解析：
解答】解: $\because \mathrm{Rt} \triangle A B C$ 绕直角边 $A C$ 的中点 $H$ 旋转, 得到 $\triangle E F D$,
$$
\begin{aligned}
& \therefore \mathrm{AH}=\mathrm{EH}=\mathrm{CH}=\mathrm{DH}=\frac{1}{2} \mathrm{AC}, \angle A=\angle E, \\
& \therefore \angle A=\angle A D H, \\
& \therefore \angle G H E=\angle A+\angle A D H=2 \angle A, \\
& \because \triangle E G H \text { 恰好是以 } G H \text { 为底边的等腰三角形, } \\
& \therefore \angle E G H=\angle E H G=2 \angle A, \\
& \because \angle E G H+\angle E H G+\angle E=180^{\circ}, \\
& \therefore 2 \angle A+2 \angle A+\angle A=180^{\circ}, \\
& \therefore \angle A=36^{\circ} .
\end{aligned}
$$
故答案为: $36^{\circ}$.