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_. !Related diagram show that: $$\frac{Area(\triangle CGF)}{Area(\triangle AGF)} = \frac{Area(\triangle BGC)}{Area(\triangle BGA)}$$

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.