1. As shown in Figure 1, in $\triangle A B C$, $A B=A C, G$ is the centroid, $D$ and $F$ are the midpoints of $B C$ and $A B$ respectively, $C D=D G=a, G E \perp A B$. Then $A E=$ . $\qquad$
$$
\text { II. } 1 . \frac{3 \sqrt{10} a}{5} \text {. }
$$

Since $G$ is the centroid of $\triangle A B C$, we have
$$
\begin{array}{l}
A D=3 D G=3 D B=3 a, \\
A B=\sqrt{10} a, A G=2 D G=2 a, \\
\cos \angle D A B=\frac{3}{\sqrt{10}} .
\end{array}
$$

Therefore, $A E=A G \cos \angle D A B=\frac{3 \sqrt{10} a}{5}$.