I am trying to prove that $AG = 2(GD)$, given $AD, BE$ and $CF$ are 3 medians meeting at point $G$ of a triangle $ABC$. I found this website that seems to show what I want to prove: http://jwilson.coe.uga.edu/EMAT6680Su09/Park/As4dspark/As4dspark.html
In the proof you can see that he first connected the $2$ midpoints $E$ and $F$ (is this even okay to assume you can connect both midpoints?) Then he states that $AC = 2AE$ and $AB = 2AF$ (why or how does he deduce that its $2$ times the length?) Then triangles $AEF$ and $ACB$ are similar (he did not state why but is it because of $SAS$?) How does he deduce that $CB = 2EF$? (Is it because they are similar and have the same proportions?) Then he states that $\angle GEF = \angle GBC$ and $\angle EFG = \angle GCB$ (is that by converse to alternate interior angle theorem?)
Then I get confused when he mentions angle $GDE$ ... Where is angle $GDE$? Does he mean that $\angle EGC$ and $angle FGB$ are congruent since they are opposite angles (vertical angle theorem?) I just don't get why he states that $EGF$ and $BDC$ are similar triangles .. by what? I get the rest of the proof but I just wanted to confirm what he left out or did not explain.
In summary do you think if i proved $AEF$ and $ACB$ are similar by $SAS$, then $EFG$ and $GCB$ are similar by $AAA$ would be enough in proving $AG=2(GD)$ where the common ratio is $2:1$ by the fundamental theorem of similar triangles?
Yes, it's okay to connect the mid-points. Why wouldn't it be?
$AC=2AE$ because $E$ is the mid-point of $AC$. That's what being the mid-point means: $AE = EC$, so $AC = AE + EC = 2AE$. Also $AB=2AF$ because $F$ is the mid-point of $AB$.
Yes, $\triangle AEF$ is similar to $\triangle ACB$ becase they share an angle and the adjacent sides are proportional.
$EF$ is parallel to $CB$, so the alternate interior angles are equal. This is just what you call the alternate interior angle theorem I think, not the converse.
This part is non-sense. It's not true in general that $\angle GDE = \angle GBC$. For example if $\angle BAC$ is taken to be almost $180^\circ$ and $AB=AC$, then $\angle GDE$ is close to $90^\circ$, but $\angle GBC$ is close to $0$.
They probably meant $\angle CGB = \angle FGE$, since it looks like the next line is deducing $\triangle GCB$ is similar to $\triangle EFG$. That assertion is not used at all anyway.
$\triangle GCB \sim \triangle GFE$ because they have the same angles.
If you prove $\triangle EFG \sim \triangle GCB$ then you get $CG=2GF$. If you want to get $AG=2GD$ you have to use a similar argument to prove for instance $\triangle ACG \sim \triangle DFG$. Usually we just say it's true by symmetry.
One note about names of theorems. These are not standard names, and they come from some particular textbook probably. There's very few theorems that actually have names, so when speaking to a general math audience you shouldn't expect that they know which theorem exactly you're talking about if you mention the name only (though in this case it's not hard to guess.)
Of course disregard that advice if you do this for a high-school class, because unfortunately there you're forced to pick up all kinds of strange mathematical habits that you have to later forget.