Perpendiculars from $B,D$ and $F$ to $DF \parallel AC,$ $FB \parallel CE$ and $BD \parallel EA$ obviously concur at the orthocenter $H$ of $\triangle BDF.$ Then $AB^2-BC^2=HA^2-HC^2,$ $CD^2-DE^2=HC^2-HE^2$ and $EF^2-FA^2=HE^2-HA^2.$ Adding these expressions together gives $AB^2-BC^2+CD^2-DE^2+EF^2-FA^2=0 \Longrightarrow$ $AB^2+CD^2+EF^2=BC^2+DE^2+AF^2.$

This vectorial solution captures the same phenomenon as the more synthetic one above. Associate position vectors to the vertices of the hexagon (with $v$ being the position vector associated to a point $V$). Then we can take advantage of the writing $XY^2 = \|y-x\|^2 = \left < y-x, y-x\right > = \|x\|^2 + \|y\|^2 - 2\left < y, x\right >$. If we write these relations for the sides of the hexagon, the required equality comes to \[\left < b, a\right > + \left < d, c\right > + \left < f, e\right > = \left < b, c\right > + \left < d, e\right > + \left < f, a\right >,\] that is \[\left < b, a - c\right > + \left < d, c - e\right > + \left < f, e - a\right > = 0.\] But if we take the origin on the perpendicular from $B$ on $AC$, then $b \perp a - c$, and so $\left < b, a - c\right > = 0$. Similar consequences will hold for the other two terms. Therefore, if we take the origin at the orthocentre of the triangle $BDF$, all three terms will be null, and so the required equality will be proved. We can in fact proceed without considering the orthocentre of $\triangle BDF$, by leaving the origin at an arbitrary point. The equivalent relation obtained may also be written (using $\left < b, a - c\right > = \left < b, (a - e) + (e - c)\right >$ as \[\left < b - f, a - e\right > + \left < b - d, e - c\right > = 0.\] But triangles $BDF$ and $EAC$ are similar, in a similarity ratio $\rho$, and then $\dfrac {BD} {AE} = \dfrac {FB} {EC} = \rho$, whence $b-d = \rho(a-e)$ and $b-f = -\rho(e-c)$, therefore \[\left < b - f, a - e\right > + \left < b - d, e - c\right > = -\rho\left < e-c, a - e\right > + \rho\left < a - e, e - c\right > = 0.\] One more proof, if needed, of the strength of these dot-product manipulations.

it's easy from the Carnot's theorem!