where is the 5th postulate being used at in this proof 
Given the triangle on the left, the theorem is that all the medians of each side of the triangle go through point $G$. Suppose the points do not go through a point $G$ as the picture on the right shows. Notice how I labeled each section of the triangle by a lower case letter. consider the sections where the $a$'s are first. Since $K$ is the median then $AK = KC$ and triangle $AZK$ and triangle $CZK$ share the same height. So by using the formula for the area of a triangle it should be clear that triangle $AZK =$ triangle $CZK = a$. We can make a similar argument for the $b$ sections and $c$ sections. Now suppose triangle $XYZ = \epsilon > 0$. Consider the $3$ points $x,y,z$. Each point has two lines going through them and essentially making vertical angles. So for example, point $y$ has vertical angles where in what section we have $a$ and in another section we have $b$. Similarly Look at point $x$. The two lines that go through them form vertical angles where $c$ is in one section and $b$ is in another section. Again you can see an argument like this for point $z$. Then it follows that $a < b < c < a$ which is a contradiction. Hence, $\epsilon = 0$. Its funny that I write the whole proof out but my question has nothing to do with the proof. My question is as follows : The fifth postulate of euclidean geometry is used in this proof in some way but its sort of hiding itself. Where is it being used at?
 A: I could not exactly follow your line of reasoning, since it seems unclear what you mean by the labels a,b,c. The following is a way to get a contradiction from the diagram, involving looking at areas of the seven regions triangle ABC is cut into by the medians (assumed not to be concurrent). But with this area comparison method it doesn't seem that the parallel postulate comes in anywere.
Starting from triangle KZC and going around clockwise, label the regions other than the inner triangle XYZ as 1,2,3,4,5,6. Then for example 2 denotes the four sided figure AKZX. Now since |KZC|=|KZA| and the right side of this is strictly inside region 2 we have "1<2", meaning the area of 1 is less than the area of 2. Similarly 3<4 and 5<6. Also let t denote the area of the inner triangle XYZ.
Now in the triangle ABC since AN is a median it must bisect the area of ABC. Similarly CM and BK each bisect the area of ABC. This gives three equations involving 1,2,3,4,5,6,t namely
(*) 1+2+6+t=3+4+5,
(**) 1+2+3=4+5+6+t,
(***) 2+3+4+t=5+6+1.
Now if each of these is solved for t and the three expressions for t are added together the equation 3t = (1-2)+(3-4)+(5-6) results. However 3t>0 whereas each of (1-2), (3-4), and (5-6) is negative, from the above remarks comparing each triangle area with that of the quadrilateral on that side. 
There may arguably be another case, since perhaps line crosses to the left of the point you named x in your diagram, however this case is the same after it is reflected through a vertical line going through B.
