Proof of irregularity of an octagon determined by lines from vertices to midpoints of sides of a square I stumbled upon this old question "Area of octagon constructed in a square" that involves finding the area of an octagon within a square as shown:

I found that the octagon is not regular. So, my question is this:

Prove that the octagon shown above is not regular, assuming the only length we are initially given is the side length of the square.
Note: The edges of the octagon are determined by the line joining the vertices of the square to midpoints of the square's sides.

So far, the easiest way I could think of was to use the cosine rule to get the largest angle in $\triangle ABC$ (below) and show that the total of the angles would not be equal to the expected sum of a regular octagon.


Are there any other ways of proving that this octagon is not regular?

 A: 
$$\underbrace{|OA| = \frac12|OM|}_{\text{$A$ is the midpoint of $\overline{OM}$}}=\;\; \frac14|PQ| \;\;\color{red}{\neq}\;\; \frac13\cdot\frac1{\sqrt{2}}|PQ| \;\;= \underbrace{\frac13|OP| = |OB|}_{\text{$B$ is the centroid of $\triangle PQR$}}$$
A: Here is a better way that requires absolutely no calculation at all.

If the octagon is regular, then rotating the blue triangle would yield the red triangle, but clearly the blue triangle has the shorter long side.
A: A regular octagon has all its exterior angles equal, so to demonstrate non-regularity it suffices to show that there are two exterior angles which are not equal to each other.  In the diagram below we focus on showing that $\angle EXB <\angle AYF$.

Given that $ABCD$ is a square and $E,F,G,H$ are midpoints, we can infer that $EB=BH=\frac{1}{2}AD$ and hence, using Pythagoras' Theorem, $EH=\frac{\sqrt{2}}{2}AD$ so that $EH < AD$. 
Focusing now on $\triangle EDH$ and $\triangle AHD$, we have $ED=DH=AH$, so both are isosceles and their equal sides are the same.  Hence, since $EH < AD$, $\angle EDH < \angle AHD$.
Since $F$ and $H$ are mid-points of sides $AD$ and $BC$ respectively of the square, $BF$ and $HD$ are parallel, and so $\angle EXB = \angle EDH$ and $\angle AYF = \angle AHD$.  Hence:
$$\angle EXB = \angle EDH < \angle AHD = \angle AYF$$
QED
