The medians of $△TUV$ are $\overline{TX}, \overline{UY},$ and $\overline{VW}$. They meet at a single point $Z$. In other words, $Z$ is the centroid of $△TUV$. Suppose $\overline{UY}=33$, $\overline{TZ}=8$, and $\overline{VZ}=14$. Find the lengths $\overline{ZW}, \overline{ZY},$ and $\overline{TX}$.
I'm not quite sure how to solve this type of problem so I've provided an example. Your feedback is greatly appreciated.
In order to solve this problem, you only need to use a well-known fact about medians and the way the centroid divide them into ratios of 2:1. Therefore $$\frac{\overline{VZ}}{\overline{ZW}}={\frac{14}{\overline{ZW}}}={\frac{2}{1}}$$ $$\Rightarrow {\overline{ZW}}=7$$ $$\frac{\overline{UZ}}{\overline{ZY}}={\frac{\overline{UY}-\overline{ZY}}{\overline{ZY}}}={\frac{33}{\overline{ZY}}-1}={\frac{2}{1}}$$ $$\Rightarrow {\frac{33}{\overline{ZY}}}=3\Rightarrow{\overline{ZY}}=11$$ Finally $$\frac{\overline{TZ}}{\overline{ZX}}=\frac{8}{\overline{TX}-8}={\frac{2}{1}}$$ $$\Rightarrow{8}={2}({\overline{TX}-8})={2\overline{TX}-16}\Rightarrow\overline{TX}=\frac{8+16}{2}=12$$
There's an easy way to prove this characteristic property in regards to medians and their centroid:
Let $△ABC$ be a triangle and $D, E, F$ the midpoints of the sides $\overline{AB}, \overline{BC}$ and $\overline{CA}$ respectively. Let furthermore $\overline{FH}$ be parallel to $\overline{CD}$. Note now, that by Thale's theorem (not to be confused with another theorem with that name) in triangle $△ADC$ $$\frac{\overline{DH}}{\overline{HA}}=\frac{\overline{CF}}{\overline{FA}}=1$$ The point $H$ is hence the midpoint of the segment $\overline{AD}$, which implies $$\frac{\overline{AH}}{\overline{AB}}={\frac{1}{4}}$$ Thus $${\frac{\overline{BD}}{\overline{DH}}}={\frac{2}{1}}$$ Recall now, that $\overline{FH}||\overline{GD}$. So, again, by Thale's theorem $${\frac{\overline{BG}}{\overline{GF}}}={\frac{\overline{BD}}{\overline{DH}}}={\frac{2}{1}}$$
Q.E.D.
PS: This property has a very interesting application in physics, where the centroid is called "center of mass of a triangle". This (very simplified) is related to the fact, that the point $D$ stays for the sum of masses of the points $A$ and $B$, so in order to balance the triangle, the point C has to be twice so separated from the center of mass than the point $D$.