<s>[INST] Find the ratio of lengths in a triangle with cevians, without using mass-points In $\triangle ABC$, $D$ is the midpoint of $BC$ and $E$ is the midpoint of $AD$. $CE$ is extended to meet $AB$ at $F$.
The question is to find the ratio of the lengths of $AF$ to $FB$.
This is trivial using mass-points geometry:
$$C=1\to B=1\to D=2\to A=2\to E=4\to F=3$$
and thus $AF:FB=1:2$.
But how would one solve this without using mass-point geometry?
Thanks!
 [/INST] Let the points $A,B,C$ corresponds to vectors $\mathbf a,\mathbf b,\mathbf c$.
The vector $\vec{AD}$ is
$$\begin{align*}
\vec{AB} + \vec{BD}
&= \vec{AB} + \frac12\vec{BC}\\
&= \mathbf b - \mathbf a+\frac12(\mathbf c-\mathbf b)\\
&= -\mathbf a+\frac12\mathbf b + \frac12\mathbf c
\end{align*}$$
The vector $\vec{AE}$ is half of $\vec{AD}$,
$$\vec{AE} =-\frac12\mathbf a+\frac14\mathbf b +\frac14\mathbf c$$
The vector $\vec{CE}$ is
$$\begin{align*}\vec{CA} + \vec{AE} &=\mathbf a-\mathbf c-\frac12\mathbf a+\frac14\mathbf b + \frac14\mathbf c\\
&= \frac12\mathbf a + \frac14\mathbf b -\frac34\mathbf c
\end{align*}$$
Let $AF:FB = 1:p$. Then $\vec{AF} = \frac{1}{1+p}(\mathbf b-\mathbf a)$. Since $CEF$ is a straight line, $\vec{CF}$ has the same direction as $\vec{CE}$.
$$\begin{align*}
\vec{CF} &= k\vec{CE}\\
\vec{CA} + \vec{AF} &= k\vec{CE}\\
\mathbf a - \mathbf c + \frac1{1+p}(\mathbf b-\mathbf a) &= \frac k2\mathbf a + \frac k4b - \frac{3k}4\mathbf c\\
\frac p{1+p}\mathbf a + \frac1{1+p}\mathbf b-\mathbf c &= \frac k2\mathbf a + \frac k4b - \frac {3k}4\mathbf c\\
p &= 2\\
AF:FB &= 1:2
\end{align*}$$
</s>