Can congruence or similarity be used to find the length of a triangle when some of its sides have odds proportionality?
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 $AH = 13 \>\text{feet}$.