Q:

Smallest square that can be fitted outside the regular hexagon

They have the derivation here 
http://www.drking.org.uk/hexagons/misc/deriv4.html
The figure is

In this derivation the second line is $ a^2 = b^2 + b^2 $ .How have they assumed $ AB = AC $ ? . Is this assumption based on some vision ? . Why I am asking this is , because the problem is to find smallest square fitted outside the hexagon . If I am given this question in exam how will this point  $AB= AC $ strike my brain for this case ? I will try to go with general case $AB \neq AC $ and solve . But this will make the calculation a bit tough as the number of unknowns is 1 more . So is there a vision or wisdom behind this guess . Am I missing something ?
Second question is on what basis did they start with this figure , because both are regular polygons the first image which struck my mind was something like 
Am I missing something .
Thanks 

A:

In the attached image, note that: $$\angle EBI = 180^\circ - \angle ABG - 60^\circ$$
Moreover, to minimize the square you want point $E$ to be on the square, so that $$2a\sin (\angle EBI)= x$$
Where $x$ is the square side length. The same logic works for point $D$, so that:
$$2a\sin (\angle DAK)= x$$
Leading to:
$$\angle EBI = \angle DAK \ \to \ \angle ABG = \angle GAB \ \ \blacksquare$$
Note that in the proof I did not assume a particular orientation of the hexagon, so that this also covers your second question - i.e. why can't the hexagon be parallel to the square. 

