Three, (16 points) Given that $\triangle X Y Z$ is an isosceles right triangle with legs of length 1 $\left(\angle Z=90^{\circ}\right)$, its three vertices lie on the three sides of isosceles right $\triangle A B C\left(\angle C=90^{\circ}\right)$. Find the maximum possible length of the legs of $\triangle A B C$.
(1) As shown in Figure 4, vertex $Z$ is on the hypotenuse $A B$.
Take the midpoint $M$ of $X Y$, and connect $C M$, $Z M$, and $C Z$, and draw the perpendicular $C N$ from $C$ to side $A B$, then
$$
\begin{array}{l}
C Z \leqslant C M + M Z \\
= \frac{1}{2} X Y + \frac{1}{2} X Y \\
= X Y = \sqrt{2} .
\end{array}
$$

Also, $C N \leqslant C Z$, so $C N \leqslant \sqrt{2}$, and $C A = \sqrt{2} C N \leqslant 2$.
(2) As shown in Figure 5, vertex $Z$ is on the right-angle side $C A$ (or $C B$). By symmetry, assume $Z$ is on $C A$. Let $C X = x$, $C Z = y$, and draw $Y H \perp C A$ at $H$.
It is easy to prove that $\triangle Z Y H \cong \triangle X Z C$, thus
$$
H Z = C X = x, H Y = C Z = y .
$$

Clearly, $\triangle A H Y$ is an isosceles right triangle, so $A H = y$. Let $A C = b$, then $2 y + x = b$, i.e., $x = b - 2 y$.
In $\triangle C X Z$, by the Pythagorean theorem, we have
$$
y^{2} + (b - 2 y)^{2} = 1^{2} \text {, }
$$

which simplifies to $5 y^{2} - 4 b y + b^{2} - 1 = 0$.
Since $y$ is a real number, then
$$
\Delta = 16 b^{2} - 20 (b^{2} - 1) = 20 - 4 b^{2} \geqslant 0, b \leqslant \sqrt{5} \text {. }
$$

When $b = \sqrt{5}$, $y = \frac{2 \sqrt{5}}{5}$, and $x = \frac{\sqrt{5}}{5}$.
Combining (1) and (2), the maximum value of $b$ is $\sqrt{5}$.