Let $Z=BF\cap(BEX)\neq B$ and $W=CE\cap(CFY)\neq C$, $K=BF\cap CE$. Angle chase yields $XZ\parallel FY\parallel BI$ and $YW\parallel EX\parallel CI$. Pascal on $BACEGF$ gives $X$, $Y$, $K$ colinear. Now this implies $\bigtriangleup XZE$ and $\bigtriangleup YFW$ are homothetic with homothety center $K$. Further angle chase gives $\bigtriangleup BIC\sim FYW$ so $BF$, $IY$, $CW$ concur - that is $I\in\overline{X-Y-K}$. Now $\bigtriangleup XZE$, $\bigtriangleup YFW$, $\bigtriangleup IBC$ are homothetic from $K$, so their circumcenters must be colinear with $K$, i.e. $S$, $T$, $D$ are colinear.

Let $K$ and $L$ be the midpoints of the major arcs $AB$ and $AC$. Simple angle-chasing gives us $GE//FC//LD$ and $GF//EB//KD$. Let $P$ and $Q$ be the feet of the altitudes drawn from $D$ to $FC$ and $EB$. Let $M,N,U,V$ be the midpoints of $[FY], [FC], [EX], [EB].$ We need to show that $\frac{d(T,LD)}{d(T,KD)}=\frac{d(S,LD)}{d(S,KD)}.$ As $FY//DK$ and $DL//FC$, we get $\frac{d(T,LD)}{d(T,KD)}=\frac{|DP|-|TN|}{|DF|-|TM|}.$ We know all the angles of the quadrilateral $FYCD$ so we can find each of $|DP|, |TN|, |DF|, |TM|$ in terms of $|FC|$ and some trigonometric expression. After simplifying, we find that $$\frac{|DP|-|TN|}{|DF|-|TM|}=\frac{\cos(\frac B2+C)+2\cos(\frac B2)}{\cos(\frac C2+B)+2\cos(\frac C2)}.$$ Similarly, we get $\frac{d(S,LD)}{d(S,KD)}=\frac{\cos(\frac B2+C)+2\cos(\frac B2)}{\cos(\frac C2+B)+2\cos(\frac C2)}$ and we are done.

Similar to #2 (but with more details maybe)

Let $B' \in \overleftrightarrow{AC}$ be the reflection of $B$ with respect to $\overleftrightarrow{AD}$. $2\measuredangle BB'C = \measuredangle BB'A + \measuredangle ABB' = \measuredangle BAB' = \measuredangle BDC$, so $B'$ lies on $\odot(D, \overline{DB})$. Then $\triangle BB'C \overset{+}{\sim} \triangle BAG \implies \measuredangle BB'C = \measuredangle BAG = \measuredangle BFG \implies B, F, Y, B'$ are concyclic $\implies V = \overleftrightarrow{BB'} \cap \overleftrightarrow{FY}$ lies on the radical axis of $\odot(D, \overline{DB})$ and $\odot(CFY) \implies \overleftrightarrow{CV} \perp \overleftrightarrow{DT}$. Similarly, let $C' \in \overleftrightarrow{AB}$ be the reflection of $C$ with respect to $\overleftrightarrow{AD}$ and $U = \overleftrightarrow{CC'} \cap \overleftrightarrow{EX}$. Then $\overleftrightarrow{BU} \perp \overleftrightarrow{DS}$. Let $M$ be the midpoint of $\overline{BC}$. $G(E, F; A, D) = G(F, E; D, A) = (F, E; D, A) = (C, B; A, D) = (C, B; \overleftrightarrow{GA} \cap \overleftrightarrow{BC}, M) = \infty_{\overleftrightarrow{GA}}(C, B; \overleftrightarrow{GA} \cap \overleftrightarrow{BC}, M)$, so $U, M, V$ are collinear. With $\overleftrightarrow{CU} \parallel \overleftrightarrow{BV}$, we have $CUBV$ is a parallelogram. Then $\overleftrightarrow{DS}$ and $\overleftrightarrow{DT}$ are perpendicular to parallel lines $\implies D, S, T$ are collinear.

We show that $BF, DS, CE$ are concurrent, which finishes the problem by symmetry. Let $I$ be the incenter, and $I_A, I_B, I_C$ be the excenters. First, note that $EG \parallel CF$ as $E$ is midpoint of $II_B$ and $G$ is the midpoint of $I_CI_B$. So we have, using formal sum in $(ABC)$, \[ \measuredangle BES = 90^\circ - \measuredangle EXB = 90^\circ - B - A + C + F = (A+D-E-F) - B - A + C + F = C + D - B - E = \measuredangle (BE, CD) \]so $ES \parallel CD$. Then, let $BS$ meet $(ABC)$ again at $J$ and $CJ$ meet $DE$ at $K$. The main claim is that $DIKC$ is a parallelogram. Indeed, continuing the last step we have \[ \measuredangle JCE = \measuredangle SBE = \measuredangle BES = C + D - B - E = (2E - A) + D - (2D - C) - E = C + E - A - D = \measuredangle(AD,CE) \]so $AD \parallel CJ$. Now it is easy to see that $DI_ACK$ is a parallelogram and beacuse $I_AD = DI$, we have $DIKC$ being a pallelogram as well. Finally let $L = BJ \cap DF$; Pascal on $BJCFDE$ gives $L,I,K$ collinear, so $LI \parallel CD \parallel ES$. Finally, Desargues on $\Delta LDS, \Delta ICE$ gives the desired result.

We post a complex bash using synthetic stuff to shorten it into a 1-page thing. Note that from anti-parallels it follows that $AI \parallel XS \parallel YT$, and also note that $OS \perp BE$ and $OT \perp CF$, so now set $A=a^2, B=b^2, C=c^2, D=-bc, E=-ac, F=-ab, G=bc, O=0$ where $(ABC)$ is unit circle. $$X=AB \cap EG=\frac{a^2b^2(bc-ac)+abc^2(a^2+b^2)}{a^2b^2+abc^2}=\frac{abc(b-a)+c^2(a^2+b^2)}{ab+c^2}$$Similarily we can obtain that $Y=\frac{abc(c-a)+b^2(a^2+c^2)}{ac+b^2}$, now since $OS \perp BE$ we get that $\frac{s}{b^2+ac} \in i \mathbb R$ which gives $\frac{s}{\overline{s}}=-b^2ac$, and similarily $\frac{t}{\overline{t}}=-c^2ab$, but by $AI \parallel XS$ we get that $s=\overline{s}a^2bc-\overline{x}a^2bc+x$, solving the two equations (which is 4 lines if you did this correctly), gives $s=\frac{bc(bc-a^2)}{ab+c^2}$ and similarily $t=\frac{bc(bc-a^2)}{ac+b^2}$, by colinearity criteria all we need is: $$\frac{\frac{1}{ab+c^2}+\frac{1}{bc-a^2}}{\frac{1}{ac+b^2}+\frac{1}{bc-a^2}}=\frac{\frac{c}{ab+c^2}-\frac{a}{bc-a^2}}{\frac{b}{ac+b^2}-\frac{a}{bc-a^2}} \iff \frac{bc+ab+c^2-a^2}{bc+ac+b^2-a^2}=\frac{bc^2-ac^2-a^2b-ac^2}{b^2c-a^2b-a^2c-ab^2} \iff \frac{(a+c)(b+c-a)}{(a+b)(b+c-a)}=\frac{(a+c)(bc-ab-ac)}{(a+b)(bc-ab-ac)}$$The last one being true therefore the problem is solved . (Note: if $a+b=c$ or $bc=ab+ac$ we have an equilateral triangle in which case the problem is trivial by symetry)

Since $DA\perp BC,EB\perp AC,FC\perp AB$ we can rephrase the problem like below. New Problem Statement: $ABC$ is a triangle with altitudes $AD,BE,CF$ with $D,E,F\in (ABC)$. $H$ is the orthocenter of $ABC$. $A'$ is the antipode of $A$ and $A'B,A'C$ intersect $DE,DF$ at $X,Y$ respectively. $(BEX),(CFY)$ and $(HFE)$ intersect at $K,L$ respectively. Prove that $EK\parallel FL$. Let $(BHC)\cap (HEF)=M$. Note that $A$ is the circumcenter of $(HEF)$. Let $M_B,M_C$ be the reflections of $M$ with respect to $AB,AC$. \[\measuredangle FM_BB=\measuredangle BMH=\measuredangle BCH=90-\measuredangle B=(90+\measuredangle A-\measuredangle B)-\measuredangle A=\measuredangle AKB-\measuredangle AKF=\measuredangle FKB\]Since $F,K,M_B,B$ and $F,K,M_B$ are different circles, we have $M_B=K$. Similarily $M_C=L$. \[\measuredangle LAK=\measuredangle LAC+\measuredangle A+\measuredangle BAK=\measuredangle CAM+\measuredangle A+\measuredangle MAB=2\measuredangle A=\measuredangle EAF\]Thus, $EK\parallel FL$ as desired.$\blacksquare$