Let a1,..,a6 by radius-vectors of hexagon vertices. We have abs((a1+a2)-(a4+a5))/2=sqrt(3/4)*(abs(a1-a2)+abs(a4-a5))>= sqrt(3/4)*abs(a1+a5-a2-a4). Take the square, ans sum up three such inequalities. Then simplify. We get 0>=(a1+a3+a5-a2-a4-a6)^2. So, a1+a3+a5=a2+a4+a6 and above inequalities were equalities, so opposite sides are parallel. From a1+a3+a5=a2+a4+a6 we have (projecting to the perpendicular to a1a2) that a6 and a3 are equidistant from the line a1a2. Then consider the triangle formed by continuations of sides a1a2, a3a4, a5a6. The hexagon is gotten from the triangle by cutting small similar to big and equal to each other triangles near vertives. We get for this triangle, that each of its medians equals sqrt(3/4)*respective side. The rest is easy (one of the ways is the formula for the length of the median).

We use vectors. By the triangle inequality, \[\|A+B-D-E\| = \sqrt{3}(\|A-B\|+\|E-D\|) \ge \sqrt{3}\|A-B+E-D\|,\]etc. Squaring these three inequalities and adding up yields \[0 \ge (A-D-B+E+C-F)^2\implies A+C+E=B+D+F,\]so equality must occur and thus there exist reals $r,s,t>0$ such that \[(A-B,B-C,C-D)=(rE-rD,sF-sE,tA-tF).\]WLOG let $E=0$, so $A=B-rD$ and $C=B-sF$. Then \[B+D+F = A+C = 2B-rD-sF \implies B=(1+r)D+(1+s)F\]while \[D-tF = C-tA = B-sF-tB+rtD \implies (1-t)B=(1-rt)D+(s-t)F.\]Substituting, \[(1-rt)D+(s-t)F=(1-rt+r-t)D+(1-st+s-t)F\implies 0=(r-t)D+(1-st)F.\]But $D,F$, and $0=E$ are not collinear, so $r=t$ and $s=1/t$. Solving, we obtain \[(A,B,C,D,E,F)=(D+(1+\frac{1}{t})F,(1+t)D+(1+\frac{1}{t})F,(1+t)D+F,D,0,F).\]Now returning to the original condition for pairs of sides $(AB,DE),(BC,EF)$, we obtain \[\|(1+t)D+2(1+\frac{1}{t})F\| = \sqrt{3}(1+t)\|D\| \implies \|tD+2F\| = \sqrt{3}\|tD\|\]and \[\|2(1+t)D+(1+\frac{1}{t})F\| = \sqrt{3}(1+\frac{1}{t})\|F\| \implies \|2tD+F\| = \sqrt{3}\|F\|.\]Squaring these two equations and subtracting, we find $\|F\|=t\|D\|$, whence the first equation yields \[4(tD)\cdot F=2t^2D^2-4F^2=-2\|tD\|\|F\|,\]i.e. the angle between $ED$ and $EF$ is $120^\circ$. We're done by symmetry.

Let $ c = \sqrt{3}/2$. Consider the three diagonals of the hexagon. Two must intersect at an angle $ \ge 60º$. Suppose $AD$ and $BE$ cut at $P$, and $ \angle APB = \angle DPE > 60º$. Then if $M, N$ are the midpoints of $AB, DE$ then $MP \le c \cdot AB$. This is because $M$ is inside the circle through $AB$ and radius $c \cdot AB / 2$, and this circle is internally tangent to the circle centered at $M$ with radius $c \cdot AB$. Similarly $NP \le c \cdot DE$. But $MP + NP \ge MN \ge c \cdot (DE+AB)$. Thus, both equalities occur, which implies $M,N$ are the points of tangency of the circles I mentioned. That is, $ABM, CDN$ are equilateral. Then there are another two diagonals that intersect at an angle $\ge 60º$ (since these ones intersect at exactly $60º$). We repeat the above argument. Thus we can prove that if $Q = CF \cap BE$, then $BCQ, QEF$ are equilateral. One can easily prove the problem from here.

Unsurprisingly, this is a geometric inequality. Denote the hexagon by $ABCDEF$. Then we have that \[ \left\lvert \frac{\vec D + \vec E}{2} - \frac{\vec A + \vec B}{2} \right\rvert = \sqrt3 \cdot \frac{\left\lvert \vec B - \vec A \right\rvert + \left\lvert \vec E - \vec D \right\rvert}{2} \ge \sqrt 3 \cdot \left\lvert \frac{(\vec B - \vec A) - (\vec E - \vec D)}{2} \right\rvert \]and cyclic variations. Suppose we define the right-hand sides as variables \begin{align*} \vec x = (\vec B - \vec A) - (\vec E - \vec D) \\ \vec y = (\vec D - \vec C) - (\vec A - \vec F) \\ \vec z = (\vec F - \vec E) - (\vec C - \vec B). \end{align*}Then we now have \begin{align*} \left\lvert \vec y - \vec z \right\rvert &\ge \sqrt 3 \left\lvert \vec x \right\rvert \\ \left\lvert \vec z - \vec x \right\rvert &\ge \sqrt 3 \left\lvert \vec y \right\rvert \\ \left\lvert \vec x - \vec y \right\rvert &\ge \sqrt 3 \left\lvert \vec z \right\rvert. \end{align*}We square all sides, use the dot product in the form $\left\lvert \vec v \right\rvert^2 = \vec v \cdot \vec v$ and the sum in order to get \[ \sum_{\text{cyc}} (\vec y - \vec z) \cdot (\vec y - \vec z) \ge 3 \sum_{\text{cyc}} \vec x \cdot \vec x \]which rearranges to \[ -\left\lvert \vec x + \vec y + \vec z \right\rvert^2 \ge 0. \]This can only happen if $\vec x + \vec y + \vec z =0$, and moreover all the inequalities above were actually equalities. That means that our triangle inequalities above were actually sharp (and already we have $\overline{AB} \parallel \overline{DE}$ and so on). Working with just $x$ and $y$ now we have \begin{align*} 3 (\vec x \cdot \vec x) &= (2 \vec y - \vec x) \cdot (2 \vec y - \vec x) \\ &= \vec x \cdot \vec x - 4 \vec y \cdot \vec x + 4 \vec y \cdot \vec y \\ \implies -\vec x \cdot \vec x + 2 (\vec y \cdot \vec y) &= 2 \vec x \cdot \vec y \\ 2 (\vec x \cdot \vec x) - \vec y \cdot \vec y &= 2 \vec x \cdot \vec y. \end{align*}which implies $\vec x \cdot \vec x = \vec y \cdot \vec y$, that is, $\vec x$ and $\vec y$ have the same magnitude. In this way we find $\vec x$, $\vec y$, $\vec z$ all have the same magnitude, and since $\vec x + \vec y + \vec z = 0$ they are related by $120^{\circ}$ rotations, as desired.

Claim: In triangle $\triangle XYZ$ with $\angle X \ge 60^\circ$ where midpoint of $YZ$ is $M,$ $MX \le \frac{\sqrt{3}}{2}YZ$ with equality only when $XYZ$ is equilateral. Proof: Define $W$ on same side as $YZ$ as $X$ so $WYZ$ is equilateral. Note $X$ is inside $(WYZ)$ and $(WYZ)$ is internally tangent to the circle centered at $M$ through $W$ at $W.$ So $MX \le MW = \frac{\sqrt{3}}{2}YZ$ with equality only when $X = W.$ $\blacksquare$ Label the hexagon $ABCDEF$ and assume WLOG the acute angle between diagonals $AD,BE$ is $\ge 60^\circ.$ Let $AB, DE$ have midpoints $M,N$ respectively. Let $AD,BE$ meet at $P.$ But $$MN \le PM + PN \le \frac{\sqrt{3}}{2}AB + \frac{\sqrt{3}}{2}DE$$and equality is only achieved when $ABP, DEP$ are equilateral. Now that we know the acute angle between diagonals $AD,BE$ is $\ge 60^\circ,$ then $CF$ must make an acute angle $\ge 60^\circ$ with either $AD$ or $BE.$ Using the same argument as before, $CF$ must make an exactly $60^\circ$ angle with both $AD$ and $BE$ and furthermore the acute angle between any side and any diagonal sharing a vertex with that side must be $60^\circ,$ as desired. $\blacksquare$

Let the vertices of the convex hexagon be $v_1$, $v_2$, $\dots$, $v_6$ in that order. We have \[\left|\frac{a_1+a_2-a_4-a_5}{2}\right|=\frac{\sqrt{3}}{2}\left(|a_1-a_2|+|a_4-a_5|\right)\ge \sqrt{3}\cdot \left|\frac{a_1+a_5-a_2-a_4}{2}\right|\]so $(a_1+a_2-a_4-a_5)^2\ge 3(a_1+a_5-a_2-a_4)^2$. This rearranges to \[0\ge 2a_1^2+2a_2^2+2a_4^2+2a_5^2+8a_1a_5+8a_2a_4-8a_1a_2-8a_4a_5-4a_1a_4-4a_2a_5\]and summing cyclically gives $a_1+a_3+a_5=a_2+a_4+a_6$ and that the triangle inequality was an equality and so the sides are parallel and equal. Now, extend the sides of the hexagon to triangle $ABC$. Let the hexagon by $U$, $V$ on $AB$, $W$, $X$ on $BC$, and $Y$, $Z$ on $CA$, in those orders respectively. Then, let $G$ be the centroid of $ABC$ and the centroid of $UVWXYZ$. Then clearly, $GA$ bisects $UZ$ since $AUZ$ is simply a homothety of $ABC$ with center $A$. Then, that means that $AUZ$, $VBW$ and $YVC$ are all congruent, so $UVWXYZ$ trisects each of $ABC$'s sides, and we get that each median is $\tfrac{\sqrt{3}}{2}$ of its opposite side, and the result follows.

Work in the vector plane. Let the hexagon be $ABCDEF$. First see that \[MQ=\left| \frac{\overrightarrow{D}+\overrightarrow{E}-\overrightarrow{A}-\overrightarrow{B}}{2}\right|=\frac{\sqrt{3}}{2}\cdot \left(\left|\overrightarrow{B}-\overrightarrow{A}\right|+\left|\overrightarrow{D}-\overrightarrow{E}\right|\right)\ge \frac{\sqrt{3}}{2}\cdot \left|(\overrightarrow{B}-\overrightarrow{A})-(\overrightarrow{D}-\overrightarrow{E})\right|\]Consider the following vectors \[\overrightarrow{u}=(\overrightarrow{B}-\overrightarrow{A})-(\overrightarrow{D}-\overrightarrow{E})\]\[\overrightarrow{v}=(\overrightarrow{D}-\overrightarrow{C})-(\overrightarrow{A}-\overrightarrow{F})\]\[\overrightarrow{w}=(\overrightarrow{F}-\overrightarrow{E})-(\overrightarrow{C}-\overrightarrow{B})\]Then we have the inequalities \[|\overrightarrow{v}-\overrightarrow{w}|\ge \sqrt{3}\cdot |\overrightarrow{u}|\]\[|\overrightarrow{w}-\overrightarrow{u}|\ge \sqrt{3}\cdot |\overrightarrow{v}|\]\[|\overrightarrow{u}-\overrightarrow{v}|\ge \sqrt{3}\cdot |\overrightarrow{w}|\]After squaring and summing up we are left with \[(\overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w})^2\le 0\]which means that $\overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}=0$ or $\overrightarrow{A}+\overrightarrow{C}+\overrightarrow{F}=0$. Now take the points $A',B'$ and $C'$ such that $\overrightarrow{OA'}=\overrightarrow{u}, \overrightarrow{OB'}=\overrightarrow{v}, \overrightarrow{OC'}=\overrightarrow{w}$. Pluging $\overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}=0$ in the inequalities and using the notations get us to \[\cos \angle A'OB'\le -\frac{1}{2}\]\[\cos \angle B'OC'\le -\frac{1}{2}\]\[\cos \angle C'OA'\le -\frac{1}{2}\]They sum up to $360^o$, therefore they are all equal to $120^o$, meaning that $A'B'C'$ is equilateral. Since we have equalities everywhere we have that $AB\parallel ED$ and so on.Put toghether with the previous observation solves the problem.

Wow. I got a solution no one have posted in 21 years Let two opposite sides be $AB$ and $CD$, where $ABDC$ is convex. My idea is: because midpoints of $AB$ and $CD$ are quite far from each other, our quadraliteral should be very stretched, and thus the angle between it's diagonals($AD$ and $BC$) should be very small. To formalize, we will prove that it is less or equal than $60º$ Proof: Let the midpoints of $AB$, $BD$, $DC$, $AC$ be $M,N,K,L$. It is well-known that $NL \leq \frac{AB+CD}{2}$ and equality is when $AB \parallel CD$. Now $MNKL$ is a parallelogram and $MK \geq \sqrt{3}NL$. One can easily check that from this it follows that $\angle NML \leq 60º$ (for example, use median formula in a triangle). Thus if $O$ is the intersection of $AD$ and $BC$, then $\angle AOB \leq 60º$. Rewrite this angle as the angle by which you need to rotate a directed segment $DA$ to get a directed segment $CB$. Do that for every pair of opposite sides. Now all six angles between our directed segments are $\leq 60º$, but their sum is $360º$. Thus all of them are $60º$, and because all our inequalities become equalities, $AB \parallel CD$ and $AD=BC$. From this $\angle ABC=60º$, doing the same for all other diagonals we get that all angles of our hexagon are equal to $120º$. Actually, the condition we got is both necessary and sufficient (all diagonals with opposite vertices are equal and all angles are $120º$)