IMO Shortlist

In all three solutions, we denote $\theta = \angle A - \angle D = \angle C - \angle F = \angle E - \angle B$ and assume without loss of generality that $\theta \geqslant 0$.

Solution 1. Let $x = A B = D E$, $y = C D = F A$, $z = E F = B C$. Consider the points $P, Q$, and $R$ such that the quadrilaterals $C D E P$, $E F A Q$, and $A B C R$ are parallelograms. We compute
$$\begin{aligned} \angle P E Q & = \angle F E Q + \angle D E P - \angle E = (180^\circ - \angle F) + (180^\circ - \angle D) - \angle E \\ & = 360^\circ - \angle D - \angle E - \angle F = \frac{1}{2}(\angle A + \angle B + \angle C - \angle D - \angle E - \angle F) = \theta / 2 \end{aligned}$$
Similarly, $\angle Q A R = \angle R C P = \theta / 2$.

If $\theta = 0$, since $\triangle R C P$ is isosceles, $R = P$. Therefore $A B \parallel R C = P C \parallel E D$, so $A B D E$ is a parallelogram. Similarly, $B C E F$ and $C D F A$ are parallelograms. It follows that $A D$, $B E$ and $C F$ meet at their common midpoint.

Now assume $\theta > 0$. Since $\triangle P E Q$, $\triangle Q A R$, and $\triangle R C P$ are isosceles and have the same angle at the apex, we have $\triangle P E Q \sim \triangle Q A R \sim \triangle R C P$ with ratios of similarity $y : z : x$. Thus
$$\begin{equation*} \triangle P Q R \text{ is similar to the triangle with sidelengths } y, z, \text{ and } x. \tag{1} \end{equation*}$$
Next, notice that
$$\frac{R Q}{Q P} = \frac{z}{y} = \frac{R A}{A F}$$
and, using directed angles between rays,
$$\text{(R Q, Q P) = (R Q, Q E) + (Q E, Q P) = (R Q, Q E) + (R A, R Q) = (R A, Q E) = (R A, A F) .}$$
Thus $\triangle P Q R \sim \triangle F A R$. Since $F A = y$ and $A R = z$, (1) then implies that $F R = x$. Similarly $F P = x$. Therefore $C R F P$ is a rhombus.

We conclude that $C F$ is the perpendicular bisector of $P R$. Similarly, $B E$ is the perpendicular bisector of $P Q$ and $A D$ is the perpendicular bisector of $Q R$. It follows that $A D$, $B E$, and $C F$ are concurrent at the circumcenter of $P Q R$.

Solution 2. Let $X = C D \cap E F$, $Y = E F \cap A B$, $Z = A B \cap C D$, $X' = F A \cap B C$, $Y' = B C \cap D E$, and $Z' = D E \cap F A$. From $\angle A + \angle B + \angle C = 360^\circ + \theta / 2$ we get $\angle A + \angle B > 180^\circ$ and $\angle B + \angle C > 180^\circ$, so $Z$ and $X'$ are respectively on the opposite sides of $B C$ and $A B$ from the hexagon. Similar conclusions hold for $X, Y, Y'$, and $Z'$. Then
$$\angle Y Z X = \angle B + \angle C - 180^\circ = \angle E + \angle F - 180^\circ = \angle Y' Z' X'$$
and similarly $\angle Z X Y = \angle Z' X' Y'$ and $\angle X Y Z = \angle X' Y' Z'$, so $\triangle X Y Z \sim \triangle X' Y' Z'$. Thus there is a rotation $R$ which sends $\triangle X Y Z$ to a triangle with sides parallel to $\triangle X' Y' Z'$. Since $A B = D E$ we have $R(\overrightarrow{A B}) = \overrightarrow{D E}$. Similarly, $R(\overrightarrow{C D}) = \overrightarrow{F A}$ and $R(\overrightarrow{E F}) = \overrightarrow{B C}$. Therefore
$$\overrightarrow{0} = \overrightarrow{A B} + \overrightarrow{B C} + \overrightarrow{C D} + \overrightarrow{D E} + \overrightarrow{E F} + \overrightarrow{F A} = (\overrightarrow{A B} + \overrightarrow{C D} + \overrightarrow{E F}) + R(\overrightarrow{A B} + \overrightarrow{C D} + \overrightarrow{E F})$$
If $R$ is a rotation by $180^\circ$, then any two opposite sides of our hexagon are equal and parallel, so the three diagonals meet at their common midpoint. Otherwise, we must have
$$\overrightarrow{A B} + \overrightarrow{C D} + \overrightarrow{E F} = \overrightarrow{0}$$
or else we would have two vectors with different directions whose sum is $\overrightarrow{0}$.

This allows us to consider a triangle $L M N$ with $\overrightarrow{L M} = \overrightarrow{E F}$, $\overrightarrow{M N} = \overrightarrow{A B}$, and $\overrightarrow{N L} = \overrightarrow{C D}$. Let $O$ be the circumcenter of $\triangle L M N$ and consider the points $O_1, O_2, O_3$ such that $\triangle A O_1 B$, $\triangle C O_2 D$, and $\triangle E O_3 F$ are translations of $\triangle M O N$, $\triangle N O L$, and $\triangle L O M$, respectively. Since $F O_3$ and $A O_1$ are translations of $M O$, quadrilateral $A F O_3 O_1$ is a parallelogram and $O_3 O_1 = F A = C D = N L$. Similarly, $O_1 O_2 = L M$ and $O_2 O_3 = M N$. Therefore $\triangle O_1 O_2 O_3 \cong \triangle L M N$. Moreover, by means of the rotation $R$ one may check that these triangles have the same orientation.

Let $T$ be the circumcenter of $\triangle O_1 O_2 O_3$. We claim that $A D$, $B E$, and $C F$ meet at $T$. Let us show that $C$, $T$, and $F$ are collinear. Notice that $C O_2 = O_2 T = T O_3 = O_3 F$ since they are all equal to the circumradius of $\triangle L M N$. Therefore $\triangle T O_3 F$ and $\triangle C O_2 T$ are isosceles. Using directed angles between rays again, we get
$$\text{(T F, T O 3 ) = (F O 3, F T ) and (T O 2, T C ) = (C T, C O 2 ) . 2}$$
Also, $T$ and $O$ are the circumcenters of the congruent triangles $\triangle O_1 O_2 O_3$ and $\triangle L M N$ so we have $\text{(T O 3, T O 2 ) = (O N, O M)}$. Since $C O_2$ and $F O_3$ are translations of $N O$ and $M O$ respectively, this implies
$$\text{(T O 3, T O 2 ) = (C O 2, F O 3 ) . 3}$$
Adding the three equations in (2) and (3) gives
$$\text{(T F, T C) = (C T, F T) = - (T F, T C)}$$
which implies that $T$ is on $C F$. Analogous arguments show that it is on $A D$ and $B E$ also. The desired result follows.

Solution 3. Place the hexagon on the complex plane, with $A$ at the origin and vertices labelled clockwise. Now $A, B, C, D, E, F$ represent the corresponding complex numbers. Also consider the complex numbers $a, b, c, a', b', c'$ given by $B - A = a$, $D - C = b$, $F - E = c$, $E - D = a'$, $A - F = b'$, and $C - B = c'$. Let $k = |a| / |b|$. From $a / b' = -k e^{i \angle A}$ and $a' / b = -k e^{i \angle D}$ we get that $\left(a' / a\right)\left(b' / b\right) = e^{-i \theta}$ and similarly $\left(b' / b\right)\left(c' / c\right) = e^{-i \theta}$ and $\left(c' / c\right)\left(a' / a\right) = e^{-i \theta}$. It follows that $a' = a r$, $b' = b r$, and $c' = c r$ for a complex number $r$ with $|r| = 1$, as shown below.

We have
$$0 = a + c r + b + a r + c + b r = (a + b + c)(1 + r) .$$
If $r = -1$, then the hexagon is centrally symmetric and its diagonals intersect at its center of symmetry. Otherwise
$$a + b + c = 0 .$$
Therefore
$$A = 0, \quad B = a, \quad C = a + c r, \quad D = c(r - 1), \quad E = -b r - c, \quad F = -b r .$$
Now consider a point $W$ on $A D$ given by the complex number $c(r - 1) \lambda$, where $\lambda$ is a real number with $0 < \lambda < 1$. Since $D \neq A$, we have $r \neq 1$, so we can define $s = 1 / (r - 1)$. From $r \bar{r} = |r|^2 = 1$ we get
$$1 + s = \frac{r}{r - 1} = \frac{r}{r - r \bar{r}} = \frac{1}{1 - \bar{r}} = -\bar{s} .$$
Now,
$$\begin{aligned} W \text{ is on } B E & \Longleftrightarrow c(r - 1) \lambda - a \| a - (-b r - c) = b(r - 1) \Longleftrightarrow c \lambda - a s \| b \\ & \Longleftrightarrow -a \lambda - b \lambda - a s \| b \Longleftrightarrow a(\lambda + s) \| b . \end{aligned}$$
One easily checks that $r \neq \pm 1$ implies that $\lambda + s \neq 0$ since $s$ is not real. On the other hand,
$$\begin{aligned} W \text{ on } C F & \Longleftrightarrow c(r - 1) \lambda + b r \| -b r - (a + c r) = a(r - 1) \Longleftrightarrow c \lambda + b(1 + s) \| a \\ & \Longleftrightarrow -a \lambda - b \lambda - b \bar{s} \| a \Longleftrightarrow b(\lambda + \bar{s}) \| a \Longleftrightarrow b \| a(\lambda + s), \end{aligned}$$
where in the last step we use that $(\lambda + s)(\lambda + \bar{s}) = |\lambda + s|^2 \in \mathbb{R}_{>0}$. We conclude that $A D \cap B E = C F \cap B E$, and the desired result follows.