It is clear that $(CDEU)$ is cyclic so let $P=(CDE)\cap AB$ other than $U$. So now we have, $\angle APD=\angle UCD=\angle UAC=\angle PAD$. So $DA=DP$ and therefore $AF=FP$. Similarly $BG=GP$. From this, $AB=AF+AP+GP+BG=2(FP+PG)=2FG$. so we are done. *Note that if $U=P$, we have, $\angle APD=\angle PCA=\angle PAC=\angle PAD$ so we get $DA=DP$ again and the rest of the solution goes in the same way.

Let's Complex Bash! Set $a$, $b$, $c$ on the unit circle. Furthermore let $a=-1$, $b=1$ Now to deal with the conditions: $a,c,e$ are collinear corresponds to the shoelace determinant being = 0: \[ 0 = \begin{vmatrix} a & \overline{a} & 1 \\ c & \overline{c} & 1 \\ e & \overline{e} & 1 \\ \end{vmatrix} = \begin{vmatrix} -1 & -1 & 1 \\ c & \frac{1}{c} & 1 \\ e & \overline{e} & 1 \\ \end{vmatrix} = \begin{vmatrix} 0 & 0 & 1 \\ c+1 & \frac{1}{c}+1 & 1 \\ e+1 & \overline{e}+1 & 1 \\ \end{vmatrix} = (e+1)\left(\frac{1}{c}+1\right) - (\overline{e}+1)(c+1) \]This implies: \[ \overline{e} = \frac{e+1}{c}+1 \]and analagously: \[ \overline{d} = \frac{1-d}{c} + 1 \] Now note that the points $f$, $g$ are the real parts of $d$ and $e$: \[ f = \frac{d+\overline{d}}{2} = \frac{(c-1)(d+1)}{2c} + \frac{1}{c} \] \[ g = \frac{e+\overline{e}}{2} = \frac{(c+1)(e+1)}{2c} - 1\] Now lastly we want to prove that $FG = \frac{1}{2}AB = \frac{1}{2}2 = 1$: \[ f - g = 1 \]\[ \frac{(c-1)(d+1)}{2c} + \frac{1}{c} - \frac{(c+1)(e+1)}{2c} + 1 = 1 \]\[ (c-1)(d+1) + 2 = (c+1)(e+1) \]\[ cd = d + e + ec \]However, this follows from the perpendicularity criterion: \[ DU \perp UG \Longleftrightarrow \frac{d-u}{e-u} = -\overline{\frac{d-u}{e-u}} \implies \frac{d}{u} = \frac{\frac{d-1-c}{2c}}{\frac{e+c}{c}-c} \implies dc = d+e+ec\]