In the triangle ABC, D and E are points of trisection of segment AB; F is the midpoint of segment AC. What is the ratio: MN/BF

Question

This is a euclidean geometry problem. No angles measures are given. There are no right angles given. DE/AB = 1/3; AF = FC. I have tried countless extensions and constructions betond what is shown to find something to prove a ratio of MN to anything. I have been stumped.

Answer 1

The given data are not enough to determine $\frac{MN}{BF}$. See the picture:

We need more information, for example the ratio $\frac{BD}{BA}$. If we assume that $D,E$ trisect $AB$, then we can proceed as follows.

$EF\parallel CD$ implies $$BM=\frac{BF}{2},EF=\frac{CD}{2},DM=\frac{EF}{2}.$$ From that, we get $EF:CM=2:3$ and $$\frac{MN}{NF+MN}=\frac{CM}{EF+CM}=\frac{3}{5}$$ That means $$\frac{MN}{BF}=\frac{MN}{2MF}=\frac3{10}.$$

Answer 2

The problem is invariant under affine transformations, so choose any coordinates, e.g. like this:

\begin{align*} A &= (0, 6) & D &= (-4, 2) \\ B &= (-6, 0) & E &= (-2, 4) \\ C &= (0, 0) & M &= (-3, 3/2) \\ F &= (0, 3) & N &= (-6/5, 12/5) \end{align*}

Answer 3

Menelaus Theorem is probably the simplest tool to solve this problem.

First you can apply it in $\Delta ABF$ taking CE as a transversal as:$$\frac{CF}{CA}*\frac{DA}{DB}*\frac{NB}{NF}=1$$

Then you can apply it in $\Delta ABF$ taking CD as a transversal as:$$\frac{CF}{CA}*\frac{EA}{EB}*\frac{MB}{MF}=1$$

Above equations give the values of $\frac{MF}{MB}$ and $\frac{NF}{NB}$ as 4 and 1 which easily give the value of $\frac{MN}{FB}$ as $\frac{3}{10}$

A proof of Menelaus theorem can easily be shown using the Law of Sines.

Answer 4

Let $ABC$ be given with sides $a=11$, $b=8$, and $c=8$. Assume that $D$ and $E$ are on side $BC$ such that $AD$, $AE$ trisect $BAC$. Show that $AD=AE=6$