In my book there is this problem but I don't know if congruence or similarity can be used as means to solve it.
The problem is as follows:
A CTV broadcasting tower is resting atop a square flat base in the highest hill of Taipei. The tower is supported by five cables which are held tight to the ground by four retention screws labeled $\textrm{A}$, $\textrm{C}$, $\textrm{D}$, $\textrm{E}$, $\textrm{F}$ as indicated in the picture. The radio technician notices that the length between $\textrm{HC = 3 feet}$, meanwhile $3AB=5BC$ and $m\angle ABH= 3m \angle HBC$ From this information find the distance between $\textrm{H}$ and $\textrm{A}$.
The alternatives given on my book are:
$\begin{array}{ll} 1.&\textrm{13 feet}\\ 2.&\textrm{11 feet}\\ 3.&\textrm{9 feet}\\ 4.&\textrm{10 feet}\\ \end{array}$
What I attempted to do is shown in the figure from below.
But other than just putting what is stated in the problem I could not reach any further in my observations. I tried hardly to imagine different ways to arrange the triangles in a manner that they could be used as congruence (from which I noticed one). But other than that I coudn't find more.
Therefore can somebody help me to find the answer using mainly simple euclidean geometry?. i.e using constructions and tying it with similarity or congruence or something like this? This is what basically gets me stuck.
Although I welcome an answer using plane trigonometry this question was supposedly been able to be solved without requiring trig or any advanced math tools, hence I hope somebody could help me with a trig free approach.
Draw the line BD such that $\angle DBH = \angle CBH$. Then, $DH = HC = 3$ feet. Apply the sine rule to triangles BAD and BDC,
$$\frac{\sin 2\alpha}{\sin(180-\beta)} = \frac{AH - 3}{AB},\>\>\>\>\> \frac{\sin 2\alpha}{\sin\beta} = \frac{6}{BC}$$
Use the fact $\sin(180-\beta)=\sin\beta$ to get $$\frac{AH-3}{6}=\frac{AB}{BC}=\frac 53$$
which yields $AH = 13 \>\text{feet}$.