I am trying to help my brother who's taken up Math as a hobby with a problem he can't solve, but I'm stuck on it myself.
It goes like this:
We take a point D on the side AB of an ABC triangle such that AD:DB = 3:5, and a point E on the side CD such that CE:ED = 2:3. If AE crosses BC at the point F, what is the relation BF:FC.
The answer is 5:12, but I can't figure out how to use Thales' theorem to get it.
Hint: Menelaus' theorem for $\,\triangle BCD\,$ and transversal $\,AF\,$ gives $\,\dfrac{BF}{FC} \cdot \dfrac{CE}{ED} \cdot \dfrac{DA}{AB} = 1\,$.
You know $\,\dfrac{CE}{ED}=\dfrac{2}{3}\,$ and $\,\dfrac{DA}{AB} = \ldots\,$