In the following figure, $D$ is a point on $BC$ such that $\angle ABD = \angle CAD$ and $\frac{BD}{AC}=\frac{8}{3}$. If $\frac{\text{Area of }\Delta ABD}{\text{Area of }\Delta ADC} =k$, find the value of $k$.
So what I need to find is just $\frac{BD}{DC}$ or $\frac{AB \cdot BD}{AD \cdot AC}$ but either way I cannot see how I should proceed. Any hints? Thanks in advance.
Using $\triangle ABC \sim \triangle CAD$, we have $$\frac{8+DC}{3}=\frac{3}{DC}$$, so $DC=1$, which shows that $k = 8$