In the acute-angled triangle $\triangle ABC$, we mark the middle with $M$ side $BC$ and with $D$ the foot of the height in $A$. Let $G$ be a point on the median $AM$ and $E$ is its projection on the $MD$ segment. If $BD + BC = 3BE$, prove that $G$ is the centroid of $\triangle ABC$..
Okey. You can see below the drawing and the ideas I had... I thought of the formula for the median. And then show that $MG$ is $1/3$ of $AM$ or $AG$ is $2/3$ of $AM$. Another idea was to put $BG$ intersect $AC$ in point $N$. And then show that $N$ is half of $AC$. Anyways, I don't know which idea is the right one and with which one to start and where to start. Any ideas are welcome!
You would like to "show that $MG$ is $1/3$ of $AM$", which will imply $G$ is the centroid as $AM$ is an median. This is the right approach.
Since $GE\parallel AD$, it is enough to show that $ME$ is $1/3$ of $MD$. Now all given conditions and our goal are on the segment $BC$.
Consider the equality $BD +BC = 3BE$. While segment $BM$ is shared among $BD, BC$ and $BE$, it appear neither in $ME$ nor in $MD$. Let us try eliminating it from that equality, using the condition $BC = 2BM$.
$$\begin{aligned} BD +BC &= (BM+MD)+ 2BM \\ 3BE&= 3(BM+ME) \end{aligned}$$
The difference of the two equalities above is $0=3ME-MD$, which is what we want.