Answer: $\sqrt{\sqrt{2}-1}$.

Solution. Let $A D=C F=x, B C=C E=y$. Then, from the right triangles $A B C$, $A C E$ and $A C D$, by the Pythagorean theorem, we find $A B=\sqrt{1+y^{2}}, A E=\sqrt{1-y^{2}}, C D=\sqrt{1+x^{2}}$. The area of triangle $A B C$ is $S_{A B C}=\frac{1}{2} A C \cdot B C=\frac{1}{2} C F \cdot A B$, from which we get

$$
y=x \sqrt{1+y^{2}} \Rightarrow y^{2}=x^{2}\left(1+y^{2}\right) \quad \Rightarrow \quad y^{2}=\frac{x^{2}}{1-x^{2}}
$$

Similarly, the area of triangle $A C D$ is $S_{A C D}=\frac{1}{2} A C \cdot A D=\frac{1}{2} A E \cdot C D$, so $x=$ $\sqrt{1-y^{2}} \sqrt{1+x^{2}}$. Substituting the previously obtained expression for $y^{2}$ in terms of $x^{2}$ here, considering the condition $0<x<1$, we find

$$
\begin{gathered}
x=\frac{\sqrt{1-2 x^{2}} \sqrt{1+x^{2}}}{\sqrt{1-x^{2}}} \Rightarrow x^{2}\left(1-x^{2}\right)=\left(1-2 x^{2}\right)\left(1+x^{2}\right) \quad \Rightarrow \quad x^{2}-x^{4}=1-x^{2}-2 x^{4} \Rightarrow \\
\Rightarrow \quad x^{4}+2 x^{2}-1=0 \quad \Rightarrow \quad\left(x^{2}+1\right)^{2}=2 \quad \Rightarrow \quad x=\sqrt{\sqrt{2}-1}
\end{gathered}
$$
15.1. $[9.3$ (15 points)] In trapezoid $ABCD$, diagonal $AC$ is equal to 1 and is also its height. Perpendiculars $AE$ and $CF$ are drawn from points $A$ and $C$ to sides $CD$ and $AB$ respectively. Find $AD$, if $AD=CF$ and $BC=CE$.