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Let $(O)$ be unit circle with $A(1), B(b^2), C(c^2)$. Let $W$ be the reflection of $J$ in $A$. We get \begin{align*}i&=-bc - b - c,\\ \bar{i}&=\frac{-b - c - 1}{bc},\\ d&=\frac{b^2c + b^2 + bc^2 - b + c^2 - c}{2},\\ \bar{d}&=\frac{-bc^2 - b^2c + b^2 + c^2 + b + c}{2b^2c^2},\\ e&=\frac{bc^2 - b^2c - b^2 + c^2 + b + c}{2b},\\ \bar{e}&=\frac{bc^2 + b^2c + b^2 - c^2 + b - c}{2bc^2},\\ f&=\frac{-bc^2 + b^2c + b^2 - c^2 + b + c}{2c}.\\ \bar{f}&=\frac{bc^2 + b^2c - b^2 + c^2 - b + c}{2b^2c}\\ t&=\frac{2b^2c^2}{b^2 + c^2},\\ \bar{t}&=\frac{2}{b^2+c^2},\\ j&=\frac{-2bc}{bc^2 + b^2c + b^2 + c^2 + b + c},\\ w&=2-j=2+\frac{2bc}{bc^2 + b^2c + b^2 + c^2 + b + c}. \end{align*}Since $M$ is the intersection of ray $EF$ and unit circle $(O)$, we get $$m=\frac{bc^3 + b^3c + b^3 + c^3 - bc^2 - b^2c + b^2 + c^2 + u}{4bc}$$and $$\bar{m}=\frac{bc^3 + b^3c + b^3 + c^3 - bc^2 - b^2c + b^2 + c^2 - u}{4b^2c^2}$$where $$u=\sqrt{b^2c^6 + 2b^4c^4 + b^6c^2 + 2bc^6 - 2b^2c^5 - 2b^5c^2 + 2b^6c + b^6 + c^6 - b^2c^4 - 8b^3c^3 - b^4c^2 + 2b^5 + 2c^5 - 2bc^4 - 2b^4c + b^4 + c^4 + 2b^2c^2}.\quad (1)$$Since the tangentss at $A$ and $M$ meet at $S$, we have $$s=\frac{2m}{1+m}=2\frac{u + c^3b + c^3 - c^2b + c^2 + cb^3 - cb^2 + b^3 + b^2}{u + c^3b + c^3 - c^2b + c^2 + cb^3 - cb^2 + 4cb + b^3 + b^2}.$$Let $K$ be the intersection of $DM$ and $ST$, we get $$k=\frac{b^2c^5 + b^3c^4 + b^4c^3 + b^5c^2 + 2bc^5 + 4b^3c^3 + 2b^5c + b^5 + c^5 + bc^4 - 2b^2c^3 - 2b^3c^2 + b^4c + 2b^4 + 2c^4 - 4b^2c^2 + b^3 + c^3 + bc^2 + b^2c+u\left(bc^2 + b^2c + b^2 + c^2 + 2bc + b + c \right) }{4(bc^3 + b^3c + bc^2 + b^2c)}.$$In order to prove that $\angle ASJ=\angle IST$, we will prove that two triangles $SWA$ and $SIK$ are (dirrectly) similar, so $$\angle IST=\angle ISK=\angle WSA=\angle ASJ.$$Ideed, consider $$\frac{s-w}{s-a}-\frac{s-i}{s-k}=\frac{-2bc\left(b + 1 \right) \left(c + 1 \right)\left(b+c\right)\left(u + c^3b + c^3 - c^2b + c^2 + cb^3 - cb^2 + 4cb + b^3 + b^2 \right)}{\left(c^2b + c^2 + cb^2 + c + b^2 + b \right)\left(u + c^3b + c^3 - c^2b + c^2 + cb^3 - cb^2 - 4cb + b^3 + b^2 \right)}\cdot\frac{X}{Y}$$where $$X=\left(u^2 - c^6b^2 - 2c^6b - c^6 + 2c^5b^2 - 2c^5 - 2c^4b^4 + c^4b^2 + 2c^4b - c^4 + 8c^3b^3 - c^2b^6 + 2c^2b^5 + c^2b^4 - 2c^2b^2 - 2cb^6 + 2cb^4 - b^6 - 2b^5 - b^4\right)$$and $$\begin{aligned}Y=(u^2c^2b + u^2c^2 + u^2cb^2 + 2u^2cb + u^2c + u^2b^2 + u^2b + 2uc^5b^2 + 4uc^5b + 2uc^5 + 2uc^4b^3 + 2uc^4b^2 + 4uc^4b + 4uc^4 + 2uc^3b^4 + 4uc^3b^3 + 2uc^3b^2 - 2uc^3b + 2uc^3 + 2uc^2b^5 + 2uc^2b^4 + 2uc^2b^3 + 4uc^2b^2\\ - 2uc^2b + 4ucb^5 + 4ucb^4 - 2ucb^3 - 2ucb^2 + 2ub^5 + 4ub^4 + 2ub^3 + c^{8}b^3 + 3c^{8}b^2 + 3c^{8}b + c^{8} + c^{7}b^4 + 4c^{7}b + 3c^{7} + 2c^6b^5 + 5c^6b^4 + 6c^6b^3 - 4c^6b^2 - 4c^6b + 3c^6 + 2c^5b^6 + 3c^5b^4 + 4c^5b^3 - 2c^5b^2 - 8c^5b\\ + c^5 + c^4b^{7} + 5c^4b^6 + 3c^4b^5 + 6c^4b^4 - 9c^4b^3 - 3c^4b^2 - 3c^4b + c^3b^{8} + 6c^3b^6 + 4c^3b^5 - 9c^3b^4 - 16c^3b^3 - 2c^3b^2 + 3c^2b^{8} - 4c^2b^6 - 2c^2b^5 - 3c^2b^4 - 2c^2b^3 + 3cb^{8} + 4cb^{7} - 4cb^6 - 8cb^5 - 3cb^4 + b^{8} + 3b^{7} + 3b^6 + b^5\end{aligned}.$$From $(1)$, we get $X=0$. Therefore $$\frac{s-w}{s-a}=\frac{s-i}{s-k}$$which means two triangles $SWA$ and $SIK$ are (dirrectly) similar. We complete the proof.

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Bump. Any synthetic solution ?

Here is a synthetic solution. Lemma: Given points $A,B,C,D$ lying on a circle. The tangents from $A$ and $B$ intersect at $K$ and the tangents from $C$ and $D$ intersect at $L$. Then, the points $K, L, AC\cap BD, AD\cap BC$ are collinear.

Proof

Let $EF\cap (ABC)=\{M,N\}$. Let the tangents from $A$ and $N$ to $(ABC)$ intersect at $V$. Let $X$ and $Y$ be the intersection of $MD$ and $ND$ with $(I)$. Let $BN\cap AC=K, CM\cap AB=L$. Using our lemma on the quadrilateral $FEDX$ gives us that $LX$ is tangent to $(I)$. Let $AC\cap MB=U$. Using the lemma on $AMBC$, one can show that $U, L\in ST$. Now, let's do something different and use our lemma on $EXDF$ to show that the tangent from $X$ to $(I)$ (so the line $LX$) is also passing through $U$. Hence, the line $ST$ is tangent to $(I)$ at $X$. Similarly, the line $UT$ is tangent to incircle at $Y$. Let $H$ be the foot of the altitude from $I$ to $SV$. We want to show that $I$ and $J$ are isogonal conjugates in $\triangle TSV$. We know that $TI$ is the angle bisector of $\angle STV$ so proving $AHXY$ is cyclic is sufficient. Applying Pascal on $AACMNB$ gives us that $P=KL\cap EF$ lies on $AH$. We know that $EFHA$ and $EFXY$ are cyclic, hence it suffices to show that $AH, EF$ and $XY$ are concurrent (or $P\in XY$). Just notice that this is a result of using the lemma on $EFXY$.

Remark

Let $ST$ meet $\odot ABC$ at $G,H$. Maybe $NG,NH$ are also two tangents of $\odot I$.