5. Answer: (A).
Let $x=\frac{D E}{A E}$. Then $x=\frac{E F}{E B}$. Note that
$$
1=\frac{D C}{A B}=\frac{D F}{A B}+\frac{F C}{A B}=\frac{E F}{E B}+\frac{F G}{G B}=\frac{E F}{E B}+\frac{E B-2 E F}{E F}=x+\left(\frac{1}{x}-2\right)
$$

Then $x+\frac{1}{x}-3=0$, and thus $x=\frac{3-\sqrt{5}}{2}(0<x<1)$.
5. The diagram below shows that $A B C D$ is a parallelogram and both $A E$ and $B E$ are straight lines. Let $F$ and $G$ be the intersections of $B E$ with $C D$ and $A C$ respectively. Given that $B G=E F$, find the quotient $D E A E$.
(A) $\frac{3-\sqrt{5}}{2}$;
(B) $\frac{3-\sqrt{6}}{2}$;
(C) $\frac{3-\sqrt{6}}{4}$;
(D) $\frac{3-\sqrt{7}}{2}$;
(E) $\frac{3-\sqrt{7}}{4}$.