<s>[INST] 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?

 [/INST] 
Let $H$ be the midpoint of $AC$ and $\angle EIC= 90^{\circ}$. We can observe that $$FH+3=HG+4,\quad FH+HG=x.$$ So we obtain $HG=\frac{x-1}2$. Since two corresponding angles are congruent; $\angle FEG =\angle EHG=45^{\circ}$ and $\angle EGF=\angle HGE$, we have that $\triangle FEG$ and $\triangle EHG$ are similar to each other. This gives $$FG:EG=EG:HG\implies EG^2 = FG\cdot HG=\frac{x(x-1)}2.$$ Now, note that $EI=\frac14 AC=\frac{x+7}4$ and $IG=IH-GH=\frac{x+7}{4}-\frac{x-1}2=\frac{9-x}{4}$. Since $\triangle EIG$ is a right triangle, by Pythagorean theorem, we find that
$$
EG^2=\frac{x(x-1)}{2}=EI^2+IG^2=\frac{(x+7)^2}{16}+\frac{(9-x)^2}{16},
$$ which implies $x=5  $ or $x=-\frac{13}3$. Since $x>0$, we get $x=5$.
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