Working with Centroids

Question

The question is as follows:

Points $D$ and $E$ are marked on segments $AB$ and $BC$, respectively. When segments $CD$ and $AE$ are drawn, they intersect at point $T$ inside triangle $ABC$. It is found that segment $AT$ is twice as long as segment $TE$, and that segment $CT$ is twice as long as segment $TD$. Must $T$ be the centroid of triangle $ABC$?

I don't know where to start from. Please help!

Answer 1

Yes, since $ATC\sim ETD$ with coefficient of similarity $k=2$ we have that $AC||DE$ and $DE=AC/2$ so $DE$ is middle line and thus conclusion.

If it is easier to you, look at homothety at $T$ which takes $A$ to $E$. Then it takes $C$ to $D$ and thus $AC||DE$ and $DE=AC/2$ again...

Answer 2

hint:

let $[ABC]$ denote are of triangle

$[ABT] = \frac{[ABC]}{3}$ similarly $[BTC] = \frac{[ABC]}{3}$

So, we get $[ATC] = \frac{[ABC]}{3}$