a1267ab wrote: Let $ABCDEF$ be a hexagon inscribed in a circle $\Omega$ such that triangles $ACE$ and $BDF$ have the same orthocenter. Suppose that segments $BD$ and $DF$ intersect $CE$ at $X$ and $Y$, respectively. Show that there is a point common to $\Omega$, the circumcircle of $DXY$, and the line through $A$ perpendicular to $CE$. Proposed by Michael Ren and Vincent Huang Let $H$ be the common orthocenter. Pick any two vertices $X,Y$ of either $\triangle ACE$ or $\triangle BDF$ and notice that $\triangle XYH$ has circumradius equal to the radius of $\Omega$. Now invert at $H$. We obtain the following proposition: ELMO SL 2018 G4 inverted wrote: Let $ABCDEF$ be a cyclic hexagon with $\triangle ACE$ and $\triangle BDF$ sharing a common incircle $\omega$ centered at point $H$. Let $\odot(HBD), \odot(HFD)$ meet $\odot(CHE)$ again at points $X$ and $Y$ respectively. Let $M$ be the midpoint of arc $CE$ not containing $A$. Then $\odot(DXY)$ passes through point $M$. Let $\omega$ touch $\overline{CE}$ at point $N$ and $L=\overline{AD} \cap \overline{CE}$. Let $P=\overline{DB} \cap \overline{CE}$ and $Q=\overline{DF} \cap \overline{CE}$. By Dual of Desragues Involution Theorem on circumscribed $ACEN$ and point $D$; we conclude $(\overline{DN}, \overline{DL}), (\overline{DC}, \overline{DE}), (\overline{DP}, \overline{DQ})$ are pairs of an involution. Notice that $P$ has equal powers in $\odot(HBD), \odot(CHE)$ hence $P$ lies on $\overline{XH}$. Similarly, $Q$ lies on $\overline{YH}$. Let $\overline{HN}, \overline{HL}$ meet $\odot(CHE)$ again at $S,T$. Project through $H$ to conclude that $(C,E), (X,Y), (S,T)$ are pairs of an involution on the circle $\odot(CHE)$. Thus, we conclude that lines $\overline{CE}, \overline{XY}, \overline{ST}$ concur. Lemma. $\overline{CE}, \overline{ST}, \overline{DM}$ concur. (Proof) Animate $D$ on $\odot(ACE)$; then $D \mapsto L \mapsto T$ is projective. Let $U=\overline{DM} \cap \overline{CE}$ and $V=\overline{ST} \cap \overline{CE}$ then $D \mapsto U$ and $D \mapsto V$ are also projective. Thus to show $W \overset{\text{def}}{:=} U \equiv V$ we need to verify for three choices of point $D$; namely we pick $\{C, E, M\}$. These are all clearly true and the lemma is proved. $\blacksquare$ Finally, notice $WX \cdot WY=WC \cdot WE=WD \cdot WM$ proving $DXYM$ is cyclic. $\blacksquare$

Observe that $\triangle ACE$ and $\triangle BDF$ share a common inellipse whose foci are the common circumcenter and orthocenter between the two triangles. Hence, by Dual of Desargues' Involution Theorem, $\{\overline{DX}, \overline{DY}\}, \{\overline{DC}, \overline{DE}\}, \{\overline{DA}, \overline{DZ}\}$ form pairs of an involution, where $Z$ is the tangency point of the ellipse on $\overline{CE}$. Thus it suffices to show that $\odot(DD'Z)$ passes through a fixed point on the circumcircle, where $D' \equiv \overline{AD} \cap \overline{CE}$; we claim this point is $H'$ with $\{A, H'\} \equiv \overline{AH} \cap \odot(ABC)$. But note that $\overline{H'ZO}$ are collinear by the reflection property of ellipses, so simple angle chasing yields $\angle D'DH' = \angle D'ZH$, whence $H' \in \odot(DD'Z)$ for any choice of $D$ as desired.

Way too easy compared to G3. Let $A_1, D_1$ be the reflection of $H$ across $CE, BF$ and let $H_A, H_D$ be the feet from $A$ to $CE$ and $D$ to $BF$. By Reim's Theorem, $A,D,H_A,H_D$ are concyclic. Let $T = CE\cap BF$. By angle chasing, $$\angle A_1TY = \angle HTY = \angle HH_DH_A = \angle DAH_A = \angle A_1FY$$so $A_1\in\odot(TFY)$ hence $A_1$ is the Miquel's point of $BFYX$ so $A_1\in\odot(DXY)$ and we are done.

The tangent in $D$,$DA'$,$FB$, $AD$ intersect $CE$ in $N,M,S,P$, and $X$ be $DH\cap \omega$. since $M$ is the center of $XHA'$, we have $\angle PMA'=\angle A'XD=\angle PAA'$. But $\angle A'DN=\angle A'DE +\angle EDN=180-\angle A'AD=180-\angle A'MN$, so $MA'DN$ is cyclic. So $SM\cdot SN=SA'\cdot SD=SC\cdot SE$. By Desargues Involution Theorem on quadrilateral $BDDF$ and line $MN$, we get $(M,N);(C,E);(X,Y)$ are pairs of involution with center $S$. This yelds $SC\cdot SE=SX\cdot SY=SA'\cdot SD$. The conclusion follows.

Simple angle chasing and power of point suffices. Let the foot of $H$ on a side $MN$ be $H_{mn}$. Then due to power of $H$ with respect to $\Omega$, we have $ADH_{bf}H_{ce}$ is cyclic. Let the reflection of $H$ in $CE$ be $Z$, and $BE \cap CF = W$. We claim that $WFYZ$ is cyclic, which does the problem. For this, note that we have $\angle ZWY = \angle HWY = \angle HH_{bf}H_{ce} = \angle HAD = \angle ZFY$.

Here's a complex bash. Denote by lowercase letters the coordinates of the points in uppercase letters. The problem implies that $a+c+e = b+d+f$, so we also have $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \overline{(a+b+c)}=\overline{(d+e+f)} = \frac{1}{d} + \frac{1}{e} + \frac{1}{f}$. Let $(DXY)$ and $(ACE)$ intersect at $K \neq D$. Since $K$ is the center of the spiral similarity taking $XY$ onto $BF$, it has coordinates, $$k = \frac{by-xf}{b+y-x-f}.$$Let the perpendicular from $A$ onto $CE$ intersect $(ACE)$ at $K'$. It's easy to see that $k' = -\frac{ce}{a}$. By complex intersection the coordinates of $X$ and $Y$ are, \begin{align*} x = \frac{bd(c+e) - ce(b+d)}{bd-ce}\\ y = \frac{df(c+e) - ce(d+f)}{df - ce} \end{align*}Now, we can compute: \begin{align*} by-xf &= b\left( \frac{df(c+e) - ce(d+f)}{df - ce} \right) - f\left( \frac{bd(c+e) - ce(b+d)}{bd-ce}\right) \\ &= \frac{b^2d^2f(c+e) - b^2ced(d+f) - bcedf(c+e) + bc^2e^2(d+f) - bd^2f^2(c+e) + cedf^2(b+d) + bcedf(c+e) - fc^2e^2(b+d)}{(df-ce)(bd-ce)}\\ &= \frac{b^2d^2f(c+e) - bd^2f^2(c+e) - b^2ced(d+f) + cedf^2(b+d) + bc^2e^2(d+f) - fc^2e^2(b+d)}{(df-ce)(bd-ce)}\\ & = \frac{bd^2f(b-f)(c+e)- ced(b-f)(db+bf+fd) + dc^2e^2(b-f) }{(df-ce)(bd-ce)}\\ & = \frac{d(b-f)(bdf(c+e) - ce(db+bf+fd)+c^2e^2)}{(df-ce)(bd-ce)}\\ & = \frac{d(b-f)(bdfce\left(\frac{1}{c} + \frac{1}{e} - \frac{1}{d} - \frac{1}{b} - \frac{1}{f}\right) + c^2e^2)}{(df-ce)(bd-ce)}\\ & = \frac{d(b-f)ce(-\frac{bdf}{a} + ce)}{(df-ce)(bd-ce)}\\ & = \frac{-ce}{a}\left(\frac{d(b-f)(bdf-ace)}{(df-ce)(bd-ce)}\right) \end{align*}Next, we compute: \begin{align*} b+y-x-f &= b+ \frac{df(c+e) - ce(d+f)}{df - ce} - f - \frac{bd(c+e) - ce(b+d)}{bd-ce}\\ & = \frac{(b-f)(df-ce)(bd-ce)+bd^2f(c+e) - bced(d+f) - cedf(c+e) + c^2e^2(d+f) - bd^2f(c+e) + cedf(b+d) + becd(c+e) -c^2e^2(b+d)}{(df-ce)(bd-ce)}\\ & = \frac{(b-f)(df-ce)(bd-ce)+ced(c+e)(b-f) - ced^2(b-f) -c^2e^2(b-f)}{(df-ce)(bd-ce)}\\ &= \frac{(b-f)(bd^2f - bced - dcef + c^2e^2 + c^2ed+ce^2d - ced^2 - c^2e^2)}{(df-ce)(bd-ce)}\\ &= \frac{d(b-f)(bdf - ce(b+d+f-c-e))}{(df-ce)(bd-ce)}\\ &= \frac{d(b-f)(bdf-ace)}{(df-ce)(bd-ce)} \end{align*}Therefore, $k = \frac{-ce}{a} = k' \implies K = K'$, as desired.

Let the altitude $BH$ of $\bigtriangleup BDF$ meets $\Omega$ again at $M$, while $AH$ meets $\Omega$ again at $L$. We know that $CE$ is the perpendicular bisector of $HL$ and $DF$ is the perpendicular bisector of $HM$. So, $YM=YH=YL$. It means there exist a circle centered at $Y$ passing through $H,L,M$. Therefore we get: $$ 180-\angle XDL = 180-\angle BDL = \angle BML = \angle HML = \frac{\angle HYL}{2} = \angle XYL $$ So, $X,D,Y,L$ are concyclic. So, $\Omega$, $(DXY)$ and line through $A$ perpendicular to $CE$ has common point $L$.

Let $\Omega$ be the unit circle. We see that $a+c+e=b+d+f$ and $1/a+1/c+1/e=1/b+1/d+1/f$. Let $G$ be the intersection of $\Omega$ and the perpendicular from $A$ to $CE$. Note that \[\frac{g-a}{c-e}=-\frac{1/g-1/a}{1/c-1/e}\implies ce/ag=-1\implies g=-ce/a.\]Note that \[x=\frac{bd(c+e)-ce(b+d)}{bd-ce}=\frac{bd(h-a)-ce(h-f)}{bd-ce} = \frac{h(bd-ce)+(cef-abd)}{bd-ce}\]and \[y=\frac{df(c+e)-ce(d+f)}{df-ce}=\frac{df(h-a)-ce(h-b)}{df-ce} = \frac{h(df-ce)+(bec-adf)}{df-ce}.\]Let $Z$ denote the intersection of $(DXY)$ and $\Omega$. We have from that one spiral similarity lemma (Lemma 6.18 in Evan's Geo book) that \[z = \frac{xf-yb}{x+f-y-b}.\]We wish to show that $z=g$, or \[\frac{xf-yb}{x+f-y-b}+\frac{ce}{a}=0.\]It suffices to show that the left hand side multiplied by $a(x+f-y-b)(bd-ce)(df-ce)$ is $0$. The left hand side multiplied by that factor is \[(bd-ce)(df-ce)\left[a(xf-yb)+ce(x+f-y-b)\right],\]or \begin{align*} & af(df-ce)(bd(c+e)-ce(b+d)) \\ -& ab(bd-ce)(df(c+e)-ce(d+f)) \\ +& ce(df-ce)(bd(c+e)-ce(b+d)) \\ -& ce(bd-ce)(df(c+e)-ce(d+f)) \\ +& ce(f-b)(bd-ce)(df-ce). \end{align*}It is not hard to check that \[(df-ce)(bd(c+e)-ce(b+d))-(bd-ce)(df(c+e)-ce(d+f))=-c^2e^2(f-b)(e-d)(c-d),\]so it suffices to show that \begin{align*} S:=& af(df-ce)(bd(c+e)-ce(b+d)) \\ -& ab(bd-ce)(df(c+e)-ce(d+f)) \\ +&ce(f-b)(bd-ce)(df-ce) \\ -&c^2e^2(f-b)(e-d)(c-d) \end{align*}is $0$. We expand the first term to \[af\left[bd^2cf+bd^2fe+c^2e^2b+c^2e^2d-bdc^2e-bdce^2-becdf-ced^2f\right].\]Now the first and second terms together clearly have a factor of $b-f$ since the second term is negative of the first term under a $b\leftrightarrow f$ swap. We see then that the first and second terms give \[a(f-b)\left[cd^2fb+d^2ebf+c^2e^2d-cedbf-ced^2(f+b)\right],\]or \[ad(f-b)\left[cdfb+debf+c^2e^2-cebf-cedf-cedb\right].\]However, note that \[cdfb+debf-cebf-cedf-cedb=bcdef\left(\frac{1}{e}+\frac{1}{c}-\frac{1}{b}-\frac{1}{d}-\frac{1}{f}\right)=-\frac{bcdef}{a},\]so the first two terms of $S$ give \[(f-b)(-bcd^2ef+c^2e^2ad)=ced(f-b)(-bdf+cea).\]We now have \[S=ce(f-b)(d(-bdf+cea)+(bd-ce)(df-ce)-ce(e-d)(c-d))=:ce(f-b)W.\]Note that \[W = ce(ad-bd-df+ce-ec+ed+cd-d^2)=ced(a-b-f+e+c-d)=0,\]so $S=0$, as desired.

Working in conjugates is much neater. a1267ab wrote: Let $ABCDEF$ be a hexagon inscribed in a circle $\Omega$ such that triangles $ACE$ and $BDF$ have the same orthocenter. Suppose that segments $BD$ and $DF$ intersect $CE$ at $X$ and $Y$, respectively. Show that there is a point common to $\Omega$, the circumcircle of $DXY$, and the line through $A$ perpendicular to $CE$. Proposed by Michael Ren and Vincent Huang Let $\Omega$ be the unit circle. Now, $a+c+e=b+d+f.$ Let $R$ be the point on $\Omega$ with $AR \perp CE.$ Then $ar+ce=0,$ and so $$\overline r=-\frac{a}{ce}$$Now, using the intersection formulae: $$\overline x=\frac{b+d-c-e}{bd-ce}, \qquad \overline y=\frac{f+d-c-e}{fd-ce}$$Now, we must show that $X,Y,R,D$ are concyclic. Calculate: $$\overline x-\overline d=\frac{bd+d^2-cd-ed-bd+ce}{d(bd-ce)}=\frac{(d-c)(d-e)}{d(bd-ce)}$$Also, $$\overline x-\overline r=\frac{ce(b+d-c-e)+abd-ace}{ce(bd-ce)}=\frac{abd-fce}{ce(bd-ce)}$$By symmetry, $\overline y-\overline d, \overline y-\overline r$ are the same, except for we swap $b,f.$ Thus $$\frac{\overline x-\overline d}{\overline y-\overline d} \div \frac{\overline x-\overline r}{\overline y-\overline r}=\frac{fd-ce}{bd-ce} \cdot \frac{afd-bce}{fd-ce} \cdot \frac{bd-ce}{abd-fce}=\frac{afd-bce}{abd-fce} \in \mathbb{R}$$so done. $\blacksquare$

ELMO SL 2018 G4 wrote: Let $ABCDEF$ be a hexagon inscribed in a circle $\Omega$ such that triangles $ACE$ and $BDF$ have the same orthocenter. Suppose that segments $BD$ and $DF$ intersect $CE$ at $X$ and $Y$, respectively. Show that there is a point common to $\Omega$, the circumcircle of $DXY$, and the line through $A$ perpendicular to $CE$. Proposed by Michael Ren and Vincent Huang Let $A',D'$ be the reflections of $H$ over $CE,AC$ respectively and let $H_A,H_B$ be the foot of perpendiculars from $A$ to $CE$ and $E$ to $AC$ respectively. It's well known that $\{D',A'\}\in\odot(ABC)$. So, $$AH\cdot HA'=HD\cdot HD'\implies HH_A\cdot HA=HH_D\cdot HD\implies A,H_D,H_A,D\text { are concyclic.}$$Now let $BF\cap EC=K$. Then $$\angle A'KH_A=\angle HKH_A=\angle H_AH_DD=\angle H_AAD=\angle A'BX\implies A'\in\odot(KBX)$$So, $A'$ is the Miquel Point of $BXYF$. Hence, $A'\in\odot(DXY)$. $\blacksquare$

No one has mentioned this yet, but one can destroy this problem by noting that the two triangles share a $9$-point circle, and that this is the pedal circle from both $H$ and $O$ to quadrilateral $BXMF$. With this done we now have that $H$ and $O$ are isogonal conjugates with relation to $BXYF$, which by angle chasing and stuff gives us that $\angle{XHY} + \angle{BHF} = 180^\circ$ (which is equivalent to the problem).

Huh ratio lemma too powerful. (i'm taking uncitable things from https://artofproblemsolving.com/community/c6h2357938 for granted here) Lemma: Let $ABC$ be a triangle with orthocenter $H$, let $P$ be a point on $(ABC)$, let $M$ be the midpoint of the chord formed by the perpendicular bisector of $HP$. Then, $B,C,H$, and the reflection $R$ of $A$ over $M$ are concyclic. Proof: $R$ is the point you get when you take the point $Q$ on $(ABC)$ with $\angle HPQ = 90^{\circ}$, dilate by 0.5 at $H$, and dilate by $2$ at $A$, implying the result. Let $H$ be common orthocenter, let $H'$ be reflection of $H$ over $CE$, and let $D'$ be the reflection of $D$ over the midpoint of $CE$. Let $f(Z)=\frac{CZ}{EZ}$. The problem is equivalent to $$f(D)f(H')=f(X)f(Y).$$Some ratio lemma gives $$f(D)f(H')=f(X)f(Y) \Leftrightarrow f(H')\frac{1}{f(D)}=f(B)f(F).$$Now, by the lemma, we have that $BFHD'$ is cyclic, and by ratio lemma, this gives $$f(B)f(F)=f(H)f(D')=f(H')\frac{1}{f(D)}$$as desired.

Use complex numbers. Let $T$ be the other intersection between $(DXY)$ and $\Omega$. Because it is the center of the spiral similarity sending $XY$ to $BF$, it is expressed by \[t = \frac{by-xf}{b+y-x-f}.\]We note that \begin{align*} x &= \frac{bd(c+e) - (b+d)ce}{bd-ce}\\ &= \frac{bd(h-a) - ce(h-f)}{bd-ce}\\ &= h + \frac{cef-abd}{bd-ce}, \end{align*}and similarly \[y = \frac{df(c+e) - (d+f)ce}{bf-ce} = h + \frac{bce-adf}{df-ce}.\]Substitute $x$ and $y$ into the expression of $t$: \begin{align*} t &= \frac{by-xf}{b+y-x-f}\\ &= \frac{b(h + \frac{bce-adf}{df-ce}) - f(h + \frac{cef-abd}{bd-ce})}{b + \frac{bce-adf}{df-ce} - f - \frac{cef-abd}{bd-ce}}\\ &= \frac{\frac{b}{df-ce}(cdf+def-cde-cef) - \frac{f}{bd-ce}(bcd+bde-bce-cde)}{\frac{1}{(df-ce)(bd-ce)}d(b-f)(bdf-ace)}\\ &= \frac{(bcdf+bdef-bcef)(\frac{1}{df-ce} - \frac{1}{bd-ce}) + \frac{cdef}{bd-ce} - \frac{bcde}{df-ce}}{\frac{1}{(df-ce)(bd-ce)}d(b-f)(bdf-ace)}\\ &= \frac{(bcdf + bdef - bcef)(b-f)d + cde(b-f)(ce-bd-df)}{d(b-f)(bdf-ace)}\\ &= \frac{bcdef(\frac1e + \frac1c - \frac1d + \frac{ce}{bdf} - \frac1f - \frac1b)}{bdf-ace}\\ &= \frac{\frac{ce}{a}(ace-bdf)}{bdf-ace}\\ &= -\frac{ce}{a} \end{align*}as desired.

Use complex numbers with $\Omega$ the unit circle. Let $S=(DXY)\cap (DBF)$. We want to show $S=-\tfrac{ce}{a}$, since that is where the foot from $A$ to $CE$ hits $\Omega$. By the condition in the problem, \[ a+c+e=b+d+f, \quad \text{hence also}\quad \tfrac{1}{a}+\tfrac{1}{c}+\tfrac{1}{e}=\tfrac{1}{b}+\tfrac{1}{d}+\tfrac{1}{f}.\]Note \[ x = \frac{bd(c+e)-ce(b+d)}{bd-ce}, \qquad y = \frac{fd(c+e)-ce(f+d)}{fd-ce}.\]To compute $S$, note $S$ is the center of spiral similarity $XY\mapsto BF$, so \[ S = \frac{xf-yb}{x+f-y-b}. \]Multiply the numerator and denominator above by $(bd-ce)(fd-ce)$. The numerator then is \begin{align*} &\quad \ (fx-by)(bd-ce)(fd-ce) \\ &= f(fd-ce)[bd(c+e)-ce(b+d)] \ - \ b(bd-ce)[fd(c+e)-ce(f+d)] \\ &= ffdbdc + ffdbde - ffdceb - ffdced + fceced -bbdfdc \\ &\quad - bbdfde + bbdcef + bbdced - bceced\\ &= fbd(fdc+fde-fce-bdc-bde+bce) \ + \ ced(-ffd+fce+bbd-bce)) \\ &= fbd(b-f)(ec-ed-cd) \ + \ ced(b-f)(bd-ec+df) \\ &= d(b-f)\cdot (fbec-fbed-fbcd+cebd-cece+cedf) \\ &= d(b-f)\cdot [ce(fb+bd+df)-fbd(c+e) \ - \ c^2e^2] \\ &= d(b-f) \cdot \left[bdfce\left( \tfrac{1}{b}+\tfrac{1}{d}+\tfrac{1}{f}-\tfrac{1}{c}-\tfrac{1}{e} \right) - c^2e^2\right] \\ &= d(b-f)\cdot \left[\tfrac{bcdef}{a}-c^2e^2\right] \\ &= \frac{ced(b-f)(bdf-ace)}{a}, \end{align*}and the denominator then is \begin{align*} &\quad \ (x-y + f-b)(bd-ce)(fd-ce) \\ &= (fd-ce)[bd(c+e)-ce(b+d)]-(bd-ce)[fd(c+e)-ce(f+d)] \\ &\qquad +(f-b)(bd-ce)(fd-ce)\\ &= fdbdc+fdbde-fdceb-fdced-cebdc-cebde+ceceb+ceced \\ &\qquad-(bdfdc+bdfde-bdcef-bdced-cefdc-cefde+cecef+ceced) \\ &\qquad + (f-b)(bd-ce)(fd-ce) \\ &= (-fdced-cebdc-cebde+ceceb+bdced+cefdc+cefde-cecef) \\ &\qquad + (fbdfd-fbdce-fcefd+fcece-bbdfd+bbdce+bcefd-bcece) \\ &= d(-fced-cebc-cebe+bced+cefc+cefe \\ &\qquad + fbdf-fbce-fcef-bbdf+bbce+bcef) \\ &= d(b-f)\cdot (-bdf + ce(b+d+f-c-e)) \\ &= d(b-f)(-bdf+ace). \end{align*}Their quotient is $-\tfrac{ce}{a}$, as desired.

Standard complex numbers solution Let $\Omega$ be the unit circle. We have $a + c + e = b + d + f$ since they're orthocenter are the same. Let $T$ be the intersection of $\Omega$ and the line through $A$ perpendicular to $CE$. We have $t = -\frac{ce}{a}$. By complex intersection, we have \[x = \frac{bdc + bde - ceb - ced}{bd - ce}, y = \frac{fdc + fde - cef - ced}{fd-ce}\]We will now prove that $(DXYT)$ is cyclic, so we need to show that $\frac{d-x}{d-y}\cdot \frac{t-y}{t-x}$ is real. We have \[\frac{d-x}{d-y} \cdot \frac{t-y}{t-x} = \frac{\frac{bd^{2} + ceb - bdc - bde}{bd - ce}}{\frac{fd^{2} + cef -fdc - fde}{fd-ce}}\cdot \frac{\frac{a(fdc + fde - cef - ced) + ce(fd-ce)}{a(fd-ce)}}{\frac{a(bdc + bde - cef - ced) + ce(bd-ce)}{a(bd-ce)}}=\frac{b}{f}\cdot \frac{a(fdc + fde - cef - ced) + ce(fd-ce)}{a(bdc + bde - cef -cde) + ce(bd-ce)}\]Now, to prove that it's real, we take it's complement. If it's complement is real, then we also know that $\frac{d-x}{d-y}\cdot\frac{t-y}{t-x}$ is real. The complement is \[\frac{\frac{1}{b}}{\frac{1}{f}}\cdot \frac{\frac{1}{a}(\frac{1}{fdc} + \frac{1}{fde} - \frac{1}{cef} - \frac{1}{cde}) + \frac{1}{ce}(\frac{1}{fd} - \frac{1}{ce})}{\frac{1}{a}(\frac{1}{bdc} + \frac{1}{bde} - \frac{1}{cef} - \frac{1}{ced}) + \frac{1}{ce}(\frac{1}{bd} - \frac{1}{ce})}\cdot \frac{abcdef}{abcdef}= \frac{f}{b}\cdot \frac{be + bc - bd - bf + ba - \frac{abdf}{ce}}{fe + fc - fd - fb + fa - \frac{abdf}{ce}} = \frac{a + c + e - d - f - \frac{adf}{ce}}{a + c + e - b - d - \frac{abd}{ce}}\]Since $a + c + e = b + d + f$, this means $a + c + e - d - f = b, a + c + e - b - d = f$. Substituting, we get \[\frac{b - \frac{adf}{ce}}{f-\frac{abd}{ce}} = \frac{bec - adf}{fec - abd}\]Now, this is a real number, because, when we take the complement, we get \[\frac{\frac{1}{bec} - \frac{1}{adf}}{\frac{1}{fec} - \frac{1}{abd}} \cdot \frac{-abcdef}{-abcdef} = \frac{bec - adf}{fec - abd}\]Therefore, $\frac{d-x}{d-y} \cdot \frac{t-y}{t-x}$ is real, so $(DXYT)$ is cyclic.

Moving points Fix $\triangle BDF$ and animate $A$ along $\Omega$. Let $H$ be the common orthocenter ($\deg = 0$), let $P$ be the foot from $A$ to $CE$, and let $AP$ meet $\Omega$ again at $Q$. Notice that $\deg Q = 2$ and that $P$ is the midpoint of $HQ$, so $\deg P=2$. Now, line $CE$ is $P\infty_{\perp AH}$, which would have degree $2+2=4$, but very surprisingly, point $P$ and $\infty_{\perp AH}$ coincide twice! To see why, we take the case that $Q=(1:i:0)$, one of the circle points. Then, since $P$ is the midpoint of $HQ$, $P=Q$ as well. Thus $\infty_{AH} = Q$. Moreover, since $Q$ is the fixed point of the orthogonal involution, we must have $\infty_{\perp AH} = Q$ as well. Similarly, $P$ and $\infty_{\perp AH}$ coincide another time at the other circle point, so $\deg CE = 2$. Now, we compute $\deg X$ and $\deg Y$. Normally, we would have $\deg X = \deg CE = 2$, but notice that when $A=F$, line $CE$ coincide with $DB$, so actually, $\deg X=1$. Similarly, $\deg Y=1$. To compute the degree of the problem, let $O$ and $S$ be the centers of $\Omega$ and $\odot(DXY)$. The perpendicular bisector of $DX$ connects the midpoint of $DX$, which has degree $1$, with $\infty_{\perp DB}$, so it has degree $1$. Thus, $\deg S=1+1=2$. However, when $A=D$, we have $S=O$, so $\deg OS = 1\implies \deg \infty_{OS}=1$. Moreover, we clearly have $\deg DQ=1 \implies \deg\infty_{DQ}=1$. Thus, it suffices to check three cases. Consider when $A,D,H$ are colinear. Then, $Q=D$. Moreover, $XY\parallel BF$, so $\odot(DXY)$ and $\Omega$ are tangent, done. When $A,B,H$ are colinear, we have $CE\parallel DF$, so $Y=\infty_{DF}$. Thus, $S=\infty_{\perp DX}$. Moreover, $Q=B$, so $DQ = DB \perp O\infty_{\perp DB} = OS$, done. Similarly, we are done when $A,F,H$ are colinear. Having checked three cases, we are done.

Slightly ugly complex bash. Let $\Omega$ be the unit circle. We see that the line through $A$ perpendicular to $\overline{CE}$ hits $\Omega$ at $-\frac{ce}{a}$. Now it suffices to show \[\frac{y-x}{d-x} \div \frac{y+\frac{ce}{a}}{d+\frac{ce}{a}}\in \mathbb{R} \iff \frac{e-c}{d-b} \div \frac{ay+ce}{ad+ce} \in \mathbb{R}\]by scaling the vectors $\overrightarrow{YX}$ and $\overrightarrow{DX}$. By the complex intersection formula, it suffices to show \[\frac{e-c}{d-b} \cdot \frac{ad+ce}{\frac{ace(f+d)-afd(c+e)}{ce-df}+ce} = \frac{e-c}{d-b} \cdot \frac{(ad+ce)(ce-df)}{ce(af+ad+ce-df)-afd(c+e)}\in \mathbb{R}\]\[\iff \frac{(ad+ce)(ce-df)}{ce(af+ad+ce-df)-afd(c+e)} = \frac{bd}{ce} \cdot \frac{\frac{ad+ce}{acde} \cdot \frac{df-ce}{cdef}}{\frac{1}{ce}\frac{cde+cfe+adf-ace}{acdfe} - \frac{c+e}{afdce}}\]\[\iff \frac{-1}{ce(af+ad+ce-df)-afd(c+e)} = \frac{b}{ce} \cdot \frac{1}{cde+cfe+adf-ace-c^2e-ce^2} = \frac{b}{ce} \cdot \frac{1}{(ce)(d+f-a-c-e)+adf}\]Since $-b = d+f-a-c-e$, it suffices to show \[acdef-bc^2e^2 = b(afd(c+e)-ce(af+ad+ce-df))\]\[\iff -bc^2e^2 = abcdf+abdef-abcef-abcdd-bc^2e^2+bcdef-acdef\]Since $\frac{1}{a}+\frac{1}{c}+\frac{1}{e} = \frac{1}{b}+\frac{1}{d}+\frac{1}{f}$, the above expression is true, so we're done.