How can I find the length of a segment formed by the shadow of a ladder?

Question

The problem is as follows:

Roger is an interior designer and one morning he sees that his ladder is making a shadow with respect to the floor in the backyard as indicated in the diagram from below. He's curious and found that the angles formed between the corners are the following: $\frac{\angle CAB}{2}=\frac{\angle ADC}{2}=\frac{\angle ABC}{3}=\frac{\angle ACD}{8}$. He also notices that $AB=CD$ and $AC=8$. Find the length of the segment AD which is seen by Roger.

$\begin{array}{ll} 1.&16\,\text{cm}\\ 2.&20\,\text{cm}\\ 3.&15\,\text{cm}\\ 4.&21\,\text{cm}\\ \end{array}$

I'm not sure exactly what sort of theorem or identity should be used here. I attempted different way to tackle this problem. Can someone help me?.

I'm slow at seeing things in space, so an answer which would help me the most is one which can be detailed the most as possible. This problem belongs to a section of euclidean geometry. So I appreciate that an answer could follow an euclidean geometry approach and a trigonometrical one as well so I can compare both methods and see which one is quicker. So can someone help me?.

Answer 1

attach triangle $ABC$ and $ACD$ such that $AB$ coincides with $CD$ and $AC$ coincides with $DA$. like in the picture.

look at the angles, triangle $CA'C'$ is an isosceles triangle. therefore $C'A'=CA'=8$. therefore $A'D'=8+8=16$

i wish i could give more detailed steps but i need to work, hope it is clear and do not hesitate to ask!

Answer 2

$\frac{\angle CAB}{2}=\frac{\angle ADC}{2} \Rightarrow \angle CAB=\angle ADC$. Let's call this $x$

$\frac{\angle CAB}{2}=\frac{\angle ABC}{3} \Rightarrow \frac{3 \angle CAB}{2}=\angle ABC \Rightarrow \angle ABC = \frac {3x}2$

$\frac{\angle CAB}{2}=\frac{\angle ACD}{8} \Rightarrow 4 \angle CAB=\angle ACD \Rightarrow \angle ACD=4x$

I've added that information and the information about the lengths to a diagram here:

Sine Rule applied to triangle ABC gives us $\frac{\sin (\frac {3x}{2})}{8}=\frac{\sin (180^{\circ}-\frac {5x}{2})}{AB} \Rightarrow \frac{AB}{8}=\frac{\sin (180^{\circ}-\frac {5x}{2})}{\sin (\frac {3x}{2})}$

Sine Rule applied to triangle ADC gives us $\frac{\sin x}{8}=\frac{\sin (180^{\circ}-5x)}{AB} \Rightarrow \frac{AB}{8}=\frac{\sin (180^{\circ}-5x)}{\sin x}$

Equating these and using the fact that $\sin (180^{\circ}-\theta)=\sin \theta$ gives us:

$\frac{\sin (\frac {5x}{2})}{\sin (\frac {3x}{2})}=\frac{\sin 5x}{\sin x}$

Let $\frac x2=y$ to give:

$\frac{\sin (5y)}{\sin (3y)}=\frac{\sin (10y)}{\sin (2y)}$

Maybe that's a way forward?