Do you know how use pythagoras?

$\triangle AEC \sim \triangle BAC \rightarrow \frac{AC}{BC}=\frac{EC}{AC} \therefore AC^2=BC$

As $\triangle ABC$ is rectangle

$AB^2 +(AD+1)^2=BC^2$

Substitute

$AB^2 +(AD+1)^2=(AC^2)^2=AC^4$

$(AB^2 +AD^2+2AD+1=AC^4$

Now, by pythagoras in $\triangle BAD, BD^2+AD^2=1$

Substitute in $(AB^2 +AD^2+2AD+1=AC^4$

$1+1+2AD=AC^4$

$2+2AD=AC^4$

$2(1+AD)=AC^4$ but $AD+1=AC$

$2(AC)=AC^4$

$2=AC^3$

$AC=\sqrt[3]{2}$

Let $\angle BCD$ be $x$ . Since $\triangle BCD$ is an isosceles triangle $(BD=DC)$ , $\angle CBD$ is also $x$ .

$\angle ADB=\angle BCD+\angle CBD=x+x=2x$

$cos \angle ADB=\dfrac{k}{1} \leftrightarrow \color{#3D99F6}{k=cos \angle ADB=cos 2x \text{ } ^{*}}$

$\begin{aligned} cos \angle BCD&=\dfrac{1}{1+\color{#3D99F6}{k}} \\ cosx&=\dfrac{1}{1+ \color{#3D99F6}{cos 2x}} =\dfrac{1}{1+ 2cos^{2}x-1} = \dfrac{1}{2cos^{2}x} \\ cos^{3} x&=\dfrac{1}{2} \\ \color{#D61F06}{cos x}&=\color{#D61F06}{2 ^{-\frac{1}{3}}} \end{aligned}$

From $\color{#3D99F6}{^*}$ , we have

$\begin{aligned} \color{#3D99F6}{k}&=\color{#3D99F6}{cos 2x} \color{#333333}{ = 2} \times \color{#D61F06}{cos^{\color{#333333}{2}}x} \color{#333333}{-1} \\ k+1&=2 \color{#D61F06}{cos}^{\color{#333333}{2}} \color{#D61F06}{x} \color{#333333}{=2 \times} \color{#D61F06}{(2 ^{-\frac{1}{3}} )^{\color{#333333}{2}}} \\ k+1&=2^{\frac{1}{3}}\end{aligned}$

We know that $AC=k+1$ , so $AC=\boxed{2^{\frac{1}{3}}}$