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 
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.
 A: 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
$$\frac{MN}{BF}=\frac{MN}{2MF}=\frac3{10}.$$
A: 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*}
A: 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.
A: 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 $AD=AE=6$
