Took 20 minutes So the main claim is that $X, E, T$ are colinear. For this proof, denote $X' = ET \cap AC$. It is easy to see that $\angle TX'C = 90-\angle C$, and we also have $SC = EC = TC$. Also, note that $B, D, E, C, T$ are trivially cyclic. This gives us $\angle TSC = \frac{1}{2}(180-2\angle C) = 90-\angle C$, so $T, X', C, S$ are cyclic. This gives us $X' \equiv X$, which gives $X, E, T$ colinear. Now $\angle OXC = 90-\angle XTC = 90-\angle ETC = 90-\angle ABC = 90-\angle ADE = 90-\angle XDE$. This gives us $\angle OXC + \angle XDE = 90$, so $XO \perp DE$ as required. $\blacksquare$

Attachments:

My solution: Let $EM, EN$ perpendicular to $BC CA$; $AH$ cut $BC$ at $R$. Not difficult to see that $MN//RD$, $MN//TR$ and $MN//SD$, then $T,R,D,S$ are collinear. We have $\angle OXC=\angle OCX=\angle C-\angle ECA=\angle BCE$, so $XO \perp DE$

Denote $\angle BAC=\alpha$, $\angle ABC=\beta$, $\angle ACB=\gamma$. Since $S,T$ are the reflections of $E$ through $AC$, $AB$ respectively, $CS=CE=CT$. Also $\angle SCT=2(\angle ACE+\angle ECB)=2\gamma$. Thus $\angle SXC=\angle STC=90^{\circ}-\gamma$. Also $\angle SXO=90^{\circ}-\frac{\angle SOX}{2}=90^{\circ}-\angle SCX=90^{\circ}-\angle ECX=\angle BAC=\alpha$. Thus $\angle DXO+\angle XDE=\angle SXO-\angle SXC+\angle ABC=\alpha-(90^{\circ}-\gamma)+\beta=90^{\circ}$. Thus $XO\perp DE$.

Let $F$ be the foot from $A$ to $\overline{BC},$ $O_1$ be the center of $\odot(ABC).$ Claim 1: $T,F,D,S$ are collinear. Proof. It's easy to see that $\angle FET=\angle FEB-\angle TEB=\angle C-(90^\circ-\angle B)=90^\circ-\angle A$ and $\angle DFE=180^\circ-2\angle A,$ so $D,F,T$ are collinear, similarly $S,D,F$ are collinear. Claim 2: $\odot(CST)$ is tangent to $\odot(ABC)$ at $C.$ Proof. By the symmetry we get $\widehat{CS}=\angle CTF=\angle CET=90^\circ-\angle C$ and $\widehat{XS}=\angle SCA=\angle ECA=90^\circ-\angle A\implies \widehat{XC}=\angle B.$ Now it's obvious that $\overline{AO_1}\perp DE,$ by the homothety $AO_1\parallel XO\implies XO\perp DE,$ as desired.$\square$ [asy][asy] size(6cm); pathpen=black; pointpen=black; pointfontpen=fontsize(9pt); void b(){ pair A=D("A",dir(76),dir(76)); pair B=D("B",dir(-155),dir(-155)); pair C=D("C",dir(-25),dir(-25)); pair D=D("D",foot(B,A,C),dir(0)); pair E=D("E",foot(C,A,B),dir(135)); pair F=D("F",foot(A,B,C),dir(-60)); pair S=D("S",2*foot(E,A,C)-E,dir(60)); pair T=D("T",2*foot(E,B,C)-E,dir(-90)); pair X=D("X",IP(circumcircle(C,S,T),(A-C)*5+A--A),dir(90)); pair O1=D("O_1",origin,dir(-90)); pair O=D("O",circumcenter(C,S,T),dir(-90)); D(A--B--C--cycle); D(unitcircle); D(circumcircle(C,S,T),red+linetype("4 4")); D(E--T,magenta); D(E--S,magenta); D(T--S,blue+dotted); D(A--X); D(X--O,dashed); D(A--O1,dashed); } b(); pathflag=false; b(); [/asy][/asy]

Making the asy image took way longer than actually solving the problem . [asy][asy] import cse5; import olympiad; size(6cm); pen bp=black; pair B=(0,0),C=(10,0),A=(3,8),D,EE,S,T,Aprime,Bprime,X,O,Oprime,H; D=foot(B,C,A);EE=foot(C,A,B); S=2*foot(EE,B,C)-EE;T=2*foot(EE,C,A)-EE; path c=circumcircle(C,S,T); X=IP(c,C--(2*A-C),0); H=extension(B,D,C,EE); Aprime=2*foot(H,B,C)-H;Bprime=2*foot(H,C,A)-H;path circ=circumcircle(A,B,C); O=circumcenter(C,S,T);Oprime=circumcenter(A,B,C); MP("X",X,N); D(MP("A",A,NNE)--MP("D",D,E)--MP("C",C,E)--MP("B",B,W)--MP("E",EE,NW)--cycle); D(C--MP("H",H,dir(270))--EE--D--B,cyan); draw(A--Aprime^^B--Bprime,cyan); D(EE--MP("T",S,dir(270))--MP("A\prime ",Aprime,dir(270))--C--MP("B\prime ",Bprime,E)--MP("S",T,NE)--cycle,orange); D(c,green);D(circ,green);D(A--X);D(A--MP("O_1",D(Oprime,bp),SSE));D(X--MP("O",O,SSE)); dot(A);dot(B);dot(C);dot(D);dot(EE);dot(H);dot(Aprime);dot(Bprime);dot(X);dot(O);dot(Oprime);dot(S);dot(T);[/asy][/asy] Let $H=BD\cap CE$, $A'$ and $B'$ be the second intersections of $AH$ and $BH$ respectively with $\odot (A,B,C)$, and $O_1$ be the circumcenter of $\triangle ABC$. Now $T,A$ are the reflections of $E,H$ respectively about $BC$, so $C,A',T$ are collinear; further, $\tfrac{CA'}{CT}=\tfrac{CE}{CH}$. Similarly, $C,B',S$ are collinear and $\tfrac{CB'}{CS}=\tfrac{CE}{CH}$, so the homothety centered at $C$ and with ratio $\tfrac{CE}{CH}$ sends $(A',B')$ to $(T,S)$ and hence $\odot(C,A',B')=\odot(C,A,B)$ to $\odot (C,S,T)$, and hence $A$ to $X$. So $XO||AO_1$. Thus it remains to prove $AO_1\perp DE$, which is obvious. $\blacksquare$

Let $H_a$ denote the reflection of H over BC, $H_b$ denote the reflection of H over AC. It is well known that $H_a$ and $H_b$ lie on (ABC). Note that C, $H_a$, T and C, $H_b$, S are collinear. Also note that CT=CE=CS, and $CH_a =CH = CH_b$, so triangle TCS and $H_aCH_b$ have a homethety about C. This homothety maps O to the center of circle (ABC), which we denote as $\lambda$ and X to A. So it suffices to show A$\lambda$ $\perp$ DE, which is trivial by angle chasing.

First off, let me define some points 1. point $G = \overline{ET} \cap \overline{BC}$ 2. point $H = \overline{ES} \cap \overline{CA}$ 3. point $F = \overline{XO} \cap \Gamma$, where $\Gamma$ is the circle around $\triangle CST$ Now let's also define some lemmas Lemma 1: The triangle $\triangle CST$ is isoceles and angle $\angle SCT = 2\angle C$ Proof: From reflections we get that $CT=CE=CS$, and $\angle TCG = \angle GCE$ and $\angle ECH = HCS$, then summing all 4 of those angles we get that $\angle TCS = 2(\angle GCE + \angle ECH) = 2\angle C$ Lemma 2: We have that the points $C,T,B,E,D$ are all on one circle. Proof: Since of reflections we get that $\angle GTC = \angle GEC = 90 - \angle GCE = 90-(90-\angle B) = \angle B$, if we take a look at the angle $\angle EDC = 180 - \angle B$, which clearly from here we see that the points $C,T,B,E,D$ are all concyclic. So now when we have these two lemmas, we continue on with some pascal and phantom point magic. So now we will use Pascal's theorem on the hexagon $BCDTET$, so from now on we will define point $X'$ as $X' = \overline{DC} \cap \overline{TE}$ What do we know about the point $X'$. Well by a angle-chase we get that $\angle HEX' = \angle C$ and that $\angle CX'S = 90- \angle C$. So now we have that $\angle CX'S = \angle CTS = 90 - \angle C$, which means that we have that $C,T,X',S$ are all concyclic points. But that means that $X' \equiv X$ So we have that the points $T,E,X$ are all colinear. (By Pascal) Once we have prove that the rest is a quick angle-chase. $$\angle B = \angle CTE = \angle CTX = \angle CFX = 90 - \angle OXD \implies \angle OXD = 90 - \angle B$$Now define points $K$ as $K=\overline{DE} \cap \overline{XO}$, from here we see that $\angle DKX = 180-(\angle EDX +\angle OXD) = 90$. Hence $XO \perp DE$ . . .

[asy][asy] size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -25.760433772373926, xmax = 18.737430145278708, ymin = -26.36714817636629, ymax = 11.432112570887025; /* image dimensions */ /* draw figures */draw((-2.155832124300127,2.8594954182926537)--(-4.029846106495175,-6.829555596034936), linewidth(1)); draw((8.928761217193989,-7.188409337306329)--(-4.029846106495175,-6.829555596034936), linewidth(1)); draw((8.928761217193989,-7.188409337306329)--(-2.155832124300127,2.8594954182926537), linewidth(1)); draw(circle((1.8504308571205477,-2.621309068773355), 8.423845057286568), linewidth(1)); draw((-3.629456035011144,-4.759453737085582)--(8.928761217193989,-7.188409337306329), linewidth(1)); draw((-4.029846106495175,-6.829555596034936)--(1.6367131299812852,-0.5783513589696334), linewidth(1)); draw((-3.3875506838240215,3.97601727800496)--(-2.155832124300127,2.8594954182926537), linewidth(1)); draw((-3.3875506838240215,3.97601727800496)--(1.8504308571205477,-2.621309068773355), linewidth(1)); /* dots and labels */dot((-2.155832124300127,2.8594954182926537),dotstyle); label("$A$", (-1.9963415726239526,3.2582217974830896), NE * labelscalefactor); dot((-4.029846106495175,-6.829555596034936),dotstyle); label("$B$", (-4.946916778633177,-6.191593389330239), NE * labelscalefactor); dot((8.928761217193989,-7.188409337306329),dotstyle); label("$C$", (9.646468699736772,-8.583951664472854), NE * labelscalefactor); dot((1.6367131299812852,-0.5783513589696334),linewidth(4pt) + dotstyle); label("$D$", (2.309903322632754,-0.21069770147370195), NE * labelscalefactor); dot((-3.629456035011144,-4.759453737085582),linewidth(4pt) + dotstyle); label("$E$", (-4.388699847766567,-3.998598303782842), NE * labelscalefactor); dot((-3.7446335941509665,-8.918643372690303),linewidth(4pt) + dotstyle); label("$T$", (-4.627935675280829,-9.740258164125118), NE * labelscalefactor); dot((5.2822994392476,5.071768571818907),linewidth(4pt) + dotstyle); label("$S$", (5.260478528641979,6.089179089735183), NE * labelscalefactor); dot((-3.3875506838240215,3.97601727800496),linewidth(4pt) + dotstyle); label("$X$", (-4.189336658171349,4.294910383378222), NE * labelscalefactor); dot((1.8504308571205477,-2.621309068773355),linewidth(4pt) + dotstyle); label("$O$", (1.7516863917661438,-3.6796172004304935), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] This is just straightforward angle chasing. $$\angle DXO=\angle OCX=\angle OCS-\angle ACS=90^{\circ}-\angle STC-\angle ACE=90^{\circ}-(90^{\circ}-\angle ACB)-\angle ACE=\angle ECB=90^{\circ}-\angle ADE$$

Let $H$ be the orthocenter and notice $$\angle OXD=90-\tfrac{1}{2}\angle COX=90-\angle CTE=90-\angle CBE$$and $$\angle ADE=\angle AHE=\angle CBA.$$$\square$

To easy problem