Let $\Gamma,O$ be the circumcircle and circumcenter of $\triangle ABC$, and let $A_0$ be the midpoint of $BC$. Let $XB_0,XC_0$ meet $\Gamma$ again at $B',C'$, and let $E=BB' \cap CC'$. Pascal's Theorem on hexagon $ABB'XC'C$ gives $E \in B_0C_0$. Since the dilation carrying $\omega$ to $\Gamma$ carries $B_OC_0$ to $B'C'$, $B'C' \parallel B_0C_0 \parallel BC$. Then $B'C'BC$ is an isosceles trapezium, so $E$ is the foot of the perpendicular of $A_0$ on $B_0C_0$. The dilation centered at $G$ taking $\triangle ABC$ to $\triangle A_0B_0C_0$ takes $D$ to $E$, so $D,G,E$ are collinear. It suffices to show $B'B_0, C'C_0, ED$ are concurrent. Since lines $BB_0, CC_0, ED$ concur at $G$, $\triangle BCE$ and $\triangle B_0C_0D$ are perspective. Let $U=BE \cap B_0D$ and $V=CE \cap C_0D$. Desargue's Theorem implies $UV \parallel B_0C_0$. By Desargue's Theorem again, this implies $\triangle B'C'E$ and $\triangle B_0C_0D$ are perspective, so $B'B_0, C'C_0, ED$ are concurrent as desired.

Alternative solution:

My solution interprets $DX$ as a radical axis, the power of G is easy to calculate (no trigo).

I believe my solution is along similar lines to daniel's solution but I didn't read that fully... Let $T$ be on $BC$ such that $AT$ tangents cirumcircle of $ABC$. Now by radical axis theorem on $ABC, AB_0C_0, B_0C_0X$, their radical centre is midpoint of $AT$, call it as $S$. Let $\gamma$ be circle $TAD$ with centre $S$. Since $SA\perp AO$ where $O$ is cirumcentre of $ABC$, $\omega$ and $\gamma$ are orthogonal. Also, $SX$ is tangent to $\omega$ and so, $\gamma$ passes through $X$. Now by Reim's theorem on circles $\gamma, \omega$ with lines $XD$ and $AP$, we have that if $XD$ intersects $\omega$ at $U$, then $AU\parallel BC$. The foot of the altitude on the antimedial triangle from vertex corresponding to $A$ lies on the nine point circle $\omega$ and is also the image of the point $D$ under a homothety with centre $G$ and so, we see that that point is $U$. This proves the result.

Let $G'$ and $X'$ be the reflections of $G$ and $X$ across $B_0C_0$, respectively, let $T = AA \cap B_0 C_0 \cap XX$ be the radical center of $(ABC)$, $(AB_0C_0)$, and $(B_0 C_0 X)$, let $AG'$ meet $B_0 C_0$ at $P$, let $O$ be the center of $(ABC)$, let $D'$ be where $AD$ meets $B_0 C_0$, let $A_1B_1C_1$ be the medial triangle of $\triangle A_0 B_0 C_0$, let $A'$ be the reflection of $A$ across the perpendicular bisector of $B_0 C_0$, and let $G_0$ be the centroid of $AB_0 C_0$. It is easy to see that $G'$ is the reflection of $G_0$ across the perpendicular bisector of $B_0 C_0$. We first claim that $OP \perp B_0 C_0$. Now, since $D'$ is the foot of the altitude from $A$ to $B_0 C_0$, $D'A_1 B_1 C_1$ is an isosceles trapezoid. Since a homothety centered at $G_0$ with factor -2 sends $A_1B_1C_1$ to $AB_0C_0$, and $A'AB_0C_0$ is an isosceles trapezoid, this homothety must send $D'$ to $A'$ as well, so $A'$, $G_0$, and $D'$ are collinear. Reflecting these points across the perpendicular bisector of $B_0C_0$ shows that $B_0D' = PC_0$. Since $A$ and $O$ are antipodes with respect to $(AB_0C_0)$, $AD' \perp B_0C_0$, and $B_0D' = PC_0$, $OP \perp B_0 C_0$. Since $OA \perp B_0C_0$, quadrilateral $OPAT$ must be cyclic. Thus, $\angle APT = \angle AOT$. We thus have \begin{align*} \angle G'AT &= \angle B_0 AT + \angle G'AB_0 = \angle B_0 C_0 A + \angle PA B_0 \\ &= \angle C + (180^{\circ} - \angle AB_0P - \angle B_0 PA) \\ &= 180^{\circ} + \angle C - \angle ABC - \angle TPA = 180^{\circ} + \angle C - \angle B - \angle TOA \\ &= 90^{\circ} + \angle C - \angle B + \angle ATO = 90^{\circ} + \angle C - \angle B + \frac{\angle ATX}{2} \\ &= 90^{\circ} + \angle C - \angle B + \frac{\angle ATB_0 + \angle B_0 T X}{2} \\ &= 90^{\circ} + \angle C - \angle B + \frac{\angle ATB_0 + \angle X' T B_0}{2} \\ &= 90^{\circ} + \angle C - \angle B + \frac{\angle ATB_0 + \angle X'AT + \angle ATB_0}{2} \\ &= 90^{\circ} + \angle C - \angle B + \angle ATB_0 + \frac{\angle X'TA}{2} \\ &= 90^{\circ} + \angle C - \angle B + 180^{\circ} - \angle TB_0 A - \angle B_0 AT) + 90^{\circ} - \angle TAX' \\ &= 180^{\circ} - \angle TAX' + \angle C - \angle B + \angle AB_0 C_0 - \angle AC_0 B_0 \\ &= 180^{\circ} - \angle TAX' + \angle C - \angle B + \angle B - \angle C = 180^{\circ} - \angle TAX', \end{align*} so $X'$, $A$, and $G'$ are collinear. Reflecting these points across $B_0 C_0$ shows that $X$, $D$, and $G$ are collinear, as desired.

This very short solution is from Evan o'Dorney: Let $R$ be intersection of $B_0C_0$ and tangents at $A$ and $X$. Take $E$ s.t. $AEBC$ is an isosceles trapezoid; well-known $D$, $G$ and $E$ are collinear. $RA = RD = RX \implies \angle AXD = \frac 12 \angle ARD = \angle ARB_0 = \pi - \angle RAE = \angle AXE$, done.

Here are some facts I want to say about this problems and these will also lead to the solution..... 1. $A_0, B_0, C_0$ are the midpoints of the sides $BC, CA, AB$. Now we define the points $Y,Z$ such that $\odot YA_0C_0$ us tangent to $\odot ABC$ at $Y$ and similar for $Z$. Then the lines $AX, BY, CZ$ concur at the isogonal conjugate of the isotomic conjugate of orthocente rof $\triangle ABC$. 2. Let $D'$ be the isotomic point of $D$ wrt $BC$. Then $GD$ bisects $AD'$. Proof of (1):- Inverting with respect to $A$ with some power, $B_0', C_0'$ becomes the reflection of $A$ wrt $B', C'$ respectively. The point $X'$ is such that the line $B'C'$ is tangent to $\odot X'B_0'C_0'$ at $X'$ Hence $X'$ is the midpoint of the arc $B_0'C_0'$. So $X'$ lies on the perpendicular bisector of $B_0'C_0'$. Hence $X'$ is the reflection of the foot of A-altitude of the triangle $\triangle AB'C'$ wrt the midpoint of $B'C'$. So $AX, BY, CZ$ concur at the isogonal conjugate of isotomic conjugate of the orthocenter of $\triangle ABC$. Proof of (2):- Reflect $G$ wrt the midpoint of $BC$. Let it be $G'$.Then clearly $G'D'\parallel GD$. Now $G$ is the midpoint of $AG'$. So $GD$ bisects $AD'$. Main Proof:- Suppose the isogonal ray of $AD'$ cuts $\odot ABC$ at $D_2$. And $AD'$ cuts the circumcircle at $D_1$. Then clearly $D_1D_2\parallel BC$.So the perpendicular bisector of $D_1D_2$ passes through the midpoint of $BC$ which again passes through the midpoint of $AD'$ (easy to see). So $D_2, D, \text{midpoint of} AD'$ are collinear. So $D_1, D, G$ are collinear. Hence proven.

Posted on my blog a while ago, didn't know it was from the shortlist. http://www.artofproblemsolving.com/Forum/blog.php?u=112643&b=72551 $3. \blacktriangleright$ (ISL 2011 G4) Let $\mathcal{C}$ denote the circle passing through $M_B, M_C$ tangent to the circumcircle $\omega$ of $ABC$ at a point $T_A \neq A.$ Then $T_A, H_A, G$ are collinear. [asy][asy] import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -3.59, xmax = 13.85, ymin = -4.12, ymax = 4.54; /* image dimensions */ pen ccwwqq = rgb(0.8,0.4,0); pen ttttff = rgb(0.2,0.2,1); pen ttzzqq = rgb(0.2,0.6,0); pen ffcctt = rgb(1,0.8,0.2); pen cccccc = rgb(0.8,0.8,0.8); draw((2.14,3.44)--(0.66,-1.34)--(7.36,-1.36)--cycle, ccwwqq); /* draw figures */ draw((2.14,3.44)--(0.66,-1.34), ccwwqq); draw((0.66,-1.34)--(7.36,-1.36), ccwwqq); draw((7.36,-1.36)--(2.14,3.44), ccwwqq); draw(circle((4.01,0.24), 3.71), ttttff); draw(circle((3.07,0.25), 1.85), ttttff); draw((4.75,1.04)--(4.01,-1.35), ttzzqq); draw((4.01,-1.35)--(1.4,1.05), ttzzqq); draw((2.14,3.44)--(5.91,3.43), red); draw((2.14,3.44)--(2.12,-1.34), ffcctt); draw((4.01,-1.35)--(2.14,3.44), cccccc); draw((1.4,1.05)--(7.36,-1.36), cccccc); draw((4.75,1.04)--(0.66,-1.34), cccccc); draw((5.91,3.43)--(1.34,-2.33), linetype("4 4") + red); draw((4.02,1.04)--(4.01,-1.35)); draw(circle((3.07,-0.67), 2.4), cccccc); draw(circle((3.08,1.84), 1.85), cccccc); draw((2.14,3.44)--(-1.92,1.06)); draw((-1.92,1.06)--(1.34,-2.33)); draw((-1.92,1.06)--(7.64,1.03), ttzzqq); draw((5.91,3.43)--(0.4,1.05), red); draw((5.91,3.43)--(7.64,1.03), red); /* dots and labels */ dot((2.14,3.44),dotstyle); label("$A$", (2.18,3.51), NE * labelscalefactor); dot((0.66,-1.34),dotstyle); label("$B$", (0.54,-1.56), W); dot((7.36,-1.36),dotstyle); label("$C$", (7.43,-1.59), NE * labelscalefactor); dot((4.75,1.04),dotstyle); label("$M_B$", (4.93,0.91), N*1.5); dot((1.4,1.05),dotstyle); label("$M_C$", (1.11,0.96), N); dot((4.01,-1.35),dotstyle); label("$M_A$", (4,-1.54), S); dot((3.39,0.25),dotstyle); label("$G$", (3.18,0.34), NE * labelscalefactor); dot((2.12,-1.34),dotstyle); label("$H_A$", (2.16,-1.57), S*0.25); dot((1.34,-2.33),dotstyle); label("$T_A$", (1.23,-2.07), S*4); dot((5.91,3.43),dotstyle); label("$T'_A$", (5.88,3.19), NE*2); dot((4.01,0.24),dotstyle); label("$O$", (4.06,0.32), NE * labelscalefactor); dot((4.02,1.04),dotstyle); label("$H'_A$", (3.74,1.16), N); dot((-1.92,1.06),dotstyle); label("$P_A$", (-1.98,1.22), NE * labelscalefactor); dot((7.64,1.03),dotstyle); label("$T_2$", (7.71,1.18), NE * labelscalefactor); dot((0.4,1.05),dotstyle); label("$T_1$", (0.17,1.19), W); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Proof: Let $\mathcal{C}_2$ denote the circumcircle of $AM_CM_B$, which is also tangent to $\omega.$ Let $P_A$ be the radical center of $\mathcal{C}, \mathcal{C}_2, \omega$, and let $T_1, T_2$ be the intersections of $M_BM_C$ and $\omega$, as above. Then since $P_AA$ and $P_AT_A$ are tangent to $\omega$, $AT_1T_AT_2$ is a harmonic quadrilateral. Let $T_A^*$ denote the second intersection of $T'_AG$ and $\omega$, and $H'_A$ denote the image of $H_A$ under the homothety $\mathcal{H}$ which sends $\triangle ABC$ to $\triangle M_AM_BM_C.$ Then $M_AH'_A \perp T_1T_2$, so $OH'_A \perp T_1T_2$ and $H'_A$ is the midpoint of $T_1T_2.$ We have shown in $(2)$ that $AT'_A \parallel T_1T_2$, so $T'_A(T_2, T_1; H'_A, A)$ is a harmonic pencil, which implies that $T'_A(T_2, T_1; T_A^*, A)$ is a harmonic pencil, and $AT_1T_A^*T_2$ is a harmonic quadrilateral $\implies T_A^* \equiv T_A$, as desired.

Let $\odot AB_0C_0=\alpha,\odot XB_0C_0=\omega,M=AD \cap B_0C_0$. Let $AD' \bot B_0C_0,D' \in B_0C_0$,$O$ be the centre of $\Omega$. Let $K$ be the intersection of the tangents to $\Omega$ from $A,X$. $AK$ is the radical axis of $\Omega,\alpha$;$KX$ is the radical axis of $\Omega,\omega$ and $B_0,C_0$ is the radical axis of $\alpha,\omega$. So $K \in B_0C_0$. $\angle OAK+\angle OXK=90^{\circ}+90^{\circ}=180^{\circ} \rightarrow AKXO$ are cyclic. A homothety of ratio $-2$ and centre $G$ sends $O$ to $H$,the orthocentre of $ABC$ and $\triangle A_0B_0C_0$ to $\triangle ABC$. So $O$ is the orthocentre of $\triangle A_0B_0C_0$. $\angle OD'K=\angle OAK=90^{\circ} \rightarrow AKXOD'$ is cyclic. $AB_0C_0 \cong A_0B_0C_0 \rightarrow AM=A_0D'$ $AM \bot B_0C_0,A_0D' \bot B_0C_0 \rightarrow AM \parallel A_0D'$ which implies $AMA_0D'$ is a parallelogram. Note that a homothety of ratio $-2$ and centre $G$ sends $D'$ to $D$. Hence $D,G,D'$ are collinear. So it suffices to prove that $X,D,D'$ are collinear. $\angle KD'X=\angle KAX=\angle KXA=\angle KD'A=\angle AD'M$ $=\angle A_0MD'=\angle MA_0D=\angle MD'D$[The last arguement comes from the fact that $MDA_0D'$ is a rectangle.] which implies $X,D,D'$ are collinear.

Lemma: If $DG$ intersects the circumcircle at $E$, then $ABCE$ is an isosceles trapezoid.

Now we note that the tangent from $X$, $C_0B_0$, and the tangent from $A$ are concurrent at the radical center $M$ of circles $AB_0C_0,B_0C_0X,ABC$. Then we note that $MX=MA$ from PoP and $MD=MA$ since $C_0B_0$ is the perpendicular bisector of $AD$. Then $M$ is the center of circle $ADX$ and $2\angle AXE=\angle DMA=2\angle AXD$, thus $XD$ and $XE$ are equivalent lines and $X,D,E$ are collinear, implying $X,G,D$ are collinear.

new elementary proof : Let the circumcircles of $XC_0B$ and $XB_0C$ meet at point $K$ different from $X$,because circle $\omega$ and circumcircle of $ABC$ are tangent,it follows that $K$ is on $B_0C_0$. Let $XB_0$ and $XC_0$ intersect the circumcircle of $ABC$ again at $M$,$N$. We have $B,K,M$ and $C,K,N$ collinear, all else we should prove is $K,G,X$ collinear, equivalent to proving that $G$ lies on Radical axes of circumcircles of $XB_0C$ and $XC_0B$. Let $XB_0C$ and $XC_0B$ meet lines $BB_0$ and $CC_0$ again at $B_1$,$C_1$ respectively,hence we should prove that $BB_1C_1C$ is a cyclic quadrilateral,which is obvious as we have $ \angle BB_1C = \angle CC_1B = 180 - \angle CBM = 180 - \angle BCN$. and it's done

another way $E\in BC$ $CE=BD$ $R\in \odot ABC$ $\angle BAR=\angle CAE(R,A$ are on opposite sides of $BC$) $Y$-midpoint of $AE$.Since $ARC\sim ABE$,$ARB\sim ACE$ and $Y$ is on perp bisector of $ED$(and $BC$) then if $RB_{0},RC_{0}$ meet $\odot ABC$ in $B',C'$ then $\angle C'RB=\angle YCB=\angle YBC=\angle B'RC$ so $B'C'||BC||B_{0}C_{0}$ and $\odot RB_{0}C_{0}$ and $\odot ABC$ touch so $X=R$ but if $AE$ meets $\odot ABC$ again at $P$ then $XB=CP$ $BD=CE$ $\angle XBD=\angle PCE$ so $PCE\cong XBD$ and $\angle XDB=\angle PEC=\angle AED$ but $G$ is centroid of $ADE$ so $\angle AED=\angle GDE$ and $G,D,X$ are collinear.

A less ingenious solution.

Let $B',C'$ denote the points of intersections of $\omega$ with $DB_0,DC_0$.Then by Pascal's theorem on hexagon $ABB'XC'C$ we see that $P=BB' \cap CC' \in B_0C_0$.Also $B'C' \parallel BC \parallel B_0C_0$ since $\angle{B_0C_0X}=\angle{B'C'X}=\angle{\omega,l}$ where $l$ is the tangent line through $X$.Thus $BC'B'C$ is an isosceles trapezoid $\Rightarrow A_0P \perp B_0C_0$.So the homothety with center $G$ that carries $\Omega$ to $\omega$ takes $D$ to $P \Rightarrow D,G,P$ are collinear.Now it is to note that $B_0C_0 \cap B'C'=\infty,DC_0 \cap PC',DB_0 \cap PB'$ are collinear so $\triangle{DB_0C_0},\triangle{PB'C'}$ are perspective.So by Desargue's theorem $DP,C'C_0,B'B_0$ are concurrent,or,in other words $D,G,X$ are collinear,as desired.

I have a solution with appolonius circles.WLOG AC>BC.Now,let XBo and XCo meet the circle of ABC at B' and C'.Now,since circumsircles of XB'C' and XBoCo are tangent and X,Bo,B' and X,Co,C' are respectively colinear,we have BoCo//B'C' and BoB'*BoX=BBo*BBo and C'Co*CoX=CCo*CCo and XCo/XC'=XBo/XB',so we have XCo/XBo=DCo/DBo=AB/AC=ABo/ACo,so we have that X is on the appolonius circle of DBoCo.Now,let interal angle bisector of <BAC intersects BoCo at S.Now,we have that XDSA is a cyclic,and we have that <ASD = 2<BCA+<BAC(Angle chase and use the fact that D is the reflection of A over BoCo),so we obtain <DXA=180-<ASD=<ABC-<BCA.Now,let XD intersects the circumcircle of ABC at A'.Since <AXA'=<ACA',we obtain <BCA'A is an iscolles trapezoid.Now,we want to prove that A',D and G are collinear.Let Ao be the midpoint of BC.Since A,G and Ao are collinear,it will be enough to prove that AA'=2*DAo,but this is obvious,let A1 be the projection of A' on BC.It is easy to see that DA1A'A is a rectangle and AAo=AA',(since BCA'A is an iscolles trapezoid),so we are done.

thecmd999 wrote:

Can you elaborate how $AX$ is the symmedian ??? Coz it doesn't look like it is

let the line GD intersect arc BC not contains A at p and arc BAC at L we khnow that G is the homogenous center so GL=2DG we know that AG=2GM so AL is parraled to DM and we find that angle BPD=angle B now we find that P is the out homogenous point of B0C0A0 and we find that the circle who passed from C0 and B0 and is tangant to the circume circele of ABC passed from P

took so long for no reason i see its similar to evan's, satisfiedmagma's, and minusonetwelth as pointed out. @below !!

The strangest type of config geo: apparently Russia really likes $\overline{DG}$. I believe this solution is impossible to motivate with no prior knowledge. Relabel the midpoints, and instead let $X'$ be the intersection ray $\overline{GD}$ with $\Omega$. By Russia 2015/9.7, $X',D,G,A'$ are collinear, where $A'$ is the point such that $A'ABC$ is an isosceles trapezoid, and $(BDX')$ is tangent to $\overline{AB}$. Now apply $\sqrt{bc}$ inversion. $D$ gets sent to the $A$-antipode $A_1$, and $(BDX')$ is sent to the circle trhough $A_1$ and $C$ tangent to $\overline{AC}$. Since $\angle ACA_1=90^\circ$, this circle just has diameter $\overline{A_1C}$. $X'$ gets sent to $(A_1C) \cap \overline{BC}$, which is just the foot from $A_1$ to $\overline{BC}$, i.e. the reflection of $D$ over the midpoint of $\overline{BC}$. If this point is $D'$ and the reflections of $A$ over $B$ and $C$ are $B'$ and $C'$ respectively (the images of $N$ and $M$ upon $\sqrt{bc}$-inversion), it suffices to show that $(B'C'D')$ is tangent to $\overline{BC}$ (whence $X'=X$). By reflecting $A$ over the midpoint of $\overline{BC}$, it is clear that $D'$ lies on the perpendicular bisector of $\overline{B'C'}$, hence $\angle BD'B'=\angle D'B'C'=\angle D'C'B'$, so the desired tangency holds and we are done. $\blacksquare$ Remark: CT17 points out that the introduction of $(BDX')$ is not entirely necessary: if we instead view $X'=\overline{A'D} \cap (ABC)$, a $\sqrt{bc}$ inversion sends $A'$ to $\overline{AA} \cap \overline{BC}:=T$, and the circle $(AA_1T)$ has diameter $\overline{A_1T}$ since $\angle A_1AT=90^\circ$, hence $X'$ gets sent to the foot from $A_1$ to $\overline{BC}$.

Let $E$ be the foot from the midpoint of $\overline{BC}$ to $\overline{B_0C_0}$. It's well known that $D$, $G$ and $E$ are collinear so it suffices to show that $X$, $D$ and $E$ are collinear. Let $P$ be the intersection of the tangent to $(ABC)$ through $A$ and line $B_0C_0$. Note that $X$ is simply the intersection of the other tangent from $P$ to $(ABC)$, due to the radical axis theorem. We have $APOE$ cyclic since $\angle PAO = \angle PEO = 90^{\circ}$. Thus, \[\angle APO = 180^{\circ} - \angle AEO = \angle EAD,\]so \[\angle AED = 180^{\circ} - \angle EDA - \angle EAD = 180^{\circ} - 2 \angle APO = 180^{\circ} - \angle APX.\]Thus, $\angle AED = \angle AEX$ so $X$, $D$ and $E$ are collinear as desired.

If $AB=AC$ then $D,G,$ and $X$ are all on the altitude from $A$ to $\overline{BC}$, so we are done. Otherwise, assume $AB<AC$ without loss of generality. By the existence of the radical center, the tangents to $\Omega$ at $A$ and $X$ and the line $\overline{B_0C_0}$ concur. Let these points concur at $P$. By similar triangles $\triangle{}PB_0A$ and $\triangle{}PAC_0$ we get that $\tfrac{AB_0}{AC_0}=\tfrac{PA}{PC_0}$ and $\tfrac{PB_0}{PA}=\tfrac{PA}{PC_0}$, implying that $$\frac{PB_0}{PC_0}=\frac{PB_0}{PA}\cdot\frac{PA}{PC_0}=\left(\frac{PA}{PC_0}\right)^2=\left(\frac{AB_0}{AC_0}\right)^2.$$We similarly get that $\tfrac{PB_0}{PC_0}=\left(\tfrac{XB_0}{XC_0}\right)^2$, so $\tfrac{AB_0}{AC_0}=\tfrac{XB_0}{XC_0}$. Let $E$ be the intersection of the angle bisector of $\angle{}CAB$ with $B_0C_0$. Then, we have that $\tfrac{EB_0}{EC_0}=\tfrac{AB_0}{AC_0}$ by the Angle Bisector Theorem. Notice that we also have that $\tfrac{DB_0}{DC_0}=\tfrac{AB_0}{AC_0}$ since $DB_0=AB_0$ and $DC_0=AC_0$. We then get that $$\frac{AB_0}{AC_0}=\frac{EB_0}{EC_0}=\frac{DB_0}{DC_0}=\frac{XB_0}{XC_0},$$implying that $AEDX$ is cyclic since the locus of points $H$ for which $\tfrac{HI}{HJ}$ is a constant for some fixed points $I$ and $J$ is a circle. This means that \begin{align*} \angle{}AXD&=180^\circ-\angle{}AED\\ &=180^\circ-2\angle{}B_0EA\\ &=180^\circ-2\left(180^\circ-\angle{}AB_0C_0-\angle{}EAB_0\right)\\ &=180^\circ-2\left(180^\circ-\angle{}ABC-\frac{\angle{}C_0AB_0}{2}\right)\\ &=180^\circ-2\left(180^\circ-\angle{}ABC-\frac{\angle{}CAB}{2}\right)\\ &=180^\circ-360^\circ+2\angle{}ABC+\angle{}CAB\\ &=2\angle{}ABC+\angle{}CAB-180^\circ\\ &=\angle{}ABC+\angle{}ABC+\angle{}CAB+\angle{}BCA-\angle{}BCA-180^\circ\\ &=\angle{}ABC+180^\circ-\angle{}BCA-180^\circ\\ &=\angle{}ABC-\angle{}BCA. \end{align*}Let $A'\ne{}A$ be the point on $\Omega$ such that $\overline{AA'}\parallel\overline{BC}$. We have that $\angle{}AXA'=\angle{}ABA'=\angle{}ABC-\angle{}A'BC=\angle{}ABC-\angle{}BCA$, so $D,X,$ and $A'$ are collinear. Then, notice that $\overrightarrow{A'}=\left(\overrightarrow{B}+\overrightarrow{C}-\overrightarrow{D}\right)+\left(\overrightarrow{A}-\overrightarrow{D}\right)=\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}-2\overrightarrow{D}=3\overrightarrow{G}-2\overrightarrow{D}$, so $D,G,$ and $X$ are collinear.

Let $X'$ be the point such that $ABCX'$ is an isosceles trapezoid. Then by homothety from the medial triangle to $\triangle ABC,$ we have that $D,G,X'$ are collinear. Thus we must show that $D,X', X$ are collinear. Now force-overlay about $A.$ Then the point $D$ gets mapped to $A',$ the $A$-antipode with respect to $\Omega.$ Also, $X'$ gets mapped to $T,$ the intersection of the tangent to $\Omega$ at $A$ with $BC.$ Finally, $X$ gets mapped to $D',$ the reflection of $D$ across the midpoint $M$ of $BC.$ We wish to show that $ATA'D'$ is cyclic. But we now realize that $A',D',X'$ are collinear, so $\angle TD'A' = 90^\circ.$ However, since $TA$ is a tangent and $AA'$ passes through the center of $\Omega,$ we have that $\angle TAA' = 90^\circ$ as well. We are done. For those who want to know all the details of the force-overlay and the motivation behind the solution, I have a video solution that explains all of this:

A real beauty for once. Let $A'$ be the reflection of $A$ over the perpendicular bisector of $BC$. Let $Y_A$ be the $A-$Why Point. We first have the following claim. Claim : $D-G-A'$ Let $A_0$ denote the midpoint of $BC$. Then, clearly $A-G-A_0$. Now, consider the homothety centered at $G$ with scale factor $-\frac{1}{2}$ mapping $AA'$ to $D'A_0$ (since this clearly maps $A$ to $A_0$). As $AA' \parallel BC$, and $A_0$ lies on $BC$, $D'$ must also lie on $\overline{BC}$. This homothety maps $A,B,C$ to $A_0,B_0,C_0$ and thus it must maps $(ABC)$ to the nine-point circle of $\triangle ABC$. Thus, $A'$ must map to the second intersection of the nine-point circle and $BC$ and thus it maps to the foot of the perpendicular from $A$ to $BC$ - $D$. From this, we can conclude that indeed $D-G-A'$ as claimed. Now, our claim is that $Y_A=X$. First, consider the following. Let $P$ be the foot of the altitude from $O$ to $\overline{B_0C_0}$. Then, the homothety centered at $G$ with scale factor $-\frac{1}{2}$ mapping $\overline{B_0C_0}$ to $\overline{BC}$ also maps $O$ to $H$. Thus, $P$ maps to the foot of the perpendicular from $H$ to $BC$ - $D$. Thus, $D-G-P$. Now, \[\measuredangle BY_AP = \measuredangle BY_AA' = \measuredangle A'AB = \measuredangle B_0C_0B = \measuredangle PC_0B\]Thus, $Y_ABC_0P$ is cyclic. Similarly, $Y_ACB_0P$ is also cyclic. Let $B_0'$ and $C_0'$ denote the intersections of $Y_AB_0$ and $Y_AC_0$ respectively. Now, note that \[\measuredangle BY_AC_0 = \measuredangle BPC_0 = 90 + \measuredangle BPO = 90 + \measuredangle OPC = \measuredangle B_0PC = \measuredangle B_0Y_AC \]Further, \begin{align*} \measuredangle C_0'BC &= \measuredangle C_0'Y_AC \\ &= \measuredangle C_0Y_AC\\ &= \measuredangle BY_AC + \measuredangle C_0Y_AB\\ &= \measuredangle BY_AC + \measuredangle CY_AB_0 \\ &= \measuredangle BY_AB_0\\ &= \measuredangle BY_AB_0'\\ &= \measuredangle BCB_0' \end{align*}Thus, $C_0'BCB_0'$ must be an Isosceles Trapezoid and thus $C_0'B_0'\parallel CB \parallel C_0B_0$. Notice that this means, there exists a homothety centered at $Y_A$ mapping $C_0B_0$ to $C_0'B_0'$ and thus $(ABC)$ and $(Y_AB_0C_0)$ are tangent at $Y_A$ implying that indeed $Y_A=X$. Thus, by connecting this with our first claim the points $D,G$ and $X$ are indeed collinear.

Denote the point $Y$ as the intersection of ray $DG$ with $(ABC)$. Nine-point circle homothety at $G$ tells us this is the antipode of the reflection of the orthocenter $H'$ over $BC$. Now it suffices to show that $X$, $D$, and $Y$ are collinear. Consider the radical center $S$ of $(AB_0C_0)$, $(XB_0C_0)$, and $(ABC)$, or the intersection of $B_0C_0$ and the tangents at $A$ and $X$. Noting that $SX = SA = SD$, we have \[\measuredangle AXD = \measuredangle ASC_0 = \measuredangle OAD = \measuredangle AH'O = \measuredangle ACY = \measuredangle AXY. \quad \blacksquare\]

WLOG, assume $AC > AB$, then $\angle ABC > \angle ACB$. Since the radical axes of $(AB_0C_0), \Omega, \omega$ are concurrent, so the tangent of $\Omega$ at $A$, the tangent of $\Omega$ at $X$, $B_0C_0$ are concurrent and call this point $R$. Then $RA = RX$ and since $D$ is the reflection of $A$ wrt $B_0C_0$, hence $RA = RD$. Therefore, $R$ is the center of $(ADX)$, so $\angle AXD = \angle ARC_0 = \angle ABC - \angle ACB$. Now assume ray $DG$ meets $\Omega$ at $Y$. Then by homothety, we get $AY \parallel BC$, which means $\angle AXY = \angle ABC - \angle ACB$, hence $X, D, Y$ are collinear, thus $D, G, X$ are collinear. Hence we're done. $\blacksquare$

Let $E = \overline{XX} \cap \overline{AA} \cap \overline{B_0C_0}$ by radical axis, and let $K$ be on $(ABC)$ with $\overline{AK} \parallel \overline{BC}$; negative homothety at $G$ implies $D, G, K$ collinear. Set $K' = \overline{DX} \cap (ABC)$; we wish to show $K'=K$. Let $N$ be on $(ABC)$ so that $\overline{XN} \parallel \overline{BC}$. Notice that as the nine-point circle is congruent to $(AB_0C_0)$, $E$ is the center of $(AXD)$. Then $$\angle XNK' = \angle EXD = 90^\circ - \angle XAD = \angle NXA$$hence $AK'NX$ is an isosceles trapezoid and $K=K'$.

Nice, cute problem . Here's my solution. All angles are directed. Note that taking homothety at $A$ with factor 1/2, we get the circumcircle becomes $(AB_0C_0)$, which is tangent to the circumcircle. Let $Y$ be the radical centers of the circles $(B_0C_0X),(ABC),(AB_0C_0)$. Then, $Y,B_0,C_0$ are collinear and $YA=YX$. Let $T$ be the point on $AY$ such that $AT=2AY$. From homothety at $A$ with factor 2, we get that $T,B,C$ are collinear. Consider the circle with center $Y$ and radius $\overline{YA}$. Clearly, $T$ lies on this circle. Moreover, $\triangle TDA$ is a right-angle triangle, so $YA=YT=YD$. Hence $D$ also lies on this circle. Let $\angle XYO=\theta\Longrightarrow\angle AOX=180^\circ-2\theta\Longrightarrow\angle ACX=90^\circ-\theta$ and $\angle ADX=180^\circ-\theta$. Hence, $\angle CXD$ and $\angle DAC$ are complementary. Hence, $\angle CXD=\angle ACD=\angle ACB$. Let $E$ be the second intersection of $XD$ with circumcircle $(ABC)$. Then, $\angle ACB=\angle CXE$. So, $AB=CE$. It follows that $AE\parallel BC$. Let $G'$ be the intersection of $AA_0$ and $DX$, where $A_0$ is the mid-point of $\overline{BC}$. We know that $E$ is the reflection of $A$ in the line $A_0O$. Since $A_0O$ and $AD$ are both perpendicular to $AE$, it follows that $AE=2DA_0$. Moreover, $\angle DG'A_0=\angle EG'A$ and $\angle AEG'=\angle G'DA_0$. Hence, $\triangle AEG'\sim-\triangle A_0DG'$. So, $\frac{AG'}{G'A_0}=\frac{AE}{DA_0}=2$. Hence, $G'$ is the centroid. So, $G'=G$, as required. $\blacksquare$

Note that $\Omega$ and $(AB_0C_0)$ are tangent at $A$ by homotehty. Using radax on $\Omega, \omega, (AB_0C_0)$, we see that the tangent to $A, X$ at $\Omega$ and line $B_0C_0$ are concurrent. Call this point $S$. Let $T$ be the intersection between the tangent to $\Omega$ at $A$ and line $BC$. Then $S$ is the midpoint of $AT$. Let $A'$ be the reflection of $A$ across the perpendicular bisector of $BC$, recall that $D,G,A'$ are collinear. Let $Y$ be the antipode of $A$ in $\Omega$, $E$ be the point on $BC$ such that $BD=CE$. Note that $A',E,Y$ lie on a line perpendicular to $BC$. Also let $A_1$ be the point such that $A'$ is the midpoint of $AA_1$. Take $\sqrt{bc}$-inversion. Recall that $(A',T)$ invert to each other, so $(A_1,S)$ invert to each other. Also, $(D,Y)$ invert to each other. Note that $(AA_1E)$ is tangent to $BC$ at $E$. After inversion, this means that $SE'$ is tangent to $(ABC)$, so $(X,E)$ invert to each other. Note that $\measuredangle TEY = 90^\circ = \measuredangle TAY$, so $A,E,T,Y$ are concyclic. Inverting, $X,D,A'$ are collinear. Hence $D,G,X$ are collinear.

It is well known that $\overline{D-G-A_1}$ are collinear where $A_1$ is the point on $(ABC)$ such that $AA_1 || BC$. Now we do $\sqrt{\frac{bc}{2}}$ inversion for all points(ignoring $G$) and then apply homothety with center $A$ and factor $2$ for all points except $B_0$ and $C_0$ (So we almost do a $\sqrt{bc}$ inversion but not exactly.) Now the new problem is : Quote: In $\Delta ABC$, $X$ is the point on the line $BC$ such that $AX$ is tangent to $(ABC)$ at $A$ , $A'$ is the $A$ antipode, $Y$ is on the $A$ midline such that the midline is tangent to $(BCY)$ at $Y$. $AY$ meets $BC$ at $Z$. Show that $A,X,A' , Z$ are concyclic. Note that by trivial angle chasing, $Y$ lies on the perpendicular bisector of $BC$, so $\overline{Y-O-A_0}$ are collinear where $O$ is the circumcenter of $(ABC)$ and $A_0$ is midpoint of $BC$. Also notice that since $YO \perp B_0C_0$ $\implies$ due to homothety, $ZA_0 \perp BC$ and $\angle A'AX=90°$ due to tangency, hence $A,X,A',Z$ are concyclic. Now inverting back we get the desired result.

$\sqrt{bc/2}$ invert; then $(BCX')$ is tangent to $B_0C_0$ at $X'$. Therefore $X'B=X'C$, so if $M$ is the midpoint of $BC$, then $X'$ is the foot of the perpendicular from $M$ to $B_0C_0$. Thus $AX'$ and the $A$-altitude are isotomic. Therefore, \[AX'=(?:S_B:S_C)\]\[AX=(?:\frac{b^2}{S_B}:\frac{c^2}{S_C})\]\[AX=(a^2t:b^2S_C:c^2S_B).\] Plug this into the circumcircle formula to get \[a^2b^2c^2 S_BS_C+a^2b^2c^2 S_Bt+a^2b^2c^2 S_Ct=0\]\[S_BS_C+a^2t=0\]\[a^2t=-S_BS_C.\]Thus \begin{align*} X =& (-S_BS_C:b^2S_C:c^2S_B) \\ D =& (0:S_C:S_B) \\ G =& (1:1:1). \end{align*}Now we want to prove the following equation: \[ 0 = \begin{vmatrix} -S_BS_C & b^2S_C & c^2S_B \\ 0 & S_C & S_B \\ 1 & 1 & 1 \end{vmatrix} \]\[ \Leftrightarrow 0 = \begin{vmatrix} -S_BS_C & (S_B+b^2)S_C & (S_C+c^2)S_B \\ 0 & S_C & S_B \\ 1 & 0 & 0 \end{vmatrix} \]\[ \Leftrightarrow 0 = \begin{vmatrix} (S_B+S_A+S_C)S_C & (S_C+S_A+S_B)S_B \\ S_C & S_B \end{vmatrix} \]which is true. $\blacksquare$

Let $E = \overline{XX} \cap \overline{MN} \cap \overline{AA}$ be the radical center of $(AMN)$, $(ABC)$, and $(B_0C_0X)$. Let $X'$ be on $\overline{MN}$ so that $\overline{AD}$ and $\overline{AX'}$ are isotomic. Then: $D, G, X'$ are collinear by homothety at $G$ with ratio $-\frac 12$; $(AX'OXE)$ is cyclic as $\overline{OX'} \perp \overline{MN}$, and $\sqrt{\frac{bc}2}$ inversion swaps $X$ and $X'$, so $X', D, X$ collinear. Thus $D, G, X$ are also collinear. Remark: I need to stop resolving geometry problems.