Sorry for the picture. I really could not draw this in Geogebra. I want to find the length of $AC$. I assumed that $C$, $D$ and $E$ are collinear and solved it trivially using similar triangles. However, is the information sufficient to prove that $C$, $D$ and $E$ are collinear? If I assume they are collinear then I get $AC=16/3$.
For example, the diagram may look like this:
On
The points $C$, $D$, $E$ are of course not collinear.
We are looking for $a:=|AC|$. Project $D$ orthogonally to $D'\in A\vee B$ and to $D''\in A\vee C$. Then $$|AD''|=|D'D|=\sqrt{5^2-3^2}=4\ .$$ Let $\angle(ACD)=:\alpha$. Then $\angle(ACB)=2\alpha$. From the figure we obtain $$\tan\alpha={1\over a-4},\qquad\tan(2\alpha)={8\over a}\ ,$$ which leads to a quadratic equation for $a$ with solutions ${10\over3}$ and $6$. As ${10\over3}<4$ we necessarily have $a=6$.
In fact $C,D,E$ can never be collinear.
The reason is that $CD$ is an angle bisector and $CE$ is a median, they will only coincide when triangle $ABC$ is isosceles where $CA=CB$. However since $\angle CAB=90$ degrees, $CA$ and $CB$ cannot be equal.
To solve your problem:
First $AH=4$ by Pythagorean on the blue triangle. Next $HB=4\sqrt{5}$ by Pythagorean on triangle $HAB$.
Notice the two green triangles are similar with ratio $1:4$ so $B,H,G$ are collinear, further more $HG:GB=1:4$. Therefore $GB={16\over 5}\sqrt{5}$.
Now since $FB^2=BG\times BH=64$, we know $FB=8$.
Let $CH=CF=x$.
We know $${1\over 2}(x+4)\cdot 8 = S_{ABC}=S_{CDA}+S_{CDB}+S_{ADB}={1\over 2}(x+4)\cdot 1 + {1\over 2}(x+8)\cdot 1 + {1\over 2}\cdot4\cdot8$$ where $S$ denotes area.
Simplify, $$8x+32=2x+12+32, x=2$$
Therefore $AC=6$