# 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

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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: $$\angle EBI = \angle DAK \ \to \ \angle ABG = \angle GAB \ \ \blacksquare$$