10.2.7 Find all positive pairs $a, b$ that satisfy: there exists a right triangle $\triangle CDE$ and points $A$, $B$ on its hypotenuse $DE$, such that $\overrightarrow{DA}=\overrightarrow{AB}=\overrightarrow{BE}$, and $AC=a$, $BC=b$.
Let $\overrightarrow{C A}=x, \overrightarrow{C B}=y$. Then $\overrightarrow{D A}=\overrightarrow{B E}=\overrightarrow{A B}-y-x, \overrightarrow{C E}=2 y-x, \overrightarrow{C D}=2 x-y$, and the condition $C D \perp C E$ is equivalent to $(2 \boldsymbol{x}-\boldsymbol{y}) \cdot(2 \boldsymbol{y}-\boldsymbol{x})=0$, which simplifies to $5(\boldsymbol{x}-\boldsymbol{y})^{2}=\boldsymbol{x}^{2}+\boldsymbol{y}^{2}$. The latter condition indicates that the sides of $\triangle A B C$, $B C=b, A C=a, A B=c$, should satisfy $5 c^{2}=a^{2}+b^{2}$. Such a triangle exists if and only if the values of $a$ and $b$ satisfy the triangle inequality
$$
|a-b|<c<a+b
$$
for some $c>0$, and the left inequality is equivalent to $(2 a-b)(2 b-a)>0$. Therefore, the answer is $\frac{1}{2}<\frac{a}{b}<2$.