Please ignore the pencilled 4m in the diagram but I really need to know what the length of the bottom line - line DC - is. A procedure or tips on how to calculate this would be useful. Also, is the pencil line from point C dividing right angle ACD exactly in half? I figured this would be useful to calculate the solution to this problem but I doubt that it is dividing right angle ACD exactly in half. Please use correct rules and explain all working and reasons for this.
Developed hints:
Put $\;\angle BDC=x\;$ , and now
$$\begin{cases}I&\text{On triangle}\;\Delta ABC:&\;\;\frac{\tan 35^\circ+\tan x}{1-\tan35^\circ\cdot\tan x}=\tan(35+x)=\frac4{CD}\\{}\\II&\text{On triangle}\;\;\Delta BCD:&\;\;\tan x=\frac1{CD}\end{cases}$$
Now develope I using II:
$$\frac4{CD}=\frac{\tan35^\circ+\frac1{CD}}{1-\tan35^\circ\cdot\frac1{CD}}$$
If you want now put $\;y:=CD\;,\;\;a:=\tan35^\circ\;$ to make things simpler, so that you have to solve the equation
$$\frac4y=\frac{a+\frac1y}{1-a\frac1y}=\frac{ay+1}{y-a}\iff4y-4a=ay^2+y\iff ay^2-3y+4a=0$$
Now solve the above quadratic and that's all...