The triangle ABC is given. You get B' by mirroring point B over point A. Let point M lie on the side BC such that | BM | : | BC | = k, where k∈(0,1). Let N be the intersection of the line MB' and the side AC. Calculate the ratio | AN | : | NC |.
I know that M lies somewhere on the line BC, but you do not know where specifically and that makes it so much more confusing. If the k would be given, I would need no further help. But now that k is not given, how do you calculate the ratio?
Please use vectors in your answers.
Let $D$ be a midpoint of $BM$.
Thus, $AD||B'M$ because $$\vec{AD}=\vec{AB}+\vec{BD}=\frac{1}{2}\vec{B'B}+\frac{1}{2}\vec{BM}=\frac{1}{2}\vec{B'M}$$ and $$\frac{AN}{NC}=\frac{DM}{MC}=\frac{BM}{2MC}=\frac{BM}{2(BC-BM)}=\frac{k}{2(1-k)}.$$