We start by observing that a positive homothety centre $ A'$ takes the circumcircle to $ K_a$, and that a positive homothety centre $ A$ takes $ K_a$ to the incircle. Thus the composition of these two homotheties takes the circumcircle to the incircle, is positive, and has homothetic centre $ K$ on the line $ AA'$ (it is well known that the fixed point of a composition of two homotheties lies on the line containing the fixed points of the two individual homotheties). But $ K$ is clearly the external centre of similitude of the incircle and circumcircle, since the composed homothety is positive and takes the circumcircle to the incircle, and so also (because this characterisation is symmetric in $ A,B,C$) $ K$ lies on lines $ BB'$,$ CC'$. Having established this, we can define $ A',B',C'$ in terms of $ K$ and finish the problem with (not necessarily normalised) areal co-ordinates. $ K$, the external centre of similitude of the circumcircle and incircle, is Kimberling centre $ X_{56}$ which is known to have areal co-ordinates $ (\frac {a^2}{b + c - a},\frac {b^2}{c + a - b},\frac {c^2}{a + b - c})$ (I shall attach a proof of this fact at the end if it is not considered sufficient to simply quote this from the Encyclopedia of Triangle Centers). Thus $ AA'$ has parametrisation $ (\frac {a^2}{b + c - a},\frac {tb^2}{c + a - b},\frac {tc^2}{a + b - c})$ with $ A$ at $ t = 0$. Intersecting this with the circumcircle $ a^2yz + b^2zx + c^2xy = 0$, clearing denominators, and dividing by $ a^2b^2c^2$ gives: \[ t^2(b + c - a) + t(c + a - b) + t(a + b - c) = t(t(b + c - a) + 2a) = 0 \] . Thus the value of $ t$ corresponding to $ A'$ is $ t = \frac { - 2a}{b + c - a}$. This gives the co-ordinates of $ A'$ as (after multiplying through by common things and -1) $ A'( - a,\frac {2b^2}{c + a - b}, \frac {2c^2}{a + b - c})$. It may now be easily verified that $ A,A',X$ ($ X(0,b,c)$ well known - equivalent to angle bisector theorem) all lie upon the circle \[ \Gamma_A: a^2yz + b^2zx + c^2xy + a^2(x + y + z)(\frac {c^2(a + c - b)y - b^2(a + b - c)z}{(b^2 - c^2)(a + b + c)}) = 0 \] By cyclic permutation of letters, we obtain the equations for other two circles in the question, containing $ B,B',Y$ and $ C,C',Z$ respectively. \[ \Gamma_B: a^2yz + b^2zx + c^2xy + b^2(x + y + z)(\frac {a^2(b + a - c)z - c^2(b + c - a)x}{(c^2 - a^2)(a + b + c)}) = 0\] \[ \Gamma_C: a^2yz + b^2zx + c^2xy + c^2(x + y + z)(\frac {b^2(c + b - a)x - a^2(c + a - b)y}{(a^2 - b^2)(a + b + c)}) = 0\] It is known that the areal radical axis of two circles is obtained by subtracting the equations of the circles from one another and dividing by $ x+y+z$. Doing this for $ \Gamma_A,\Gamma_B$, we get that their radical axis is described by the equation \[ \frac {a^2c^2(a + c - b)y - a^2b^2(a + b - c)z}{(b^2 - c^2)(a + b + c)}=\frac {b^2a^2(b + a - c)z - b^2c^2(b + c - a)x}{(c^2 - a^2)(a + b + c)}\] This simplifies, clearing denominators and grouping coefficients on the RHS, to \[ b^2c^2(c^2-b^2)(b+c-a)x+c^2a^2(a^2-c^2)(c+a-b)y+a^2b^2(a+b-c)(b^2-a^2)z = 0\] But this line is symmetric in $ ((a,x),(b,y),(c,z))$, and so the same line would have been obtained by intersecting $ \Gamma_B,\Gamma_C$ or $ \Gamma_C,\Gamma_A$. Therefore $ \Gamma_A,\Gamma_B,\Gamma_C$ are coaxal, and since the point $ K$ is interior to all of them, they must all intersect one another at two points, so, being coaxal, they must all intersect one another at the same two points, so they have two common points, and we are done. Deriving the areal co-ordinates of $ K=X_{56}$ The homothety centre $ K$ factor $ r/R$ takes the line $ BC$ to a line tangent to the circumcircle at the midpoint $ L$ of the arc $ BC$. Let $ L$ projected onto $ BC$ be $ D$. Let the areals of $ K$ be $ (ax,by,cz)$ where $ x$ is the perpendicular distance from $ K$ to $ BC$, etc. Then $ x = \frac{r}{R-r}LD=\frac{r}{R-r}\frac{a}{2}\tan{\frac{A}{2}}$ So $ x: y: z = a\tan{\frac{A}{2}}: b\tan{\frac{B}{2}}: c\tan{\frac{C}{2}}$ But $ \tan{\frac{A}{2}}: \tan{\frac{B}{2}}: \tan{\frac{C}{2}} = \frac{r}{s-a}: \frac{r}{s-b}: \frac{r}{s-c} = \frac{1}{b+c-a}: \frac{1}{c+a-b}: \frac{1}{a+b-c}$ considering the triangles formed dropping perpendiculars from the incentre. Thus $ ax: by: cz = \frac{a^2}{b+c-a}: \frac{b^2}{c+a-b}: \frac{c^2}{a+b-c}$, yielding the areal co-ordinates required.

For brevity in writing: $ A^{\prime} = A_1$ Lemma1: $ A_1I$ is the angle bisector of $ BA_1C$ Proof:http://www.mathlinks.ro/Forum/viewtopic.php?search_id=657778495&t=22925 Let $ R_a$ the midpoint of the minimum arc BC of the circumcircle ABC, and $ S_a$, the midpoint of the arc BC containing A, and O the circumcenter, $ R_a,O,S_a$ are collinear. from the lemma $ S_a,I,A_1$ are collinear, (wlog a>b>c)set $ X_a$, the point of BC such that $ X_aI$ is perpendicular to $ AI$, because $ < AXX_a = C + A/2$, then $ < IX_aX = B/2 - C/2$, and $ < BIC = C/2 = < ICX_a$ then $ X_aB.X_aC = (X_aI)^2 = (X_aT)(X_aR_a)$, where T is the intersection point of $ X_aR_a$ and the circumcircle BAC, then $ < ITR_a = 90$, implying: $ I,T,S_a$ are collinear, then $ T = A_1$, then: $ (R_aX)(R_aX) = (R_aC)^2 = (R_aI)^2 = (R_aA_1)(R_aX_a)$, so $ XA_1X_aA$ is inscriptible. Now use inversion with circle: Incircle of triangle BAC, this map the problem to another one: In a triangle ABC(wlog a>b>c), let $ A_1,A_2,A_3,X_a$, the midpoint of BC, the foot of the perpendicular from O to the altitude passing throught $ A$, the foot of the perpendicular from A to the bisector of BC and the foot of the altitude throught A, I have to prove the circles $ A_1A_2A_3 = w_a$ concur in two points. Because $ A_1A_2A_3X_a$ is a rectangle the center of the circle $ w_a$ is in the bisector of $ AX_a$, then for simmetry, it pass througt a point in $ AX_a$ such that $ AM_a = X_aA_2 = RcosA = AH/2$ where H is the ortocentre of BAC and R the circumradius. Then $ M_a$ is the midpoint of AH, because $ OA_1HM_a$ is a paralelogram $ M_AA_1$ intersect $ OH$ in W: center of the 9 point circle(w). Then the power of W respect to $ w_a$ is the same of $ W$ respect to $ w = - R^2/4$, similarly for the other circles. Then W is the intersections of the radical axis of this 3 circles: radical center(if they dont intersect in 2 points, either is just a point in the radical axis). If I prove the center of this circles are collinear then supposing they dont have 2 points in common, then the 3 radical axis are parallel(perpendicular to the line joining the centers), and the dont have an intersection, but W is this point, a contradiction. Finally in order to prove this: the center of $ w_a$ = $ O_a$ is in the line passing throught W and penpendicular to $ M_aA_1$, because this line is the perpendicular bisector of this segment, let $ R_a$ the circuncenter of $ A_1X_aM_a$, then letting $ 2 < x$ the measure of the arc $ A_1M_a$ of $ w_a$, then $ 2R_asenx = R$, and $ O_aW = R_acosx = R/2.cotx$, but for simmetry in the rectangle $ x = < A_2A_1O = < X_aA_2A_1$ then by easy computacion $ cotx = cosA/sen(B - C) = l_a$, similarly $ l_b,l_c$ angle between l_a and l_b is $ 180 - 2C$, because l_a is perpendicular to $ M_aA_1$ then is perpendicular to $ AO$, similarly l_b, and $ < AOB = 2A$, similarly the angle between $ l_b,l_c$ is $ 180 - 2A$, then the angle between $ l_a,l_c$ is $ 2B$, now to prove the collineality, use areas, because if $ O_b$ is in the interior of triangle $ WO_aO_c$ the sum of the areas will be least, and if the $ O_b$ is outside the sum of the areas will be greater, so i have to prove: $ R^2/4. \frac {cosAcosB}{sen(B - C).sen(A - C)}.sen2C + R^2/4. \frac {cosBcosC}{sen(A - C)sen(A - B)}.sen2A = R^2/4. \frac {cosAcosC}{sen(B - C)sen(A - B)}.sen2B$, which is equivalent to $ senC.sen(A - B) + senA.sen(B - C) = senB.sen(A - C)$, because $ sen2A = 2senAcosA$, finally this is equivalent to $ sen(A + B).sen(A - B) + sen(B + C).sen(B - C) = sen(A + C).sen(A - C)$, this is true because $ sen(x + y).sen(x - y) = sen^2x - sen^2y$