Let $2k$ be the common sum of the lengths of pairs of opposite sides. Let $X,Y$ be points on $EF,ED$ such that $BX \parallel B_1E_1, BY \parallel A_1D_1$. Then $EX=EY=k$. Construct point $Z$ such that $EXZY$ is a rhombus. A little vector addition shows that $BZ \parallel CD, AF$, and has length $k$. Hence $B,X,Y$ all lie on a circle with center $Z$ and radius $k$. The desired result quickly follows.

Here is a horrible solution. Let $AF\cap BC=X,AF\cap DE=Y, DE\cap BC=Z.$ We first extract a metric condition: Lemma: $\triangle XAB\sim \triangle FYE\sim \triangle CDZ\sim \triangle XYZ,$ and the similarity ratios to $\triangle XYZ$ is given by $1-c\left(\frac{y+z}{yz}\right),1-c\left(\frac{z+x}{zx}\right),1-c\left(\frac{x+y}{xy}\right)$ for some constant $c.$ Proof: Let the similarity ratios be $k_1,k_2,k_3.$ Then $AB+DE=k_1x+x-k_2x-k_3x=x(k_1+1-k_2-k_3),$ and now setting all of these equal to each other and solving gives the lemma. We now use vectors with $X$ at the origin. Assume capital letters stand for the vector of the corresponding point. The main claim is as follows: Lemma: $\overrightarrow{A_1D_1}$ is parallel to the angle bisector of $\angle YXZ.$ Proof: We compute \[A_1=\frac{A+B}{2}=\frac{Y+Z}{2}\left(1-c\frac{|Y|+|Z|}{|Y||Z|}\right)\]\[E=Y+(Z-Y)\left(1-c\frac{|Y|+|Y-Z|}{|Y-Z||Y|}\right),D=Z+(Y-Z)\left(1-c\frac{|Z|+|Y-Z|}{|Y-Z||Z|}\right)\]\[\implies E_1=\frac{Y+Z}{2}+\frac{(Z-Y)}{2}\left(c\frac{|Y|-|Z|}{|Y||Z|}\right),\]so \[\overrightarrow{A_1E_1}=E_1-A_1=c\left(\frac{Y|Z|+Z|Y|}{|Y||Z|}\right).\]But the angle bisector of $\angle YXZ$ is parallel to $Y|Z|+Z|Y|$ as seen by constructing a rhombus, so we are done. Let $I$ be the incenter of $\triangle XYZ.$ By the lemma, it follows that \[\angle D_1OE_1=\angle(A_1D_1,B_1E_1)=\pi-\angle YIX=\frac{\pi-\angle Z}{2}=\frac{\pi-\angle YEF}{2}=\frac{\angle DEF}{2},\]as desired.

Great practice for constructing parallelogram! Construct parallelograms $ABEX, AFCY$ and $BAFZ$. Then $AB\stackrel{\parallel}{=} FZ\stackrel{\parallel}{=} EX$ which means $EXFZ$ is parallelogram. Thus $E_1$ is midpoint of $XZ$. Analogously, $B_1$ is midpoint of $YZ$ thus $YZ\parallel B_1E_1$. But the sum of lengths condition gives $DX=DY$ thus $B_1E_1$ is perpendicular to angle bisector of $\angle CDE$. Analogously, $A_1D_1$ is perpendicular to angle bisector of $\angle BCD$. Thus if $P=BC\cap DE$ and $I$ is incenter of $\triangle PCD$, then $CI\parallel A_1D_1$ and $D_1\parallel B_1E_1$. Hence $$\angle D_1OE_1 = 180^{\circ} - \angle CID = 90^{\circ} - \frac{\angle CPD}{2} = \frac{1}{2}\angle DEF$$as desired.

Identical to #3; posting for storage. Let \(X=\overline{EF}\cap\overline{CD}\), \(Y=\overline{CD}\cap\overline{AB}\), \(Z=\overline{AB}\cap\overline{EF}\), and denote \(x=YZ\), \(y=ZX\), \(z=XY\). Claim: [Interpret length condition] Let \(r_1\), \(r_2\), \(r_3\) be the ratios of similarity from \(\triangle XDE\), \(\triangle CYB\), \(\triangle FAZ\) to \(\triangle XYZ\). Then for some \(c>0\) we have \[r_1=1-c\left(\frac1y+\frac1z\right)\quad\text{and so on}.\] Proof. Let \(AB+DE=BC+EF=CD+FA=2c\). One may compute that \[2c=AB+DE=x(1+r_1-r_2-r_3),\]and solving the system of three equations gives the desired. \(\blacksquare\) The following is the key idea, from which the conclusion readily follows: Claim: [Main claim] \(\overline{A_1D_1}\) is parallel to the bisector of \(\angle YXZ\). Proof. We use vectors. Situate \(X\) at the origin. First, \[D_1=r_1\cdot\frac{Y+Z}2=\frac{Y+Z}2\left(1-c\left(\frac1y+\frac1z\right)\right).\] Now we turn to computing \(A_1\). Evidently \begin{align*} A&=Z+(Y-Z)r_3=Y\cdot r_3+Z\cdot(1-r_3)\\ B&=Y+(Z-Y)r_2=Y\cdot(1-r_2)+Z\cdot r_2. \end{align*}Combining these, \begin{align*} A_1&=\frac{A+B}2=\frac{Y(1-r_2+r_3)+Z(1+r_2-r_3)}2\\ &=\frac{Y\left(1+c\left(\frac1z-\frac1y\right)\right)+Z\left(1+c\left(\frac1y-\frac1z\right)\right)}2. \end{align*} At last, \[A_1-D_1=c\left(\frac Yz+\frac Zy\right)=c\left(\frac Y{|Y|}+\frac Z{|Z|}\right),\]which is parallel to the bisector of \(\angle YXZ\). \(\blacksquare\) Finally if \(I\) denotes the incenter of \(\triangle XYZ\) then \[\angle D_1OE_1=180^\circ-\angle XIY=90-\tfrac12\angle XZY=\tfrac12\angle DEF,\]the end.

Same idea as #1, posting for storage. Define points $X, Y, Z, T$ such that quadrilaterals $BXD_1A_1$, $BB_1EY$, $A_1D_1EZ$ and $EE_1B_1T$ are convex parallelograms. Reflect $B$ over $ZT$ to $B'$. Hence, a translation by vector $\vec{YZ}$ maps $\triangle ZB'T$ to $\triangle YEX$. I now claim that all of the sides of hexagon $BTXEYZ$ are equal. Reflect $A$ over $Z$ to $D'$ and $F$ over $Y$ to $C'$, so that quadrilaterals $TCDX$ and $FYZA$ are directly congruent. We then learn that $$YZ=\frac{AF+D'C'}{2}=\frac{AF+CD}{2}$$as desired. To finnish, note that $B'E=YZ$, so that we must have $B'$ to be the circumcenter of $\triangle EZT$. Then $$\angle D_1OE_1=\angle TEZ=\frac{1}{2}\angle AB'Z=\frac{1}{2}\angle ABC=\angle DEF.$$

The idea is to plug complex numbers to the sides of the hexagon. Let $u,v,w$ be complex numbers in the unit circle and let $a,b,c\in \mathbb{R}$ such that $\overrightarrow{AB}=au$, $\overrightarrow{BC}=bv$, $\overrightarrow{CD}=cw$. By the length condition and scaling, we get that $\overrightarrow{DE}=(a-1)u$, $\overrightarrow{EF}=(b-1)v$ and $\overrightarrow{FA}=(c-1)w$. As $ABCDEF$ is a hexagon, we also know that \[ au+bv+cw+(a-1)u+(b-1)v+(c-1)w=0 \]hence we get that \[ au+vb+cw = \frac{u+v+w}{2} \ \ \ \ \ \ (1). \]Now let's tackle the problem. We just need to prove that complex numbers $-v/u$ and \[ \left(\frac{cw+(a-1)u+\frac{1}{2}bv+\frac{1}{2}(b-1)v}{bv+cw+\frac{1}{2}au+\frac{1}{2}(a-1)u}\right)^2 \]have the same argument (that monster is the double of the angle formed by $\overrightarrow{A_1D_1}$ and $\overrightarrow{B_1E_1}$). But by $(1)$ we have that \[ \left(\frac{cw+(a-1)u+\frac{1}{2}bv+\frac{1}{2}(b-1)v}{bv+cw+\frac{1}{2}au+\frac{1}{2}(a-1)u}\right)^2 = \left(\frac{\frac{u+v+w}{2}-u-\frac{v}{2}}{\frac{u+v+w}{2}-\frac{u}{2}}\right)^2 = \left(\frac{w-u}{w+v}\right)^2. \]Recall that $k^2$ has argument $\frac{k^2}{\vert k^2\vert} = \frac{k^2}{\vert k \vert ^2} = \frac{k}{\overline{k}}$, hence \[ \text{arg}\left[\left(\frac{w-u}{w+v}\right)^2\right] = \frac{\frac{w-u}{w+v}}{\frac{\frac{1}{w}-\frac{1}{u}}{\frac{1}{w}+\frac{1}{v}}} = \frac{-v}{u}, \]which is what we wanted to prove.

Claim: $\overline{A_1D_1}$ is perpendicular to the bisector of $\angle F$. Proof: We have \[\overrightarrow{A_1D_1}=\frac{D+E}{2}-\frac{A+B}{2}=\frac{1}{2}((E-F)+(C-B)+(F-A)+(D-C))=\frac{1}{2}(\overrightarrow{FE}+\overrightarrow{BC}+\overrightarrow{AF}+\overrightarrow{CD}).\]Since $FE+BC=AF+CD$, this vector is perpendicular to the bisector of $\angle F$, as desired. $\square$ Similarly, $\overline{B_1E_1}$ is perpendicular to the bisector of $\angle D$. Thus, the angle between $\overline{A_1D_1}$ and $\overline{B_1E_1}$ is supplementary to the angle between the bisectors of $\angle F$ and $\angle D$. Therefore \[\angle D_1OE_1=180^\circ-\left(360^\circ-\frac{\angle F}{2}-\frac{\angle D}{2}-\angle E\right)=\frac{\angle D+\angle E+\angle F}{2}-180^\circ+\frac{\angle E}{2}=\frac{\angle E}{2},\]as desired. $\blacksquare$