Show that the marked triangle is equilateral



By the symmetry of the figure with respect to the square diagonal \(AE\), the segments \(AB\) and \(AC\) have the same length.
The segments \(AB\) and \(BC\) are chords of two circular arcs with radius \(1\).
To prove that they have the same length it suffices to show that the two arcs have the same central angle.

The triangle \(BEF\) is equilateral, see .
Thus \(\widehat{DEB}=30^\circ\).
Similarly (considering the triangle \(DEC\)) we have \(\widehat{CEF}=30^\circ\).
From \[\widehat{DEB} + \widehat{BEC} + \widehat{CEF} = 90^\circ\] we deduce that \(\widehat{BEC}=30^\circ\) hence the arc \(BC\) is the twelfth of a circle.
By a similar reasoning (considering the triangle \(DEC\)) we have \(\widehat{ADC}=30^\circ\).
Thus, the arc \(AC\) is also the twelfth of a circle.