Middle lines in octagon are congruent 
Let $A_1A_2\dots A_8$ be a cyclic octagon. Suppose that $A_1A_6||A_2A_5, A_3A_8||A_4A_7$, and that $A_2A_5\perp A_3A_8$. Prove that the length of the two midline segmets, one connecting the midpoints of sides $A_8A_1, A_4A_5$, and the other connecting sides $A_2A_3, A_6A_7$ are equal.


Here is an (ugly) solution with (position) vectors: Let $M_{ij}$ be the
midpoint of a chord $A_iA_j$. Let $\vec{a}$ be a vector between midpoints of segments $A_2A_5$ and $A_1A_6$ so $$\vec{a}=\overrightarrow{M_{52}M_{16}} =M_{16}-M_{52} $$ 
and $\vec{b}$ be a vector between midpoints of segments $A_4A_7$ and $A_3A_8$ so $$\vec{b}=\overrightarrow{M_{47}M_{38}} =M_{38}-M_{47} $$ 
Notice that $\vec{a}\bot \vec{b}$ and  $\overrightarrow{A_7A_4} || \overrightarrow{A_8A_3}||\vec{a}$ and $\overrightarrow{A_2A_5} ||\overrightarrow{A_1A_6}||\vec{b}$.
Then \begin{eqnarray}
4\overline{M_{18}M_{45}}^2 -4\overline{{M_{23}M_{67}}}^2 &=&  4\overrightarrow{M_{18}M_{45}}^2 -4\overrightarrow{M_{23}M_{67}}^2 \\
 &=&  (2\overrightarrow{M_{18}M_{45}} -2\overrightarrow{M_{23}M_{67}})(2\overrightarrow{M_{18}M_{45}}+2\overrightarrow{M_{23}M_{67}}) \\
  &=&  (A_4+A_5 -A_1-A_8-A_6-A_7 +A_3+A_2)\cdot \\&&(A_4+A_5 -A_1-A_8+A_6+A_7 -A_3-A_2) \\
  &=&(\overrightarrow{A_7A_4} +\overrightarrow{A_8A_3}+2M_{52}-2M_{16})  \cdot \\&& (\overrightarrow{A_2A_5} +\overrightarrow{A_1A_6}+2M_{47}-2M_{38})\\
  &=&  (m\cdot\vec{b} +k\cdot \vec{b}-2\vec{b})\cdot (l\cdot \vec{a} +n\cdot \vec{a}-2\vec{a}) \\
&=&0
\end{eqnarray} and we are done. 
Can somebody find nicer solution, possibly with geometric transformations?
 A: If this were graphed onto a coordinate plane temporarily, focus on the y-values of all points $A_1,A_2,...,A_8$. 
We know that $A_1=A_6$ and $A_2=A_5$, while $A_3+A_8=A_4+A_7$ since they are parallel chords. So, $\frac{A_1+A_2+A_3+A_8}{4}=\frac{A_4+A_5+A_6+A_7}{4}$. This is important because if we rewrite it, it is $\frac{\frac{A_1+A_8}{2} +\frac{A_2+A_3}{2}}{2}$=$\frac{\frac{A_7+A_6}{2} +\frac{A_4+A_5}{2}}{2}$, translating into $\frac{M_{18}+M_{23}}{2}=\frac{M_{67}+M_{45}}{2}$. 
So when we draw lines $M_{18}M_{23}$ and $M_{67}M_{45}$, their midpoints $M_{1823}$ and $M_{6745}$ have the same y-values. The two lines are also parallel, so the perpendicular bisectors of both $M_{18}M_{23}$ and $M_{67}M_{45}$ is the same line. For convenience, assume that this perpendicular bisector crosses over the black point in the diagram, call it K. By the properties of a perpendicular bisector on both $M_{18}M_{23}$ and $M_{67}M_{45}$, $KM_{67}=KM_{45}$ and $KM_{18}=KM_{23}$. 
Also, angle $M_{18}KM_{67}$ is congruent to the angle $M_{23}KM_{45}$, meaning that $M_{18}KM_{67}+M_{67}KM_{45}$=$M_{23}KM_{45}+M_{67}KM_{45}$ so $M_{18}KM_{45}=M_{67}KM_{23}$. 
Thus, by SAS, triangles $M_{18}KM_{45}$ and $M_{67}KM_{23}$ are congruent, meaning that $M_{67}M_{23}=M_{18}M_{45}$ and we're done. 
A: Here is another solution motivated by Muchang Bahng writings:
Set a coordinate system so that perpendicular bisector for $A_4A_7$ (which is also for $A_3A_8$) be a x-axsis and a perpenduclar bisector for $A_1A_6$ (which is also for $A_2A_4$) be a y-axsis, in particular, let $$ M_{17}= (2b,0)\;\;\;\; {\rm and } \;\;\;\;M_{38}= (-2a,0)$$ and
$$ M_{25}= (0,-2d)\;\;\;\; {\rm and } \;\;\;\;M_{16}= (0,2c)$$ 
then $$ A_{1}= (-2k,2c)\;\;\;\; {\rm and } \;\;\;\;A_{6}= (2k,2c)$$
$$ A_{2}= (-2l,-2d)\;\;\;\; {\rm and } \;\;\;\;A_{5}= (2l,-2d)$$
$$ A_{3}= (-2a,-2m)\;\;\;\; {\rm and } \;\;\;\;A_{8}= (-2a,2m)$$
$$ A_{4}= (2b,-2n)\;\;\;\; {\rm and } \;\;\;\;A_{7}= (2b,2n)$$ 
thus $$ M_{18}= (-a-k,c+d)\;\;\;\; {\rm and } \;\;\;\;M_{23}= (-l-a,-m-a)$$
$$ M_{45}= (l+b,-n-d)\;\;\;\; {\rm and } \;\;\;\;M_{67}= (b+k,n+c)$$ 
So $$ \overline{M_{18}M_{45}}^2 = (c+m+n+d)^2+(l+b+a+k)^2 = \overline{M_{23}M_{67}}^2$$
and we are done. 
Hmm, much easier then first solution and we never used that $A_i$ are actualy on a circle!
