Let $P$ be the $A$-mixtilinear touchpoint with $\Gamma$.

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We know that $P, I$ and $T$ are collinear. First suppose that we have $IJ \parallel AP$. We have
$$\angle ITX = \angle PTX = \angle PAX = \angle IJX \implies \angle ITX = \angle IJX.$$
So, $IJTX$ is cyclic and it suffices to show that $IJ \parallel AP$. We know that $EF, AT$ and $BC$ are concurrent at some point $Q$. Let $D$ be the intersection point of lines $AI, BC$ and $N$ be the intersection point of line $AI$ and circumcircle of $\triangle ABC$. Suppose that $QN$ intersects the circumcircle of $\triangle ABC$ at $X' (X' \neq N)$ and let $J'$ be the intersection point of the lines $AX'$ and $EF$. We have
$$(AX', BC) \stackrel\{N\}\{=\} (DQ, BC) \stackrel\{A\}\{=\} -1.$$
Therefore, $X'$ lies on the $A$-symmedian of $\triangle ABC$. Since $S$ is where the circumcircles of two triangles $\triangle ABC$ and $\triangle AEF$ meet for the second time, we know that $S$ is the Miquel's point of the quadrilateral $BCEF$ and quadrilateral $SQBF$ is cyclic. So
$$\angle SQF = \angle SBF = \angle SNA \implies \angle SQK = \angle SNK,$$
and $SQNK$ is cyclic. Therefore
\begin\{aligned\} \angle AJK &= \angle ASK = \angle AS'K = \angle ASN - \angle NSK = \angle AX'N - \angle NQK \\ &= \angle AJ'K. \end\{aligned\}
Which gives us $\angle AJK = \angle AJ'K$. So $J$ and $J'$ are symmetrical with respect to line $AI$. Now let $X_A$ be the touchpoint of $A$-excircle with side $BC$. Since
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## IRN_ABooklet_2020 — Page 53
$J$ and $J'$ are symmetrical, it suffices to show that $AX_A \parallel IJ'$ to get $AP \parallel IJ$. If $M$ is the midpoint of $BC$, we know that $MI \parallel AX_A$. So we just need to prove that $M, I$ and $J'$ are collinear which is true since $SQNK$ is cyclic and
$$(EF, J'Q) = A(EF, JQ) = (BC, MQ).$$