In a triangle ABC have AB =10cm and AC=12cm. The incentro(I) and the baricenter(B) are in the same parallel to BC. The BC side measurement is equal to:
I did not calculate anymore because I do not understand baricenter and I encourage.
I already understand how Baricentro works and I encourage it, but what would their drawing look like in geogebra, according to this question?

Let $M$ be a barycenter of $\{A,B,C\}$.
Thus, $MI||BC.$
Now, let $AD$ be an altitude of $\Delta ABC$, $AF$ be a median of $\Delta ABC$ and $MI\cap AD=\{E\}.$
Thus, $ED=r$, $$\frac{AE}{ED}=\frac{AM}{MF}=2$$
and in the standard notation we have $$\frac{h_a-r}{r}=2,$$ which gives $h_a=3r.$
Thus, for the area of the triangle we obtain: $$\frac{(12+10+x)r}{2}=\frac{3rx}{2},$$ which gives $x=11.$