Problem
Let $\text{ABC}$ be a triangle, and $\text{M}$ and $\text{N}$ are points where: $\vec{\text{AM}}=\frac{1}{3}\vec{\text{AB}}$ and $\vec{\text{AN}}=\frac{1}{2}\vec{\text{AB}}$ and $\text{M'}$ and $\text{N'}$ are points in $\text{[AC]}$ such that both $\text{(NN')}$ and $\text{(MM')}$ are parallel to $\text{(BC)}$
- Write $\vec{\text{AM'}}$ and $\vec{\text{AN'}}$ with respect to $\vec{\text{AC}}$, then conclude that $\vec{\text{MM'}}=\frac{1}{3}\vec{\text{BC}}$ and $\vec{\text{NN'}}=-\frac{1}{2}\vec{\text{BC}}$
- Conclude that $\frac{\text{MM'}}{\text{NN'}}=\frac{2}{3}$
To have a better idea of what it may look like:
Here's my try:
Prove that $\vec{\text{AM'}}$ is equal to $1/3$ of $\vec{\text{AC}}$ using Thales' theorem:
- $\text{(AM')} \| \text{(AC)}$
- $\text{A is in [AC)}$
- $\|\vec{\text{AM'}}\|=\frac{1}{3}\|\vec{\text{AC}}\|$
But I can't use Thales' theorem, so how to proceed?
Hint: Start from $\vec{AB}+\vec{BC}+\vec{CA}=0$ and multiply it respectively by $1/3$ and by $1/2$ to get equations containing $\vec{AM},\vec{MM'},\vec{M'A}$ and the same with $N,N'$ replacing $M,M'$. Then use that the length of $kV$ is $k$ times the length of $V$, for a positive constant $k$, and also that $\vec{VW}=-\vec{WV}.$ For the $2/3$ ratio, note it comes from $(1/3)/(1/2)$ at the appropriate point.