I called each of the four small angles at C, also $$\alpha$$
$$\boxed {
\cos{(\alpha)} = \frac{h}{b} = \frac{h}{d} \quad | \quad h=\overline{EC} \quad \mbox{and} \quad d = \overline{FC}
}
\Rightarrow \boxed{d=b}$$
$$Areas: DCB = DCA = \frac{ \frac{c}{2} *h }{ 2 }$$
I. Areas: ACF + FCD = DCB
$$b^2\sin{(2\alpha)} +be\sin{(\alpha)} =ae\sin{(\alpha)}$$
II. Areas: FCD + DCB = FCB
$$be\sin{(\alpha)} +ae\sin{(\alpha)} =ab\sin{(2\alpha)}$$
I.+II.:
$$\begin{array}{rcl}
b^2\sin{(2\alpha)} +2be\sin{(\alpha)} &=&ab\sin{(2\alpha)} \quad | \quad :b\\
b\sin{(2\alpha)} +2e\sin{(\alpha)} &=& a\sin{(2\alpha)}
\end{array}$$
$$\Rightarrow
\boxed{
\frac { 2e\sin{( \alpha) } }
{ \sin{(2\alpha) } }
=a-b \qquad (1)
}$$
I.-II.:
$$\begin{array}{rcl}
b^2\sin{(2\alpha)} - ea\sin{(\alpha)} &=&ea\sin{(\alpha)} -ab\sin{(2\alpha)} \\
2ea\sin{(\alpha)} &=& ab\sin{(2\alpha)} + b^2\sin{(2\alpha)}
\end{array}$$
$$\Rightarrow
\boxed{
\frac { 2e\sin{( \alpha) } }
{ \sin{(2\alpha) } }
=\frac{ ab+b^2 } { a } \qquad (2)
}$$
(1)=(2):
$$a-b =\frac{ ab+b^2 } { a }$$
$$\Rightarrow
\boxed{
a^2 -2b*a -b^2 = 0
}$$
$$a_{1,2}={ 2b\pm\sqrt{4b^2-4*1*(-b^2)}\over2*1 }=b(1\pm\sqrt{2}) \quad | \quad \mbox{ b not negativ!}$$
$$\textcolor[rgb]{1,0,0}{
\boxed{
\textcolor[rgb]{0,0,0}{
a= b (1+\sqrt{2})
}
}
}$$
III. Areas: ACF + FCB = ACB
$$b^2\sin{(2\alpha)} + ba\sin{(2\alpha)} = ba\sin{(4\alpha)} \quad | \quad a=b(1+\sqrt{2})$$
$$b^2\sin{(2\alpha)} + b^2(1+\sqrt{2})\sin{(2\alpha)} = b^2(1+\sqrt{2})\sin{(4\alpha)} \quad | \quad :b^2$$
$$\sin{(2\alpha)} + (1+\sqrt{2})\sin{(2\alpha)} = (1+\sqrt{2})\sin{(4\alpha)}$$
$$(2+\sqrt{2})\sin{(2\alpha)} = (1+\sqrt{2})\sin{(4\alpha)} \quad | \quad \leftrightarrow$$
$$(1+\sqrt{2})\sin{(4\alpha)} = (2+\sqrt{2})\sin{(2\alpha)} \quad | \quad \sin{(4\alpha)}=2\sin{(2\alpha)}\cos{2\alpha}$$
$$(1+\sqrt{2})2\sin{(2\alpha)}\cos{2\alpha}= (2+\sqrt{2})\sin{(2\alpha)} \quad | \quad : \sin{(2\alpha)}$$
$$(1+\sqrt{2})2\cos{2\alpha}= (2+\sqrt{2})$$
$$(2+2\sqrt{2})\cos{2\alpha}= (2+\sqrt{2})$$
$$\cos{2\alpha}=\frac{ (2+\sqrt{2}) }{ (2+2\sqrt{2}) } \quad | \quad *\frac{ \frac{\sqrt{2}}{2} }{ \frac{\sqrt{2}}{2} }$$
$$\cos{2\alpha}=\frac{ (2+\sqrt{2})\frac{\sqrt{2}}{2} }{ (\sqrt{2}+2) } = \frac{ \sqrt{2} }{ 2 } \quad | \quad \pm \cos^{-1}()$$
$$2\alpha= \pm \cos^{-1}( \frac{ \sqrt{2} } { 2 } ) \quad | \quad \textcolor[rgb]{1,0,0}{+} solution!$$
$$2\alpha= \textcolor[rgb]{1,0,0}{+} \cos^{-1}( \frac{ \sqrt{2} } { 2 } ) =45\ensurement{^{\circ}} \quad | \quad *2$$
$$\textcolor[rgb]{1,0,0}{
\boxed{
\textcolor[rgb]{0,0,0}{
4\alpha= 90\ensurement{^{\circ}}
}
}
}$$
.
].
