Geometry/ Similar Triangles Problem Consider the trangle shown below with vertices A, B, C where point D lies on the side AB, point E lies on the side BC and point F lies on the side AC and the three lines AE, BF, and CD intersect at a common point G.

show that:
$$\frac{Area(\triangle CGF)}{Area(\triangle AGF)} = \frac{Area(\triangle BGC)}{Area(\triangle BGA)}$$
 A: DISCLAIMER: I am not a native speaker of English and I have never learned geometry in English. So my appologies, if I  use a non-standard terminology in some place. (Although I did my best to avoid it.)
I will denote the area of triangle $XYZ$ by $|XYZ|$.
You have:
$$\frac{|CGF|}{|AGF|} = \frac{|CBF|}{|ABF|} = \frac{|CGF|+|CBG|}{|AGF|+|ABG|}$$
The first equality follows from the fact that ratio of areas of the triangles with the same height is the ratio of the lengths of their sides. The second one is additivity of area.
If you compare the first and the last fraction, you get that it is also equal to $\frac{|CBG|}{|ABG|}$. (Using $\frac ab=\frac{a+c}{b+d}$ $\Rightarrow$ $\frac ab=\frac cd$.)
A: Because the height from $G$ to $\overline{AC}$ is common to both triangles, $$\frac{Area(\triangle CGF)}{Area(\triangle AGF)}=\frac{CF}{AF}.$$  Likewise, since the height from $B$ to $\overline{AC}$ is common to both triangles, $$\frac{Area(\triangle CBF)}{Area(\triangle ABF)}=\frac{CF}{AF}.$$  Now, $$\frac{Area(\triangle BGC)}{Area(\triangle BGA)}=\frac{Area(\triangle CBF)-Area(\triangle CGF)}{Area(\triangle ABF)-Area(\triangle AGF)},$$ and because both earlier ratios are $\frac{CF}{AF}$, so is this one, which gives your desired equation.
A: The calculations described by Isaac and Martin Sleziak are enough to prove a famous geometric result.  
Both solutions show that 
$$\frac{AF}{FC}=\frac{\text{Area}(\triangle GBA)}{\text{Area}(\triangle GCB)}.$$ 
The same idea applied to the other two sides of the big triangle show that
$$\frac{CE}{EB}=\frac{\text{Area}(\triangle GAC)}{\text{Area}(\triangle GBA)}$$
and
$$\frac{BD}{DA}=\frac{\text{Area}(\triangle GCB)}{\text{Area}(\triangle GAC)}.$$
Multiply the three right-hand sides.  We get $1$. It follows that 
$$\frac{AF}{FC}\cdot \frac{CE}{EB}\cdot \frac{BD}{DA}=1.  \qquad\text{(Equation 1)}$$
The above calculation is enough to prove Ceva's Theorem, which says that if we draw any three lines $BF$, $AE$, $CD$, these lines are concurrent if and only if Equation $1$ holds.  
One consequence of Ceva's Theorem is the familiar fact that the medians of a triangle are concurrent.  Ceva's Theorem has a habit of showing up as a tool in contest problems.
A: A centroid divides a triangle in six triangles having equal area.
Therefore,


$$\frac{[CGF]}{[AGF]} = \frac{1}{1}$$
    $$\frac{[BGC]}{[BGA]} = \frac{1}{1}$$ And $$[BGC] = 2[BGE] = [BGA]$$
    Thus,
    $$\frac{[CGF]}{[AGF]} = \frac{[BGC]}{[BGA]}$$


