What´s the length of the segment AC in the triangle below?

Question

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$

...????

Answer 1

Following on from $m^2+2m-2\sqrt{m+1}+1=0$, you already have $b=\sqrt{m+1}$, so substitute this in to get $b^4-2b=0 \implies b=\sqrt[3]2$.

Answer 2

$\triangle ABF\sim \triangle AFC$

We have:

$$\frac {c+m}{c+AD}=\frac {c+AD}{c}$$

So:

$$(c+AD)^2=c^2+cm\Rightarrow AD=\sqrt{c^2+mc}-c$$

Therefore:

$$AC=c+AD=\sqrt{c^2+mc}$$

Answer 3

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