Use complex numbers with $(ABC)=\mathbb{S}^1$. Then, as I already showed here, we have $t=a\frac{2bc-ca-ab}{bc-a^2}$, so $n=\frac{a}{2}\frac{3bc-ca-ab-a^2}{bc-a^2}$. Now we have that $n=\frac{2a\ell}{a+\ell}$, so$$\ell =\frac{na}{2a-n}=\frac{\dfrac{a^2}{2}\dfrac{3bc-ca-ab-a^2}{bc-a^2}}{2a-\dfrac{a}{2}\dfrac{3bc-ca-ab-a^2}{bc-a^2}}=a\frac{a^2+ca+ab-3bc}{3a^2-bc-ca-ab}.$$Redefine $S$ as the point of $(ABC)$ such that $AS\parallel BC$, then we have to prove that $L,G,S$ are collinear. We have $s=\frac{bc}{a}$ and $g=\frac{a+b+c}{3}$, so $\overline \ell =\frac{1}{\ell}=\frac{1}{a}\frac{3a^2-bc-ca-ab}{a^2+ca+ab-3bc}$, $\overline s=\frac{a}{bc}$ and $\overline g=\frac{bc+ca+ab}{3abc}$. We can immediately see that then $\overline{\left (\frac{s-\ell}{s-g}\right )}=\frac{s-\ell}{s-g}$, so $L,G,S$ are collinear.

Let $Y\in BC$ satisfy $AY\bot BC$,$YG\cap \odot(ABC)=S',LY\cap \odot(ABC)=X$ Note that $AS'//BC$ So we have to prove that $AX//BC$ Note that $AL\bot LT$ Hence $A,Y,L,T$ are cyclic So $\angle TLY=\angle TAL=\angle LXA$ Which means that $AX//BC$ Hence $X=S=S'$

Let $D$ denote the foot of altitude from $A$ to $BC$ Let $LD$ intersect $(ABC)$ at $S'\neq L$ $NT=NA=NL=ND\Rightarrow (ALDT)$ cyclic, $N$ is the center. Call this circle $\Gamma$ Let $AS'$ intersect $\Gamma$ at $K\neq A$. Let $DN$ intersect $\Gamma$ at $X\neq D$ $\measuredangle DTA=\measuredangle DTN=\measuredangle NDT=\measuredangle XDT=\measuredangle XAT$ $\measuredangle DTA=\measuredangle XAT\Rightarrow \measuredangle DTA+\measuredangle TAX=0\Rightarrow DT\parallel AX$ Is $K=X$? $\measuredangle ATK=\measuredangle ADK=\measuredangle ADN+\measuredangle NDK=\measuredangle NAD+\measuredangle NDK=\measuredangle TAD+\measuredangle NDK=\measuredangle TKD+\measuredangle NDK$ $\measuredangle ATK=\measuredangle TKD+\measuredangle NDK\Rightarrow \measuredangle NTK=\measuredangle TKD+\measuredangle NDK\Rightarrow \measuredangle TKN=\measuredangle TKD+\measuredangle NDK$ $\measuredangle TKN-\measuredangle TKD=\measuredangle NDK=\measuredangle DKN+\measuredangle KND\Rightarrow (\measuredangle TKD+\measuredangle DKN)-\measuredangle TKD-\measuredangle DKN=\measuredangle KND\Rightarrow \measuredangle KND=0$ $K;N;D$ is colinear, so $K=X\Rightarrow DT\parallel AK\Rightarrow BC\parallel AS'$ It is well known that $D;G;S'$ is colinear, and since $L;D;S'$ is colinear, we have $L;G;S'$ is colinear. $S=S'\Rightarrow BC\parallel AS$

Back when Azerbaijan used to put doable questions at p5