# 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:

$$\frac{Area(\triangle CGF)}{Area(\triangle AGF)} = \frac{Area(\triangle BGC)}{Area(\triangle BGA)}$$

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if someone could fix my code that would be nice too :) – Ross Pure Jul 10 '11 at 7:36
Looks like you intended this: $\frac{Area(\triangle CGF)}{Area(\triangle AGF)} = \frac{Area(\triangle BGC)}{Area(\triangle BGA)}$ – Brian M. Scott Jul 10 '11 at 7:41
I've added Brian's suggestion. – Zev Chonoles Jul 10 '11 at 7:44
@Zev: Thanks; it took me a while to notice the bloomin' underscores. – Brian M. Scott Jul 10 '11 at 7:46
Thanks for the fix guys. – Ross Pure Jul 10 '11 at 7:52

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.

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Thanks. It probably would've saved me alot of time spent thinking about this problem if i knew the rule for triangles with the same height... – Ross Pure Jul 10 '11 at 8:03
@Ross: It follows from the area of a triangle being half the product of the base and the height—if two triangle share a height, the ratio of their areas is the ratio of their bases; or if two triangles share a base, the ratio of their areas is the ratio of their heights. – Isaac Jul 10 '11 at 8:07
Thanks again but i looked up the proof anyway. – Ross Pure Jul 10 '11 at 8:11

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 $\frac ab=\frac{a+c}{b+d}$ $\Rightarrow$ $\frac ab=\frac cd$.)

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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.

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Almost universally, contest problems that can be done with Ceva's Theorem are more easily done with the technique of Mass Points. – Isaac Jul 10 '11 at 15:06
@Isaac Thanks for the link it was actually incredibly helpful. Mass points seem much more efficient than other methods. – Ross Pure Jul 11 '11 at 6:36

a centroid divides a triangle in six triangles having equal area.

therefore,

[CGF]/[AGF] = 1:1

[BGC]/[BGA] = 1:1 as the triangle [BGC] = 2[BGE] = [BGA]

thus ,

[CGF]/[AGF] = [BGC]/[BGA]

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