Picture of this three pentagons $\triangle FCB$ is equilateral; $F(0,0,0),C(0,1,0),B(-\frac{\sqrt{3}}{2},\frac{1}{2},0)$. $G(x_{G},y_{G},z_{G}); \angle GFC = 108^{o}; y_{G}= \cos 108^{o}$. $G'(x_{G},y_{G},0)$ lies on the bisector of $\angle BFC$ with equation $ y=-\sqrt{3}x \rightarrow x_{G}=-\frac{\cos 108^{o}}{\sqrt{3}}$. $x^{2}_{G}+y^{2}_{G}+z^{2}_{G}=1 \rightarrow z_{G}=\sqrt{1-\frac{4\cos^{2} 108^{o}}{3}}$ Equation of the plane through $F,C$ and $G : x \cdot \sqrt{1-\frac{4\cos^{2} 108^{o}}{3}} + z \cdot \frac{\cos 108^{o}}{\sqrt{3}}=0$. Normal of the $xy-$plane: $(0,0,1)$; normal of the $FCG-$plane:$(\sqrt{1-\frac{4\cos^{2} 108^{o}}{3}},0,\frac{\cos 108^{o}}{\sqrt{3}})$. Angle $\theta$ between both planes: \[\cos \theta = \left|\frac{\frac{\cos 108^{o}}{\sqrt{3}}}{\sqrt{1-\frac{4\cos^{2} 108^{o}}{3}+\frac{\cos^{2} 108^{o}}{3}}}\right|=\left|\frac{1}{\sqrt{3}\tan 108^{o}}\right|=\frac{\tan 18^{o}}{\sqrt{3}}\] Midpoint $M$ of $\left[FC\right]: MI = \frac{1}{2} \cot 18^{o}$. $II' \bot xy-$plane, $I'\in xy-$plane: $MI' = MI \cdot \cos \theta = \frac{\sqrt{3}}{6}$. $O$ centre of the circumcircle of $\triangle FCB : MO = \frac{\sqrt{3}}{6}$. Circumcircle of $\triangle HIE$ with radius $R=OI' = \frac{\sqrt{3}}{3} \rightarrow \triangle HIE$ equilateral with side $HI=IE=EH=1$.

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the quadrilaterals $ HFCE,IFBE,IHBC $ are parallelograms because have a pair of opposite sides parallel and congruent. from this we obtain immediately that $ EH=IH=HE=1 $.