Clearly we only need $\angle TIA = 90$, which follows by the usual inversion through $(I)$.

Since the circumcentres of $BIC$ and $AFMIN$ lie on $AI$, we get that $(AFI)$ and $(BIC)$ are tangent at $I$, so $AF$, $BC$ and the common tangent to $(AFI)$ and $(BIC)$ at $I$ concur at $T$ by the Radical Axis Theorem. Hence $TI\perp AD$. Since $DJ\parallel IF\perp AF\parallel AT$, we get that $J$ is the orthocentre of $ATD$, and therefore that $AJ\perp TD\parallel BC$, as desired.

Clearly $DJ\perp AT$, thus $ AJ\perp BC\iff TI\perp AI$. Notice that $T$ is the radical center of $(ABC), (AMN), (BIC)$, but $(AMN)$ and $(BIC)$ are tangent, therefore $TI$ is tangent to $(AMN)\Rightarrow TI\perp AI. \ _{\blacksquare}$

Let $L$ be the point of contact of the incircle in $BC$. Claim: $MN//TI.$ Perform an inversion $I$, $I(P)=P^*$ through the incircle of $ABC$. Clearly, $(AMNFI) \rightarrow MN$, then $F^*=MN\cap FI$, $A^*=MN\cap AI$, $BC \rightarrow$ circle with diameter $IL$ (let's call it $\omega$). Since $T\in BC$ and $T\in AF$, $T^*$ is the intersection of $(A^*F^*I)$ and $\omega$. Hence, $TI$ is the radical axis of $\omega$ and $(A^*F^*I)$. Furthermore, if $R$ is the midpoint of $F^*I$ and $S$ that of $IL$, these will be the centers of such circles, therefore $RS\bot TI$. However, by midbase $RS//F^*L\bot MN$, so we´re done. From the statement, we have that $TI\bot AI$. Thus. Let $E$ be the foot of the perpendicular from $A$ to $BC$. Define $X=TI\cap AE$ as the phantom point of $J$. Note that $AEIA$ is cyclic, therefore $\angle BAE=\angle BIE=\angle XDE$. Thus, $\triangle DEX \sim \triangle XAP$ where $P=DX\cap AT$ and then, $\angle DPA=\angle IFA= 90°$, therefore $DX//FI$ so $X\equiv J$. $\blacksquare.$

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