4. In $\triangle A B C$, $\angle C=90^{\circ}, A C=\frac{1}{2} B C$
$\angle A$ bisector $A D$ intersects $C B$ at $D$, and a point $G$ is taken on $D B$ such that $D G=C D$, then $C G: G B=$ $\qquad$
4. $\frac{\sqrt{ } \overline{5}+1}{2}$.