<s>[INST] Calculate the ratio based on the lengths of edges of triangle The problem
My problem is,
Let $\triangle ABC$ be an acute triangle, circumscribed in $(O)$ and has orthocenter $H$. Let $HO$ intersect $(O)$ at $E$ and $F$, as shown in the image. $AE$ cuts $BC$ at $K$ and $AF$ cuts $BC$ at $L$. $AO$ cuts $BC$ at $T$. Calculate the value of $\frac{TK}{TL}$ according to the side-lengths and angles of $\triangle ABC$

My approach is that I used the Anti-Steiner point and then use Menelaus Theorem. But that does not actually relates to $a,b,c$ (which are the side lengths of $BC$, $CA$, $AB$, respectively).
Any help is appreciated!
 [/INST] triangle ABC with sides a. b, and c.
$\begin{array}{} A=(\frac{a^2+b^2-c^2}{2a},\frac{2S}{a}) & B=(a,0) & C=(0,0) & S=Δ(ABC) \end{array}$
$\begin{array}{} \text{Euler line (OH)} & l·x+m·y+n=0 & l=\frac{2a^4-(b^2-c^2)^2-a^2(b^2+c^2)}{2a} & m=\frac{2S(c^2-b^2)}{a} \end{array}$
$\begin{array}{} \left\{ D,E \right\}=OH∩Circle(O,r=R)  & R=\frac{abc}{4S} & . \\ K=AD∩BC & L=AE∩BC & T=AO∩BC \\ \end{array}$
$\begin{array}{} m_{OH}=tan(ε)=\frac{-l}{m} & ε+θ=π & tan(θ)=\frac{l}{m} \end{array}$
$\begin{array}{} \text{triangle ODP} & OP=Rcos(θ) & DP=Rsin(θ) \\ x_{D}=x_{O}-Rcos(θ) & y_{D}=y_{O}+Rsin(θ) & ε+θ=π \\ cos(ϕ)=\frac{±1}{\sqrt{1+tan(ϕ)^2}} & cos(θ)=\frac{m}{ρ} & sin(θ)=\frac{l}{ρ} \\ ρ=\sqrt{l^2+m^2} & x_{O}=\frac{a}{2} & y_{O}=\frac{a(-a^2+b^2+c^2)}{8S} \\ x_{D}=\frac{a}{2}-\frac{Rm}{ρ} &y_{D}=Rcos(α)+\frac{Rl}{ρ} & D+E=2O \\ x_{E}=\frac{a}{2}+\frac{Rm}{ρ} & y_{E}=Rcos(α)-\frac{Rl}{ρ} & cos(α)=\frac{-a^2+b^2+c^2}{2bc}\end{array}$
$\begin{array}{} \text{points: K,L,T } & \left| \begin{array}{} x_{A} & y_{A} & 1 \\ x_{D} & y_{D} & 1 \\ x_{K} & 0 & 1 \\ \end{array} \right| =0  & \left| \begin{array}{} x_{A} & y_{A} & 1 \\ x_{E} & y_{E} & 1 \\ x_{L} & 0 & 1 \\  \end{array} \right|=0 & \left| \begin{array}{} x_{A} & y_{A} & 1 \\ x_{O} & y_{O} & 1 \\ x_{T} & 0 & 1 \\ \end{array} \right| =0 \end{array}$
replace everything: $x_{A}$,$y_{A}$,...$R$, $l$,$m$ and $S$. Solve $x_{K}$, $x_{L}$ and $x_{T}$. Simplify
$\begin{array}{} S=\sqrt{s(s-a)(s-b)(s-c)} & s=\dfrac{a+b+c}{2}  \end{array}$
$x_{K}=\dfrac{b (a² - b² + c²) ρ - a c [(a² - c²)² + b² (a² + c²) - 2b⁴]}{a b c [(b² - c²)² + a² (b² + c²) - 2a⁴] - [(b² - c²)² - a² (b² + c²)] ρ} a b$
$x_{L}=\dfrac{b (a² - b² + c²) ρ + a c [(a² - c²)² + b² (a² + c²) - 2b⁴]}{a b c [(b² - c²)² + a² (b² + c²) - 2a⁴] + [(b² - c²)² - a² (b² + c²)] ρ} (-a b)$
$x_{T}=\dfrac{a b² (a² - b² + c²)}{a² (b² + c²) - (b² - c²)²}$
$\begin{array}{} r=\dfrac{TK}{TL} & r=\dfrac{x_{T}-x_{K}}{x_{T}-x_{L}} & r=\dfrac{1-p}{1+p} & \text{where:} \end{array}$
$p=\dfrac{[a² (b² + c²) - (b² - c²)²] ρ}{ a b c [(b² - c²)² + a² (b² + c²) - 2a⁴]}$
$ρ=\sqrt{(abc)^2-(-a^2+b^2+c^2)(a^2-b^2+c^2)(a^2+b^2-c^2)}$

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