As $CD$ is perpendicular to $AF$ and the perpendicular to $CD$ through $A$ is $AC$, $F$ must lie on $AC$. Then is $\angle ADF=180^\circ-\angle FDE=90^\circ$ (as $A,D,E$ are collinear) and hence $\angle ADC=\angle CDF=45^\circ$ and also $\angle DCA=\angle FCD=90^\circ$, so $ACD$ and $CDF$ are congruent. Hence, through right triangle $CDF$ we get $|FD|^2=|AC|^2+|CD|^2=2|AB|^2$, so $|DE|=|FD|=\sqrt{2}|AB|$. In triangle $AEG$, we now get $(|EG|+|EG|)^2+|EG|^2=1$ and hence $5|EG|^2=1$, so $A_{DEGF}=|EG|^2=\boxed{\frac{1}{5}}$.