Prove that $E,F$ and $I$ are Collinear points.

Question

Let $\Delta ABC$ be a arbitrary triangle. And $E,F$ and $I$ points such that :

$$\vec{AE}=\dfrac{2}{3}\vec{AB}~,~\vec{BI}=-\dfrac{1}{3}\vec{BC}~,~ \vec{AF}=\dfrac{1}{3}\vec{AC}$$

Question

Prove that $E,F$ and $I$ are Collinear points.

My attempts

I want to write a vector $\vec{FE} $in terms of vector $\vec{FI}$ ( parallels)

Mean that in need find a scalar $k$ such that $\vec{FE} =k × \vec{FI}$

But I can't find it! Where is thé idea.

Answer 1

It is always a good idea to have a picture:

The solution using vectors is now easier to follow: $$ \begin{aligned} \overrightarrow{IE} &= \overrightarrow{IB} + \overrightarrow{IC} = \bbox[lightyellow]{ \frac 13\overrightarrow{BC} + \frac 13\overrightarrow{BA} } \\[3mm] \overrightarrow{EF} &= \overrightarrow{EA} + \overrightarrow{AF} = \frac 23\overrightarrow{BA} + \frac 13 \underbrace{\overrightarrow{AC}}_ {\substack{=\overrightarrow{AB}+\overrightarrow{BC}\\=-\overrightarrow{BA}+\overrightarrow{BC}}} = \bbox[lightyellow]{ \frac 13\overrightarrow{BA} + \frac 13\overrightarrow{BC} }\ , \end{aligned} $$ so $\overrightarrow{IE}=\overrightarrow{EF}$, the three points $I,E,F$ are collinear and $E$ is the mid point of the segment $IF$.