One. (20 points) As shown in Figure 2, in the isosceles right triangle $\triangle ABC$, $\angle C=90^{\circ}$, points $D$ and $E$ are on side $BC$, and point $F$ is on the extension of $AC$, such that $BE=ED=CF$. Find the tangent value of $\angle CEF + \angle CAD$.

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$\begin{array}{l}\text { I. Draw } D G \perp A B \text { at point } G . \\ \text { Let } A C=B C=x, \\ B E=E D=C F=y\left(0<y<\frac{x}{2}\right) . \\ \text { Then } G D=\sqrt{2} y \Rightarrow A G=\sqrt{2}(x-y)=\sqrt{2} C E \text {. } \\ \text { Therefore, } \mathrm{Rt} \triangle E C F \backsim \mathrm{Rt} \triangle A G D \\ \Rightarrow \angle C E F=\angle G A D \\ \Rightarrow \angle C E F+\angle C A D \\ =\angle G A D+\angle C A D=45^{\circ} \\ \Rightarrow \tan (\angle C E F+\angle C A D)=1 .\end{array}$