Prove that $\text{(BE)}\|\text{(JF)}$ using vectors.

Question

Problem

Let $\text{ABC}$ be a triangle and let $\text{I}$ , $\text{J}$ and $\text{K}$ be points such that: $\vec{\text{BI}}=\frac{1}{2}\vec{\text{IC}}$, $\vec{\text{AJ}}=2\vec{\text{JB}}$ and $\vec{\text{AK}}=\frac{1}{3}\vec{\text{AC}}$.

1. Let $\text{E}$ be the centre of $\text{[BC]}$ and $\text{F}$ the point of intersection of $\text{(AE)}$ and $\text{(IK)}$. Prove that: $\frac{1}{3}\vec{\text{EB}}=\vec{\text{EI}}$

2. Prove that $\text{(BE)}\|\text{(JF)}$

3. Show that the point $\text{F}$ is the centre of $\text{[IK]}$.

For the first question, it was relatively easy, it was sufficient to play with Chasles rule to prove it.

The second one is more complicated, I used different approaches, but no one was a success including calculations of $\vec{\text{AK}}$ with respect to $\vec{\text{IE}}$... It seems that I need a more general approach concerning what results from the intersection of two lines when dealing with vectors.

And of course I can't solve $\text{(3)}$ without $\text{(2)}$.

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The $ \| $ symbol means parallel.

Answer 1

I start by proving that $$\overrightarrow {KI}=\frac 23 \overrightarrow {AB}$$In fact $$\overrightarrow {AK}=\frac 13 \overrightarrow {AC}$$$$\overrightarrow {IB}=\frac 13 \overrightarrow {CB}$$By summing and Chasles $$\overrightarrow {AB}+\overrightarrow {IK}=\frac 13 \overrightarrow {AB}$$Now the points 2. and 3.

As to your request about a more general approach concerning what results from the intersection of two lines when dealing with vectors, in what follows I represent $\overrightarrow {AF}$ in two ways and equal them.
Because of parallelism$$\overrightarrow {AF}=\lambda\, \overrightarrow {AE}=\frac 12 \lambda \,(\overrightarrow {AB}+\overrightarrow {AC})$$$$\overrightarrow {KF}=\mu \, \overrightarrow {KI}=\frac 23\mu \,\overrightarrow {AB}$$From $\overrightarrow {AF}=\overrightarrow {AK}+\overrightarrow {KF}$, one has $$\frac 12 \lambda \,(\overrightarrow {AB}+\overrightarrow {AC})=\frac 13 \overrightarrow {AC}+\frac 23\mu \,\overrightarrow {AB}$$After some algebra $$\left(\frac 12 \lambda-\frac 23\mu \right)\,\overrightarrow {AB}+\left(\frac 12 \lambda-\frac 13 \right)\,\overrightarrow {AC}=\overrightarrow 0$$Because $\overrightarrow {AB}$ and $\overrightarrow {AC}$ are not parallel, one gets$$\begin {cases} \lambda=\frac 23 \\ \\ \mu=\frac 12 \end {cases}$$So point 3. is proved.

Now $$\overrightarrow {AF}=\overrightarrow {AJ}+\overrightarrow {JF}$$$$\frac 23\,\overrightarrow {AE}=\frac 23\,\overrightarrow {AB}+\overrightarrow {JF}$$Using Chasles$$\overrightarrow {JF}=\frac 23 \overrightarrow {BE}$$so point 2 is proved.

Answer 2

As I've suggested to you before, write all relevant vectors in terms of $x=\overrightarrow{AB}$ and $y=\overrightarrow{AC}$. For example, you should find $\overrightarrow{AF}=\frac13(x+y)$ and $\overrightarrow{JF}=\frac13(y-x)$. This should get you (2) and (3).