Find the ratio of lengths in a triangle with cevians, without using mass-points

Question

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!

Answer 1

By Menelaus' theorem in $\triangle ABD$ and transversal $C-E-F\,$: $\;\;\cfrac{AF}{FB} \cdot \cfrac{BC}{CD} \cdot \cfrac{DE}{EA} = -1$

Answer 2

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 $\vec{CE}$.

$$\begin{align*} \vec{CF} &= k\vec{CE}\\ \vec{CA} + \vec{AF} &= k\vec{CE}\\ \mathbf a - \mathbf c + \frac1{1+p}(\mathbf b-\mathbf a) &= \frac k2\mathbf a + \frac k4b - \frac{3k}4\mathbf c\\ \frac p{1+p}\mathbf a + \frac1{1+p}\mathbf b-\mathbf c &= \frac k2\mathbf a + \frac k4b - \frac {3k}4\mathbf c\\ p &= 2\\ AF:FB &= 1:2 \end{align*}$$

Answer 3

Let the area of $\triangle ABC$ be $4a$.

Since $BD : DC = 1:1$, $$\text{Area of }\triangle ABD : \text{Area of }\triangle ADC = 1:1\\ \text{Area of }\triangle ABD = \text{Area of }\triangle ADC = 2a$$

Since $AE:ED : 1:1$,

$$\text{Area of }\triangle AEC:\text{Area of }\triangle EDC = 1:1\\ \text{Area of }\triangle AEC = \text{Area of }\triangle EDC = a$$

Since $BD:DC = 1:1$, $$\text{Area of }\triangle EBD : \text{Area of }\triangle EDC = 1:1\\ \text{Area of }\triangle EBD = a$$

Consider the ratio $AF:FB$, $$\begin{align*} AF:FB &= \text{Area of }\triangle AFC : \text{Area of }\triangle FBC\\ &= \text{Area of }\triangle AFE : \text{Area of }\triangle FBE\\ &= (\text{Area of }\triangle AFC - \text{Area of }\triangle AFE):(\text{Area of }\triangle FBC-\text{Area of }\triangle FBE)\\ &= \text{Area of }\triangle AEC : \text{Area of }\triangle EBC\\ &= \text{Area of }\triangle AEC : (\text{Area of }\triangle EBD+\text{Area of }\triangle EBC)\\ &= 1:2 \end{align*}$$