let $AB\cap EF=A'$, $AF\cap BC=B'$, $AB\cap CD=C'$, $BC\cap DE=D'$, $CD\cap FE=E'$, $DE\cap AF=F'$. The condition equivalent to $ \angle AA'F=\angle CD'D$, $ angle AB'B=\angle DE'E$,and $ \angle BC'C=\angle EF'F$. Let $ABCX$, $CDEY$, $EFAZ$ are parallelogramm and the points $X,Y,Z$ are lie inside of $ABCDEF$. So we can find easily that the $AD$ is perpendicular bisector of $XZ$, the $BE$ is perpendicular bisector of $YZ$ and the $CF$ is perpendicular bisector of $XY$. Hence the diagonals $AD,BE,CF$ intersect at one point, which the circumcenter of the triangle $ABC$.

Solution with IDMasterz: Lemma: $AB||DE$, $BC || EF$, $CD||AF$ Proof: Let $P = AB \cap CD$, $Q = AF \cap CD$, and $R = BC \cap EF$ (Note, $P, Q, R$ might be at infinity...in fact they actually are. If they are, consider $\angle APE = 0$ and angle chasing will still hold). It is easy to see that $\angle APD = \angle ADE - \angle DAB$ and $\angle AQD = \angle FAD - \angle ADC$ so subtracting the equations yields $\angle P - \angle Q = \angle A - \angle D$. We can similarly arrive at $|\angle P - \angle Q| = |\angle Q - \angle R| = | \angle R - \angle P|$. Then, $\angle P - \angle Q = \angle Q - \angle R$, or $\angle P - \angle Q = \angle R - \angle Q$. If it were the latter, then $\angle P = \angle R$, so $|\angle Q - \angle R| = 0$ implying that $\angle P = \angle Q = \angle R$. Otherwise, if it were the former, let $\angle P - \angle Q = k$. That means $\angle R = \angle P - 2k$. Then, $|2k| = |k|$ implying that $k = 0$. Hence, $\angle P = \angle Q = \angle R$. Either way, we see that $\angle P = \angle Q = \angle R$. But from $\angle P = \angle Q$, we have $\angle ADE - \angle DAB = \angle FAD - \angle ADC$ so $\angle A = \angle D$. Similarly, $\angle B = \angle E$, $\angle C = \angle F$. Therefore, $\angle D + \angle E + \angle F = 360^{\circ}$. Let $X = AF \cap DE$. Then, $\angle FAB = \angle D = 360^{\circ} - \angle E - \angle F = 180^{\circ}-\angle AXD$ so we see that $AB || DE$ and similar for other pairs of sides. $\Box$ EDIT: oops Desargue doesn't work. So we now bring in the equal lengths condition and we see that $ABDE$, $BCEF$, and $ACDF$ are parallelograms. Hence, $BE, CF, AD$ bisect each other, so they are concurrent.

you can't ignore $AB=DE$ etc. see attached picture.

chen2013b wrote: you can't ignore $AB=DE$ etc. see attached picture. Sure. Then, $AB \cap EF = A_1$ and go around clockwise. Note $A_1C_1E_1 \cong B_1D_1F_1$, so they are centrally symmetric about a point $O$. This symmetry takes $A \mapsto D$, so done.

the point is not the circumcenter. see attached picture.

chen2013b wrote: the point is not the circumcenter. see attached picture. A point $O$, never mentioned circumcentre.

it is for mathuz's solution.

Sorry about that x

Also, Proglote informed me that the second angle equality should be $\angle E - \angle B$, so I haven't checked but this probably influences the solution a bit

chen2013b wrote: the point is not the circumcenter. chen2013b wrote: it is for mathuz's solution. I'm quite sure mathuz had a good solution: 1. Let $ABCX$, $CDEY$, $EFAZ$ be parallelograms, 2. Note that then $AX=AZ$, moreover by SAS congruence $\triangle CXD\equiv \triangle EDZ$. 3. Therefore $AX=AZ$ and $CX=CZ$ $\implies$ $AC$ is the perp. bisector of $XZ$, similarly $BE$ and $CF$ are perp. bisectors of $YX$ and $ZY$ $\implies$ they intersect in the circumcenter of $\triangle XYZ$. Of course here, I think we still need a bunch of easy discussion, so first of all, we want to talk about the relative position of $CX$ and $CD$ ect. for the angle chasing needed in 2. (X,Y,Z lie inside the hexagon), and then there's also the case when $XYZ$ only forms a degenerate triangle.

thanks all, ISL2013(geometry) was some easier, no-no it's very nice but there were not any very hard problems.

IDMasterz wrote: Also, Proglote informed me that the second angle equality should be $\angle E - \angle B$, so I haven't checked but this probably influences the solution a bit Yeah there was a typo in the problem when it was posted. I haven't read through your solution in any real detail but I think it doesn't work for the actual problem: in the actual problem I don't think the opposite sides have to be parallel. I found quite a nice solution using complex numbers which I will post soon if I have time.

So use complex numbers with $(ACE)$ as the unit circle. Let $a_1=b-a, a_2=c-b, a_3=d-c, a_4=e-d, a_5=f-e, a_6=a-f$. From the equal length conditions, we obtain \[ \frac{a_6}{a_1} / \frac{a_3}{a_4} = cis (\angle A - \angle D) \implies \frac{a_4}{a_1} \frac{a_6}{a_3} = cis (\angle A - \angle D). \] Similarly, \[ \frac{a_4}{a_1}\frac{a_2}{a_5} = cis (\angle E - \angle B), \frac{a_2}{a_5}\frac{a_6}{a_3} = cis (\angle C - \angle F). \] From the angle equality condition, $\frac{a_4}{a_1}\frac{a_6}{a_3}=\frac{a_4}{a_1}\frac{a_2}{a_5}=\frac{a_2}{a_5}\frac{a_6}{a_3}.$ So $\frac{a_4}{a_1}=\frac{a_2}{a_5}=\frac{a_6}{a_3}=k$, where $|k|=1$. We now solve this system of equations for $b,d,f$ in terms of $a,c,e,k$. I won't write out the details here, it's a pretty straightforward computation. But the algebra involves dividing by $k+1$ at one stage to simplify the answer, so we should quickly deal with the $k=-1$ case. This is easy, however, as we then get parallelograms and $AD,BE,CF$ bisect one another, and are thus concurrent. We get $d = \frac{k^2+k(a-c-e)+e}{k^2-k+1} \implies \overline{d} = \frac{ae+k^2ac+k(ce-ac-ae)}{(k^2-k+1)ace}.$ Let $AD,BE,CF$ intersect $(ACE)$ again at $P,Q,R$, respectively. Then $-ap = \frac{a-d}{\frac{1}{a} - \overline{d}} = \frac{a(a-d)}{1-a\overline{d}} \implies p = \frac{a-d}{a\overline{d}-1}$. So $c-p=\frac{ac\overline{d}-c-a+d}{a\overline{d}-1}, e-p=\frac{ae\overline{d}-e-a+d}{a\overline{d}-1}$. Thus \[ \frac{CP^2}{EP^2} = \frac{c-p}{e-p}\frac{\overline{(c-p)}}{\overline{(e-p)}}=\frac{e}{c} \frac{(ac\overline{d}-c-a+d)^2}{(ae\overline{d}-e-a+d)^2}. \] Now it's time to sub in our expressions for $d$ and $\overline{d}$ !! \begin{align*} \frac{ac\overline{d}-c-a+d}{ae\overline{d}-e-a+d} &= \frac{\frac{ae+k^2ac+k(ce-ac-ae)}{(k^2-k+1)e}-c-a+\frac{k^2c+k(a-c-e)+e}{k^2-k+1}}{\frac{ae+k^2ac+k(ce-ac-ae)}{(k^2-k+1)c}-e-a+\frac{k^2c+k(a-c-e)+e}{k^2-k+1}} \\ &= \frac{ace+k^2ac^2+kc(ce-ac-ae) - (k^2-k+1)ce(a+c) + k^2c^2e + kce(a-c-e)+ce^2}{ae^2+k^2ace+ke(ce-ac-ae)-(k^2-k+1)ce(a+c)+k^2c^2e+kce(a-c-e)+ce^2} \\ &= \frac{k^2(ac^2-ace)+k(-ac^2+ace+c^2e-ce^2)+ce^2-c^2e}{k^2(ec^2-ce^2)+k(-ae^2+ace+ce^2-c^2e)+ae^2-ace} \\ &= \frac{(c-e)(k^2ac+k(-ac+ce)-ce}{k^2ec+k(ae-ce)-ae} \\ &= \frac{(c-e)(kac+ce)(k-1)}{(c-e)(kec+ae)(k-1)}\\ &= \frac{c(ka+e)}{e(kc+a)}. \end{align*} Thus $\frac{CP^2}{EP^2} = \frac{e}{c} \left ( \frac{c(ka+e)}{e(kc+a)} \right )^2$. We obtain analogous expressions for $\frac{ER^2}{AR^2}$ and $\frac{AQ^2}{CQ^2}$ by permuting the variables $a,c,e$. Then they all multiply to $1$ so $CP \cdot ER \cdot AQ = EP \cdot AR \cdot CQ$. Thus $AP,CR,EQ$ are concurrent so $AD,BE,CF$ are concurrent.

IDMasterz wrote: Also, Proglote informed me that the second angle equality should be $\angle E - \angle B$, so I haven't checked but this probably influences the solution a bit Yes, it does...the solutions with the parallelograms would then work. Like above, let $ABCX, CDEY, EFAZ$ be parallelograms. You can actually get that $AXDZ$ is a rhombus. We can deduce from angle chasing that $DEZ \sim XYZ$, which implies that $DZ = BC = EF$.

It's been over a year now and the typo in the OP still hasn't been corrected. (It should be $\angle A-\angle D = \angle C -\angle F = \angle E -\angle B$.) A pretty significant typo for those trying to solve the problem, I would say...

Sorry to double post, but can a mod please fix the typo in the problem?

Triple post!

The angle condition is equivalent to the triangle formed by lines $BC, DE, FA$ being directly similar to the triangle formed by lines $EF, AB, CD$, so there exists an angle $\theta$ such that $\overrightarrow{EF}$ is $\overrightarrow{BC}$ rotated by $\theta$, and similarly for the other sides. Let $x = \overrightarrow{BC} + \overrightarrow{DE} + \overrightarrow{FA}$ and $y = \overrightarrow{EF} + \overrightarrow{AB} + \overrightarrow{CD}$. Then $y$ is $x$ rotated by $\theta$, so $x$ and $y$ have the same magnitude. However, $x + y = 0$, so either (i) $\theta = 180^{\circ}$ or (ii) $\theta \neq 180^{\circ}$ (forcing $x = y = 0$).

Case (ii): $\theta \neq 180^{\circ}$ and $x = y = 0$. Now, the vectors $\overline{BC}$, $\overline{DE}$, and $\overline{FA}$ form a triangle. Thus, there exists a point $X$ such that $BXFA$ and $CXED$ are parallelograms. Similarly, there exist points $Y$ and $Z$ such that $DYBC$, $EYAF$, $FZDE$, and $AZCB$ are parallelograms. Now, it follows from the parallelograms that $AZ = BC = EF = AY$. Similarly, $DY = DZ$, so $\overline{AD}$ is the perpendicular bisector of $\overline{YZ}$. Thus, the diagonals $\overline{AD}$, $\overline{BE}$, and $\overline{CF}$ are concurrent at the circumcenter of $\triangle XYZ$. (Note that the points $X$, $Y$, and $Z$ are indeed distinct, since because if, say, $Y = Z$, then $\overline{EF} \parallel \overline{AY} \parallel \overline{AZ} \parallel \overline{BC}$, impossible as $\theta \neq 180^{\circ}$ by assumption.)

Let $A=a,B=b,C=c,D=d,E=e,F=f$ on the complex plane. Then $\left|b-a\right|=\left|e-d\right|$, $\left|c-b\right|=\left|f-e\right|$, $\left|d-c\right|=\left|a-f\right|$ by the length condition and \[\frac{b-a}{f-a}\div\frac{e-d}{c-d}=\frac{d-c}{b-c}\div\frac{a-f}{e-f}=\frac{f-e}{d-e}\div\frac{c-b}{a-b}\]by the angle condition (these are equalities instead of proportionalities because each of these expressions have modulus $1$). Rearrange this big equality to \[\frac{b-a}{e-d}\cdot\frac{d-c}{a-f}=\frac{d-c}{a-f}\cdot\frac{f-e}{c-b}=\frac{f-e}{c-b}\cdot\frac{b-a}{e-b},\]whence \[\frac{b-a}{e-d}=\frac{f-e}{c-b}=\frac{d-c}{a-f}.\]Call this common number $\frac{1}{w}$, with $\left|w\right|=1$. Observe that \begin{align*} 0&=\left(b-a\right)+\left(c-b\right)+\left(d-c\right)+\left(e-d\right)+\left(f-e\right)+\left(a-f\right)\\ &=\left(b-a\right)+w\left(f-e\right)+\left(d-c\right)+w\left(b-a\right)+\left(f-e\right)+w\left(d-c\right)\\ &=\left(b-a+d-c+f-e\right)\left(1+w\right). \end{align*}If $w=-1$, then $ABCDEF$ is centrally symmetric and hence the long diagonals concur at the center of the hexagon. Otherwise, we have that $a+c+e=b+d+f$. First, we show that \[\overline{a}d+\overline{c}f+\overline{e}b=a\overline{d}+c\overline{f}+e\overline{b},\]equivalently that $\overline{a}d+\overline{c}f+\overline{e}b\in\mathbb{R}$. Write \begin{align*} a&=a\\ b&=a+\left(b-a\right)\\ c&=b+\left(c-b\right)=b+w\left(f-e\right)=a+\left(b-a\right)+w\left(f-e\right)\\ d&=c+\left(d-c\right)=a+\left(b-a\right)+w\left(f-e\right)+\left(d-c\right)=a-\left(f-e\right)+w\left(f-e\right)\\ e&=d+\left(e-d\right)=d+w\left(b-a\right)=a-\left(f-e\right)+w\left(f-e\right)+w\left(b-a\right)=a-\left(f-e\right)-w\left(d-c\right)\\ f&=e+\left(f-e\right)=a-w\left(d-c\right). \end{align*}Then we have \begin{align*} \overline{a}d+\overline{c}f+\overline{e}b&=\overline{a}\left(a+\left(w-1\right)\left(f-e\right)\right)\\&+\left(\overline{a}+\left(\overline{b}-\overline{a}\right)+\frac{1}{w}\left(\overline{f}-\overline{e}\right)\right)\left(a-w\left(d-c\right)\right)\\&+\left(\overline{a}-\left(\overline{f}-\overline{e}\right)-\frac{1}{w}\left(\overline{d}-\overline{c}\right)\right)\left(a+\left(b-a\right)\right) \\&=a\overline{a}+w\overline{a}\left(f-e\right)-\overline{a}\left(f-e\right)\\&+a\overline{a}-w\overline{a}\left(d-c\right)+a\left(\overline{b}-\overline{a}\right)-w\left(\overline{b}-\overline{a}\right)\left(d-c\right)+\frac{1}{w}a\left(\overline{f}-\overline{e}\right)-\left(\overline{f}-\overline{e}\right)\left(d-c\right)\\&+a\overline{a}+\overline{a}\left(b-a\right)-a\left(\overline{f}-\overline{e}\right)-\left(\overline{f}-\overline{e}\right)\left(b-a\right)-\frac{1}{w}a\left(\overline{d}-\overline{c}\right)-\frac{1}{w}\left(\overline{d}-\overline{c}\right)\left(b-a\right)\\ &=3a\overline{a}+w\overline{a}\left(f-e-d+c\right)+\frac{1}{w}a\overline{\left(f-e-d+c\right)}\\&-w\overline{\left(b-a\right)}\left(d-c\right)-\frac{1}{w}\left(b-a\right)\overline{\left(d-c\right)}\\&+a\overline{\left(b-a-f+e\right)}+\overline{a}\left(b-a-f+e\right)-\overline{\left(f-e\right)}\left(d-c+b-a\right)\\ &=3a\overline{a}+w\overline{a}\left(f-e-d+c\right)+\frac{1}{w}a\overline{\left(f-e-d+c\right)}\\&-w\overline{\left(b-a\right)}\left(d-c\right)-\frac{1}{w}\left(b-a\right)\overline{\left(d-c\right)}\\&+a\overline{\left(b-a-f+e\right)}+\overline{a}\left(b-a-f+e\right)+\overline{\left(f-e\right)}\left(f-e\right), \end{align*}which is the sum of real numbers $3a\overline{a}$, $w\overline{a}\left(f-e-d+c\right)+\frac{1}{w}a\overline{\left(f-e-d+c\right)}$, $-w\overline{\left(b-a\right)}\left(d-c\right)-\frac{1}{w}\left(b-a\right)\overline{\left(d-c\right)}$, $a\overline{\left(b-a-f+e\right)}+\overline{a}\left(b-a-f+e\right)$, and $\overline{\left(f-e\right)}\left(f-e\right)$. So \[\overline{a}d+\overline{c}f+\overline{e}b=a\overline{d}+c\overline{f}+e\overline{b},\]and thus \[\left(\overline{a}d-a\overline{d}\right)-\left(\overline{b}e-b\overline{e}\right)=-\left(\overline{c}f-c\overline{f}\right).\] Now, let $X_1=AD\cap BE,X_2=AD\cap CF$. Compute by intersection formula that \[x_1=\frac{\left(\overline{a}d-a\overline{d}\right)\left(b-e\right)-\left(\overline{b}e-b\overline{e}\right)\left(a-d\right)}{\left(\overline{a}-\overline{d}\right)\left(b-e\right)-\left(\overline{b}-\overline{e}\right)\left(a-d\right)}\]and \[x_2=\frac{\left(\overline{a}d-a\overline{d}\right)\left(c-f\right)-\left(\overline{c}f-c\overline{f}\right)\left(a-d\right)}{\left(\overline{a}-\overline{d}\right)\left(c-f\right)-\left(\overline{c}-\overline{f}\right)\left(a-d\right)}.\]Now, observe that \begin{align*} \left(\overline{a}d-a\overline{d}\right)\left(b-e\right)-\left(\overline{b}e-b\overline{e}\right)\left(a-d\right)&=\left(\overline{a}d-a\overline{d}\right)\left(a-d+c-f\right)-\left(\overline{b}e-b\overline{e}\right)\left(a-d\right)\\ &=\left(\overline{a}d-a\overline{d}\right)\left(c-f\right)+\left(\overline{a}d-a\overline{d}\right)\left(a-d\right)-\left(\overline{b}e-b\overline{e}\right)\left(a-d\right)\\ &=\left(\overline{a}d-a\overline{d}\right)\left(c-f\right)-\left(\overline{c}f-c\overline{f}\right)\left(a-d\right) \end{align*}and \begin{align*} \left(\overline{a}-\overline{d}\right)\left(b-e\right)-\left(\overline{b}-\overline{e}\right)\left(a-d\right)&=\left(\overline{a}-\overline{d}\right)\left(a-d+c-f\right)-\left(\overline{a}-\overline{d}+\overline{c}-\overline{f}\right)\left(a-d\right)\\ &=\left(\overline{a}-\overline{d}\right)\left(c-f\right)-\left(\overline{c}-\overline{f}\right)\left(a-d\right), \end{align*}so $X_1=X_2$ and hence $AD,BE,CF$ concur.

CantonMathGuy wrote: Thus, there exists a point $X$ such that $BXFA$ and $CXED$ are parallelograms. CantonMathGuy wrote: Thus, the diagonals $\overline{AD}$, $\overline{BE}$, and $\overline{CF}$ are concurrent at the circumcenter of $\triangle XYZ$ This may be an obvious/silly question, but what's the motivation/trigger behind (a) Constructing $X$ and (b) observing that the lines are concurrent at the circumcentre of $\Delta XYZ$ ?

Solved with goodbear, Th3Numb3rThr33. As shown below, let \(PQR\) be the triangle formed by lines \(DE\), \(FA\), \(BC\) and \(UVW\) the triangle formed by lines \(AB\), \(CD\), \(EF\). Also let \(FABX\), \(BCDY\), \(DEFZ\) be parallelograms. [asy][asy] size(6cm); defaultpen(fontsize(10pt)); pen pri=blue; pen sec=lightred; pen tri=purple+pink; pen qua=heavycyan; pen fil=cyan+white+white+white+white; pen sfil=invisible; pen tfil=invisible; pen qfil=invisible; pair x,y,z,O,A,B,C,D,EE,F,P,Q,R,U,V,WW; real t=10; x=dir(0); y=0.8*dir(110); z=-x-y; O=(0,0); EE=(0,0); D=EE+x; C=D-rotate(t,O)*z; B=C+y; A=B-rotate(t,O)*x; F=A+z; pair X,Y,Z; X=B+F-A; Y=D+B-C; Z=F+D-EE; P=extension(F,A,B,C); Q=extension(B,C,D,EE); R=extension(D,EE,F,A); U=extension(C,D,EE,F); V=extension(EE,F,A,B); WW=extension(A,B,C,D); draw(D--U--EE,tri); draw(F--V--A,tri); draw(B--WW--C,tri); fill(U--V--WW--cycle,tfil); draw(A--P--B,sec); draw(C--Q--D,sec); draw(EE--R--F,sec); fill(P--Q--R--cycle,sfil); filldraw(A--B--C--D--EE--F--cycle,fil,pri); fill(A--B--X--F--cycle,qfil); draw(B--X--F,qua); fill(C--D--Y--B--cycle,qfil); draw(D--Y--B,qua); fill(EE--F--Z--D--cycle,qfil); draw(F--Z--D,qua); dot("\(A\)",A,NW); dot("\(B\)",B,dir(60)); dot("\(C\)",C,E); dot("\(D\)",D,SE); dot("\(E\)",EE,SW); dot("\(F\)",F,W); dot("\(P\)",P,N); dot("\(Q\)",Q,SE); dot("\(R\)",R,SW); dot("\(U\)",U,S); dot("\(V\)",V,NW); dot("\(W\)",WW,NE); dot("\(X\)",X,NW); dot("\(Y\)",Y,E); dot("\(Z\)",Z,SW); [/asy][/asy] Claim: \(\triangle PQR\sim\triangle XYZ\). Proof. Check that \(\angle P=\angle A+\angle B-180^\circ=\angle D+\angle E-180^\circ=\angle U\). \(\blacksquare\) Observe that \(BX=AF=CD=BY\), and since \(\triangle FXE\cong\triangle DEY\) by SAS, we have \(EX=EY\). Then \(\overline{BE}\) is the perpendicular bisector of \(\overline{XY}\), so the three diagonals concur at the circumcenter of \(\triangle XYZ\). Remark: It can be shown that \(BXEY\) is a rhombus, so \(CDEX\), \(DFAY\), \(ABCZ\) are also parallelograms.

Kayak wrote: CantonMathGuy wrote: Thus, there exists a point $X$ such that $BXFA$ and $CXED$ are parallelograms. CantonMathGuy wrote: Thus, the diagonals $\overline{AD}$, $\overline{BE}$, and $\overline{CF}$ are concurrent at the circumcenter of $\triangle XYZ$ This may be an obvious/silly question, but what's the motivation/trigger behind (a) Constructing $X$ and (b) observing that the lines are concurrent at the circumcentre of $\Delta XYZ$ ? oops 2 years late but wtv The answer to the latter: I didn't. It came as a natural result from exploring enough of the diagram. Now we'll answer the former, because asking this to yourself is basically the entire question (not constructing $X$ - this is fairly natural, but realising its importance). You see, I thought that the construction of $X$ itself was pretty straightforward - it's a bunch of weird length conditions so you want to make some cool property pop out, and if IMO 2017 4 has taught us anything it's that we should always draw parallelograms. I have rather poor observation skills though, so I did not notice the double parallelogram. Even so, there's a good way to motivate this discovery (and the subsequent solving of the problem): understanding the construction of the diagram. It seems difficult to construct this diagram given the random angle and length equalities, so it would be a good idea to take a step back and scout. Let's try to parameterize stuff - a nice way to do this would be to view the sides as six individual vectors. A little bit of manipulation gives us that the angle between the opposite sides is always the same. Now, we could interpret this in two ways - try to do something with $AB \cap ED$ (which at least did nothing for me), or take a more concrete approach and just note that the complex number represented by $B - A$ (bear with my horrendous notation here) multiplied by some complex number $z$ with magnitude $1$ becomes the complex number represented by $E - D$. AIME level stuff I know, but given that this $z$ is the same for all the sides, this should be pretty useful. Let's say the six "side" numbers are $p, q, r, pz, qz, rz$. Then we clearly have that either $p + q + r = 0$ or $z = -1$ => now this is a cool property! In addition we already know how to deal with $z = -1$: that just means that it's a parallelo-hexagon? It's the easier case anyhow, so now we can just take for granted the condition that the "even-numbered sides make a triangle". Given this new discovery, let's take a side $AB$, and just consider a point $Q$ such that $BQ$ is the same "vector" as $CD$ and $AQ$ is the same "vector" as $EF$ (we know this can be done precisely from the above results). This gives us => wait, a double parallelogram! Let's try to use this. (after you fill in the rest of the diagram I think it's fairly natural to realise that it's just 3 rhombi and the concurrency point is the circumcenter) anyways that's how I did this problem today - hope this helps. Yet another example of "trying to construct the diagram => win". edit: I think I should clarify: the main idea is that we want a concrete way of drawing the diagram, and thus we explore the problem analytically (because synthetic fails). We then continually find results => update our diagram => realise we need more and dig further along the analytic path until we get something both easy to work with and substantial to the problem, which then solves it.

I will not include a diagram for personal reasons. Extend $(AF, BC), (BC, DE), (DE, AF)$ to intersect at $X, Y, Z$ and similarly $(CD, EF), (EF, AB), (AB, CD)$ at $U, V, W$. Construct points $P_A, P_C, P_E$ inside the hexagon for which $P_ABAF, P_CDCB, P_EFED$ are parallelograms. I claim that $AD$ is the perpendicular bisector of $P_CP_E$. Then, analogously, it follows that $AD, CF, EB$ concur at the circumcenter of $P_AP_BP_C$. By the parallelogram construction, note that $DP_E = EF = BC = DP_C$. It remains to show that $AP_E =AP_C$. Next, we will show that $\triangle AP_CB \sim \triangle P_EAF$, which implies the conclusion. Note that $AF = CD = BP_C$ and $AB = DE = FP_E$. Furthermore, check that\[\angle ABP_C = \angle B - (180^{\circ} - \angle C) = \angle B + \angle C - 180^{\circ} = \angle E + \angle F - 180^{\circ} - \angle F - (180^{\circ} - \angle E) = \angle P_EFA\]so we are done. $\blacksquare$ Remark: My motivation for drawing the parallelograms was because my initial diagram had opposite sides nearly parallel, so the concurrence point looked like the point that was the vertex of three parallelograms.

We use complex numbers, so $\frac{(f-a)(e-d)}{(b-a)(c-d)}=\frac{(b-c)(a-f)}{(d-c)(e-f)}=\frac{(d-e)(c-b)}{(f-e)(a-b)}$. Choose $c=0,f=1$, so that $a=\frac{b^2-be+b-e+1}{b-e+1},d=\frac{(e-1)(b-e)}{b-e+1}$. It suffices to show $A,D,BE\cap CF=\frac{\bar be-b\bar e}{\bar b-b+e-\bar e}$ collinear. $\frac{\frac{\bar be-b\bar e}{\bar b-b+e-\bar e}-\frac{(e-1)(b-e)}{b-e+1}}{\frac{b^2-be+b-e+1}{b-e+1}-\frac{(e-1)(b-e)}{b-e+1}}=\frac{(\bar be-b\bar e)(b-e+1)-(e-1)(b-e)(\bar b-b+e-\bar e)}{(\bar b-b+e-\bar e)(b^2-be+b-e+1-(e-1)(b-e))}\\=\frac{-b^2\bar e-\bar be^2+be\bar e-b\bar e+\bar be^2+b\bar b+b^2e-be^2-b^2+eb-be^2+e^3+be-e^2+be\bar e-\bar ee^2-b\bar e+e\bar e}{(\bar b-b+e-\bar e)(b-e+1)^2}\\=\lambda\frac{e\bar e-b\bar e-\bar e+be-b-e^2+1}{b-e+1},$ where we used $BC=EF\implies b\bar b=e\bar e-e-\bar e+1$ and $\lambda=\frac 1{\bar b-b+e-\bar e}\in\mathrm{i}\mathbb{R}$. Now $\frac{e\bar e-b\bar e-\bar e+be-b-e^2+1}{b-e+1}\overset !=-\frac{e\bar e-\bar be-e+\bar b\bar e-\bar b-\bar e^2+1}{\bar b-\bar e+1}\\=\frac{be\bar e-e^2\bar e+e^2-e\bar e-eb+e\bar e^2+\bar e-1-\bar e^2b+b}{(b-e+1)(\bar e-1)}=\frac{-(\bar e-1)(b\bar e-e\bar e+\bar e-eb+b+e^2-1)}{(b-e+1)(\bar e-1)},$ as desired.

Corrected solution, thanks to rafaello. Define $A'=AF\cap BC$ and $B',C',D',E',F'$ cyclically. Since $\angle A+\angle B=\angle D+\angle E$ we get $\measuredangle C'A'E'=\measuredangle F'D'B'$ and all similar arguments hold. Now let $ABCX,CDEY,EFAZ$ be parallelograms. It's clear $|CX|=|AB|=|DE|=$ $=|CY|$ and all analogous hold. Claim. $|FX|=|FY|.$ Proof. It's suffice to prove $AXF\cong EFY.$ We'll proceed by SAS. $$|AF|=|CD|=|EY|,\text{ } |AX|=|BC|=|EF|,\text{ }\measuredangle XAF=\measuredangle C'A"E'=\measuredangle F'D'B'=\measuredangle FEY\text{ } \Box$$ Thus we see that $AD,BE,CF$ are nothing but perpendicular bisectors of $XYZ$ and the conclusion follows.

As the old saying goes, "hexagon" is a synonym for "complex bash" (disclaimer: which I do NOT condone). Notice $a - f = s(d-c), c - b = s(f - e), e - d = s(b-a)$ for $s$ on the unit circle. Thus summing, we get either $s = -1$ where the concurrence is trivially the midpoint of $AD, BE, CF$ or $a + c + e = b +d + f$. Clear the value of $s$, and let $\frac{1}{s-1}, -\frac{s}{s-1}$ be $-\frac{1}{2}+gi, -\frac{1}{2} - gi$ instead. Now let $a = 0, d - c = m, f - e = n$, then let $m = \langle p, q \rangle, n = \langle r, s \rangle$. Then by coordinate bash, all three lines concur at $x = \frac{(p+q)(rq + 2rgs + 2pqg - ps)}{2(ps - qr)}$ so we are done.

@above you've become the very thing you swore to destroy... Let $P$, $Q$, $R$ be the points that make $PFAB$, $QBCD$, and $RDEF$ parallelograms. Then $DR=EF=BC=RD$. Also, $AF=CD=BQ$ and $FR=ED=AB$. We have \begin{align*} \angle AFR-\angle ABR &= \angle F - \angle RFE - \angle B - \angle RBC \\ &= \angle F - 180^\circ +\angle E - \angle B + 180^\circ - \angle C \\ &= (\angle F - \angle C) + (\angle E - \angle B)\\ &= 0 \end{align*}so $AR=AQ$. Thus, $AD$ is the perpendicular bisector of $RQ$ and so $AD$, $BE$ and $CF$ all pass through the circumcenter of $PQR$.

Toss on the complex plane. The angle (and side lengths) condition implies that \[ \frac{a-f}{d-c}=\frac{c-b}{f-e}=\frac{e-d}{b-a}=\text{cis } \theta \]for some real $\theta$. If $(d-c)+(f-e)+(b-a) \neq 0$, then \[ \text{cis } \theta = \frac{(a-f)+(c-b)+(e-d)}{(d-c)+(f-e)+(b-a)} = -1, \]which implies that $\frac{a+d}{2}=\frac{b+e}{2}=\frac{c+f}{2}$, so $\overline{AD}, \overline{BE}, \overline{CF}$ concur at their midpoints. In what follows, assume that $(d-c)+(f-e)+(b-a)=0 \Leftrightarrow a+c+e=b+d+f$. Define $X, Y, Z$ so that $ABXF, BCDY, DEFZ$ are all parallelograms. Then $x=b+f-e, y=b+d-c, z=d+f-e$. We can also write $x=c+e-d$ from our assumption, so $CDEX$ is also a parallelogram. This is analogous for $Y, Z$ as well. Note that $BX=AF$ from $BXFA$ and $BY=CD$ from $EYDC$, so $BX=BY$. Similarly, $EX=CD$ from $EXCD$ and $EY=AF$ from $EYAF$, so $EX=EY$. Thus, $\overline{BE}$ is the perpendicular bisector of segment $XY$. Applying this cyclically, $\overline{AD}, \overline{BE}, \overline{CF}$ concur at the circumcenter of $\triangle XYZ$, and we are done.

speaking on discord and seeing 13slg5 JOGGED my memory that i hadn't posted my soln on this yet! Construct G=AF\cap BC and cyclically until L, and let M,N,O be the points making FABM, BCDN,DEFO parallelograms; note that BM=AF=CD=BN and cyclically. It follows that BC=DN=DO, CD=BN=BM, CBM=KGI=HJL=CDO, so by SAS congruency we derive CM=CO, hence since both F and C lie on the perp. bisector of MO (from FM=FO,CM=CO) and analogously on the other diagonals, we have AD,BE,CF are the perp. bisectors of MNO, and concur at it's circumcenter, as desired. $\blacksquare$ Remark. Well motivated because the angles can't do anything, while the extensions of sides give natural sim triangles; the parallelograms in the making is a common trick to help transitive property equal side lengths. See for example 2001 g5; though they're not exactly brothers, they aren't too distant of a cousin either (idk i think this is how the analogy went but im not sure)

The angle condition gives that the angles between pairs of lines $AB,CD,EF$ are the same as between $DE,FA,BC.$ Thus the vector $\overrightarrow{BC}+\overrightarrow{DE}+\overrightarrow{FA}$ is a rotated copy of $\overrightarrow{AB}+\overrightarrow{CD}+\overrightarrow{EF},$ so either the rotation is by $180^\circ$ and the problem is easy (they concur at the center of the three parallelograms $ABDE,BCEF,CDFA$) or both of these vectors are $0.$ This gives $A+C+E=B+D+F,$ or $F+B-A=E+C-D,$ so there exists a point $P$ such that $ABPF,DCPE$ are parallelograms. Then angle chasing gives $\triangle PBC\cong\triangle PEF.$ Now we wish to show $AD,BE,CF$ concur. To do this, let $BE\cap CF=X.$ By homothety at $P$ with scale factor $\tfrac12$ we wish to show that the midpoints of $BF,CE,PX$ are collinear. But if $Y=BC\cap EF$ then the midpoint of $XY$ lies on this Newton-Gauss line as well. Thus we wish to show $PY$ is parallel to the newton-gauss line. This is equivalent to $PY$ being perpendicular to the Steiner line, which passes through the reflections of $P$ over $BC$ and $EF.$ In particular $Y$ is the circumcenter of these points and $P,$ but $P$ is equidistant from them by congruent triangles, so it is isosceles and we are done.