What´s the length of the segment AC in the triangle below? For reference: In the triangle $\angle A$ is right and $D$ is a point on the side $AC$ such that the segments $BD$ and $DC$ have length equal to $1 m$. Let $F$ be the point on the side $BC$ so that $AF$ is perpendicular to $BC$. If the segment $FC$ measures $1m$, determine the length of $AC$.(Answer$:\sqrt[3]{2})$

My progress:
$ AF = h\\
a=m+1\\
h^2 =m.FC = m.1 = m\\
\triangle AFC: AC^2 =b= h^2+FC^2 \implies b = \sqrt{m+1}\\
\triangle ABC: a^2 = b^2+ c^2 \implies (m+1)^2=c^2+(\sqrt{m+1})^2\\
\therefore c = \sqrt{m^2+m}\\
\triangle BAD: BD^2=AD^2+c^2\implies 1 = (\sqrt{m+1}-1)^2+m^2+m\\
\therefore: m^2+2m-2\sqrt{m+1}+1=0$
...????


Let $E$ middle of $BC$. Then triangles DEC, AFC and BAC are similar. Then $$\frac{CD}{EC}=\frac{AC}{FC}=\frac{BC}{AC}=k$$
$$EC=\frac{CD}{k}=\frac{1}{k}, AC=k FC=k, BC=k AC=k^2$$
$$BC=2 EC \Rightarrow k^2 = \frac{2}{k} \Rightarrow k=\sqrt[3]{2}$$
$$AC=k=\sqrt[3]{2}$$
