This seems like such an elemental thing to me, but I don't want to assume it's always true, since there could be an exception. Also, without knowing the length of $a$, if $\alpha=\beta$, then the lines are parallel too, right? Thank you.
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$\begingroup$On your drawing, mark the 8 points from A to H, starting with the bottom left one and then going counterclockwise. You know that if the size of angle ABG is a, then GBC is 180-a. Expand lines GB and FC over points C and B. If those two lines cross in a point X, the angle BXC will be of size 0 (since XBC is a and XCB is 180-a). Therefore, you conclude that lines GB and FC dont cross on the plane below AD.
You can conclude they dont cross above HE either similarly. Thus, GB and FC dont cross at all, which is the definition of two parallel lines.
Now, you can determine all angles in the polygon BCFG. Connect F and B, and calculate the angles created by such a line, and you'll be able to show that triangles GFB and BCF are congruent (same side BF, same angles CBF and BFG, same angles GBF and CFB), so therefore GB = FC.
Now, this is only the case if the angles a are oriented on the same side. If the initial angles a are opposite to eachother, ie not facing the same direction, then the quadrilateral will be a bilateral trapezoid, so the two lines will still be the same length, but won't be parallel (unless a = 90).
$\endgroup$ 6 $\begingroup$Construct an isosceles triangle $\triangle ABC$ with equal sides $AB$ and $BC$, and construct a line through $B$ parallel to $AC$. Then $AB$ and $BC$ are equal segments between parallel lines that are not parallel.
$\endgroup$ 2 $\begingroup$Hint:
take $\beta=180°-\alpha$, what about the two segments?
$\endgroup$ $\begingroup$Just consider the case $L_1 := \pmatrix{\mathbb{R}\\ 0}$ and $L_2 := \pmatrix{\mathbb{R}\\ 1}$. A line segment of length $a$ between those parallel lines implies $a^2 - 1= x^2$. Now construct the two vectors $\pmatrix{\sqrt{a^2 - 1}\\1}$ and $\pmatrix{-\sqrt{a^2 - 1}\\1}$, so the scalar product is $-(a^2 + 1) + 1 = -a^2 < 0$. But parallel vectors would have scalar product $> 0$.
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