Proportionallity between area and sides of a triangle. i'm working on some excercises of geometry and I found one of them that is really hard to me. The exercise is the next: 

The additional hypothesis is that the segmen $CG$ is equal to the segment $GD$ and the segment $CH$ is equal to the segment $HE$. I need to found the ratio between the area of the purple zone and the area of triangle $\Delta ABC$. 
First, we can saw that the three triangles $\Delta ADC$, $\Delta DEC$ and $\Delta EBC$ have the same height. Then, they areas are related by a ratio because their basis are related by a ratio. But, then, what can I do? I know that my ideas are very few, but, I'm really stuck and I don't know how to proced. Any idea? Any hint? I really apreciate your help. 
 A: 
It is easiest to calculate the fraction of the triangle that's pink by first calculating the fraction that's blue and the fraction that's purple.
To do so, we calculate that fraction separately for $\triangle ADC$, $\triangle DEC$, and $\triangle EBC$. For example, in $\triangle ADC$:


*

*The purple triangle is $\frac23$ of the total area of $\triangle ADC$: from the point of view of vertex $A$, one side is scaled down by $\frac23$ and the other side stays the same.

*The blue triangle is $\frac16$ of the total area of $\triangle ADC$: from the point of view of vertex $C$, one side is scaled down by $\frac13$ and the other side is scaled down by $\frac12$.

*The pink triangle is therefore the remainder: also $\frac16$ of $\triangle ADC$.


We also know that $\triangle ADC$ is $\frac39$ of the total area of the triangle. So the pink area inside $\triangle ADC$ is $\frac16 \cdot \frac39 = \frac1{18}$ of the total area.
We can handle the other two pieces similarly.

The logic going on for the blue and purple areas is always the same lemma.

Suppose you have a large triangle $\triangle XYZ$ and choose points $Y'$ on $XY$, $Z'$ on $XZ$. Then the ratio of the areas $[XY'Z'] : [XYZ]$ is the same as the product $\frac{XY'}{XY} \cdot \frac{XZ'}{XZ}$.

One of the ways to prove this lemma is to use the formula $[XYZ] = \frac12 \cdot XY \cdot XZ \cdot \sin \angle X$ for the area of a triangle. It's also possible to prove it in two steps using the ordinary half-base-times-height formula: 


*

*when comparing $\triangle XYZ$ to $\triangle XYZ'$, we think of them as having the same base $XY$ and heights in a $XZ : XZ'$ ratio by similar triangles.

*when comparing $\triangle XYZ'$ to $\triangle XY'Z'$, we think of them as having the same base $XZ'$ and heights in a $XY : XY'$ ratio by similar triangles.


Anyway, this lemma immediately tells us that, for example, the area of $\triangle CHI$ is $\frac12 \cdot \frac34 = \frac38$ of the area of $\triangle CEB$.
