Stuck on a Geometry Problem 
$ABCD$ is a square, $E$ is a midpoint of side $BC$, points $F$ and $G$ are on the diagonal $AC$ so that $|AF|=3\ \text{cm}$, $|GC|=4\ \text{cm}$ and $\angle{FEG}=45 ^{\circ}$.  Determine the length of the segment $FG$.

How can I approach this problem, preferably without trigonometry?


I do not like it. This is not a geometry problem; it is indeed a puzzle where you have to find this coincidence, which occurs only if $AF:GC=3:4$.

If we can find/prove that $GDF\angle=45^\circ$ in some different way, then we can define the point $P$ that as the reflection of $C$ about $DE$ and the reflection of $A$ about $DF$. Due to $FPD\angle=DPG\angle=45^\circ$, the triangle $FGP$ is right-angled and its legs are $PF=AF=3$, $PG=CG=4$, so $FG=5$.
