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 !Triangle ABC 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.

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

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. !enter image description here

$EF\parallel CD$ 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 $$\frac{MN}{BF}=\frac{MN}{2MF}=\frac3{10}.$$