related material to solve this problem but in vain. Any hints or helpful suggestion? Thanks!

With hint to use barycentric coordinates, here is what I have got:
Let $s=\dfrac{a+b+c}{2}, t=\sin A+\sin B+\sin C$, then
$A(1:0:0)\\
B(0:1:0)\\
C(0:0:1)\\
I(\dfrac{a}{2s}:\dfrac{b}{2s}:\dfrac{c}{2s})\\
D(0:\dfrac{s-c}{a}:\dfrac{s-b}{a})\\
E(0:1-\dfrac{c\sin C}{a\sin A}:\dfrac{c\sin C}{a\sin A})\\
F(0:\dfrac{b\sin B}{a\sin A}:1-\dfrac{b\sin B}{a\sin A})\\
G(\dfrac{\sin C}{t}:1-\dfrac{(a+c)\sin C}{at}:\dfrac{c\sin C}{at})\\
H(\dfrac{\sin B}{t}:\dfrac{b\sin B}{at}:1-\dfrac{(a+b)\sin B}{at})
$
 [/INST] $\begin{array}{} ∠(ACB)=∠(BAE)=γ & ∠(ABC)=∠(CAF)=β & ∠(AEK)=β+γ \end{array}$
$\begin{array}{} A=\left( c\,cos(β),c\,sin(β)  \right)  & B=(0,0) & C=(a,0) \end{array}$
$\begin{array}{} ΔABC & cos(β)=\frac{a^2-b^2+c^2}{2ac} & cos(γ)=\frac{a^2+b^2-c^2}{2ab} \\ sin(β)=\frac{2S}{ac} & sin(β)=\frac{2S}{ab} & S=Δ(ABC)=\sqrt{s(s-a)(s-b)(s-c)} & s=\frac{a+b+c}{2}\end{array}$
$\begin{array}{} x_{A}=\frac{a^2-b^2+c^2}{2a} & y_{A}=\frac{2S}{a}  \end{array}$
$\begin{array}{} ΔAEK & AK=c\,sin(β) & EK=\frac{AK}{tan(β+γ)} \\ tan(β+γ)=\frac{cos(β)sin(β)+cos(γ)sin(γ)}{cos(β)^2+cos(γ)^2-1}=\frac{4S}{a^2-b^2-c^2} & EK=\frac{a^2-b^2-c^2}{2a} & BE=BK-EK \\ \end{array}$
$\begin{array}{} BE=\frac{c^2}{a} & E=\left( \frac{c^2}{a},0  \right)  & ∠(AEK)=∠(AFK) & AE=AF & AE=\frac{bc}{a} \end{array}$
$\begin{array}{} BF=BK+EK & BF=\frac{a^2-b^2}{a} & F=\left(  \frac{a^2-b^2}{a},0  \right)  \end{array}$
$\begin{array}{} \text{G is the incenter ΔABE } & x_{G}=\frac{x_{A}·BE+x_{B}·AE+x_{E}·c}{BE+AE+c} & x_{G}=\frac{c(a-b+c)}{2a} & y_{G}=\frac{2Sc}{a(a+b+c)} \end{array}$
$\begin{array}{} \text{H is the incenter ΔACF} & FC=a-BF & FC=\frac{b^2}{a} \\ x_{H}=\frac{2a^2-b(a+b-c)}{2a} & y_{H}=\frac{2Sb}{a(a+b+c)} & \, \\ \text{I is the incenter ΔABC} & x_{I}=\frac{a-b+c}{2} & y_{I}=\frac{2S}{a+b+c}\end{array}$
$l.x+m.y+n=0$ is the Euler line of $ΔGHI$. $l$ and $m$ are the coefficients of the Euler line in Cartesian coordinates. With CAS we simplify the matrix and get:
$\begin{array}{} l=\left| \begin{array}{} x_{G} & y_{G}^2+2y_{I}\,y_{H}+3x_{G}^2 & 1 \\ x_{I} & y_{I}^2+2y_{G}\,y_{H}+3x_{I}^2 & 1\\ x_{H} & y_{H}^2+2y_{G}\,y_{I}+3x_{H}^2 & 1 \\ \end{array} \right|  & m=\left| \begin{array}{} y_{G} & x_{G}^2+2x_{I}\,x_{H}+3y_{G}^2 & 1 \\ y_{I} & x_{I}^2+2x_{G}\,x_{H}+3y_{I}^2 & 1\\ y_{H} & x_{H}^2+2x_{G}\,x_{I}+3y_{H}^2 & 1 \\ \end{array} \right|   \end{array}$
$\begin{array}{} l=\frac{(a-c)(a-b)(a^2-b^2-c^2+2bc)(a^2-b^2-c^2+bc)}{a^2(a+b+c)} &  \\ m=\frac{4S(a-b)(a-c)(b-c)(a^2-b^2-c^2+bc)}{a^2(a+b+c)^2} &  \\ \end{array}$
$\begin{array}{} m_{Euler}=\frac{-l}{m} & m_{Euler}=\frac{-1}{4}\frac{(a+b+c)(a-b+c)(a+b-c)}{(b-c)S} \\ (a+b+c)(a-b+c)(a+b-c)=\frac{16S^2}{-a+b+c} & m_{Euler}=\frac{4S}{(a-b-c)(b-c)} \\ \end{array}$
$\begin{array}{} \text{J is the centroid of ΔGIH} & x_{J}=\frac{3a^2-2ab-b^2+2ac+c^2}{6a} & y_{J}=\frac{2S}{3a} \end{array}$
$\begin{array}{} \text{Euler line} & y-y_{J}=m_{Euler}(x-x_{J}) & \frac{y-y_{J}}{x-x_{J}}=m_{Euler} & \text{(x,y) is any point on the Euler line }\end{array}$
$\begin{array}{} \text{checking if the points belong to the line} & \frac{y_{A}-y_{J}}{x_{A}-x_{J}}-m_{Euler} =0& A=\left( \frac{a^2-b^2+c^2}{2a},\frac{2S}{a}  \right)  & \text{true} \\ \frac{y_{D}-y_{J}}{x_{D}-x_{J}}-m_{Euler}=0 & D=\left( \frac{a-b+c}{2},0  \right)  & \text{true}  \\ \end{array}$
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