Problem with vector calculation.

Question

Problem

Let $\text{ABC}$ be a triangle and let $\text{A'}$ , $\text{B'}$ and $\text{C'}$ be respectively the center of $\text{[BC]}$ , $\text{[AC]}$ and $\text{[AB]}$.

1. Prove that: $\vec{\text{AA'}}+\vec{\text{BB'}}+\vec{\text{CC'}}=\vec{\text{0}}$

2. Let $\text{E}$ be a point in the vector plane, and $\text{G}$, $\text{F}$ are points where $\vec{\text{EF}}=\vec{\text{CC'}}$ and $\vec{\text{EG}}=\vec{\text{BB'}}$, and $\text{I}$ is the center of $\text{[FG]}$.

   Prove that: $\vec{\text{EI}}=x \cdot \vec{\text{CB}}$ or $\vec{\text{CB}}=x \cdot \vec{\text{EI}}$

For the first one, it is easy:

$$ \begin{align} \vec{\text{CA}}+\vec{\text{AB}}&=\vec{\text{CB}}\\ \vec{\text{CA}}+\vec{\text{AB}}+\vec{\text{BC}}&=\vec{\text{0}}\\ \vec{\text{CA}}+\vec{\text{AC'}}+\vec{\text{AB}}+\vec{\text{BA'}}+\vec{\text{CB}}+\vec{\text{BC'}}&=\vec{\text{AC'}}+\vec{\text{BA'}}+\vec{\text{CB'}}\\ \vec{\text{CC'}}+\vec{\text{BB'}}+\vec{\text{AA'}}&=\frac{1}{2}(\vec{\text{AB}}+\vec{\text{BC}}+\vec{\text{CA}})\\ &=\vec{\text{0}} \end{align}$$

For the second one, I failed at all the attempts, and started wondering if the problem was at those equations: $\vec{\text{EF}}=\vec{\text{CC'}}$ and $\vec{\text{EG}}=\vec{\text{BB'}}$. So, are they correct, that is, can we prove that: $\vec{\text{EI}}=x \cdot \vec{\text{CB}}$ or $\vec{\text{CB}}=x \cdot \vec{\text{EI}}$ using them?

Answer 1

Theorem:
Given a triangle $XYZ\,$, if $X'$ is the center of $\,[YZ]$, then $$\overrightarrow {XX'}=\frac 12 \cdot (\overrightarrow {XY}+\overrightarrow {XZ})$$In fact $$\overrightarrow {XX'}=\overrightarrow {XY}+\overrightarrow {YX'}$$$$\overrightarrow {XX'}=\overrightarrow {XZ}+\overrightarrow {ZX'}$$After summing side by side, we are done.

Now your question:

1.$$\overrightarrow {AA'}+\overrightarrow {BB'}+\overrightarrow {CC'}=$$$$\frac 12 \cdot (\overrightarrow {AB}+\overrightarrow {AC}+\overrightarrow {BA}+\overrightarrow {BC}+\overrightarrow {CA}+\overrightarrow {CB})=0$$2.$$\overrightarrow {EI}=\frac 12 \cdot (\overrightarrow {EF}+\overrightarrow {EG})=$$$$\frac 12 \cdot (\overrightarrow {CC'}+\overrightarrow {BB'})=$$$$\frac 14 \cdot (\overrightarrow {CA}+\overrightarrow {CB}+\overrightarrow {BC}+\overrightarrow {BA})=$$$$-\frac 14 \cdot (\overrightarrow {AC}+\overrightarrow {AB})=$$$$-\frac 12 \cdot \overrightarrow {AA'}$$