Ratio of two areas in a triangle Suppose you have the following triangle $ABC$:
with the following properties:
$$|AB|=4\cdot |AA'|,\>\>\>|AC|=4\cdot |CC'|\>\>\>|BC|=4\cdot |BB'|$$
I have to find the ratio of the total area of the triangle and the red area. I tried a lot of algebraic manipulations with the lengths of the sides, but couldnt solve it that way.
I really need hints for this one. Thanks!
 A: EDIT: Final solution.
First off, notice that $\triangle ABB',\triangle BCC', \triangle CAA'$ all have area equal to $\frac{1}{4}$ of the area of $\triangle ABC$. To see this, pick any side as the base when measuring $\triangle ABC$, say side $BC$. Then $BB'$ is a triangle with a common height and one fourth base, so its area is one fourth. This argument obviously holds for all three sides. 
What might be more surprising is that all three of the small triangles at the corners have the same area. They all equal $\frac{1}{13}\times \frac{1}{4}$ of the area of $\triangle ABC$, or rather, area equal to one thirteenth of the three aforementioned triangles. To make this explicit, look at $\triangle ABB'$ and $\triangle AA'P$, where $P$ is the intersection of $A'C$ and $AB'$. We have the following relationships:
$$AB = 4AA' \text{, and }\angle{BAB'} = \angle{A'AP} = \theta$$
Next we have the following equations for the triangles' areas:
$$\triangle AA'P = \frac{1}{2}AP\times AA'\sin\theta \text{, and } \triangle ABB' = \frac{1}{2}AB\times AB'\sin\theta $$
Therefore, the ratio of their areas is $\dfrac{\triangle AA'P}{\triangle ABB'} = \dfrac{AP}{4AB'}$. Now as soon as I can prove that this last quantity equals $\frac{1}{13}$, the proof will be complete. 
EDIT2:
Got it.
My solution requires Menelaus' Theorem. I will not prove it here.
By Menelaus' Theorem, we have that
$$\begin{align*} BC \times B'P \times AA' &= B'C \times AP \times BA' \\
BC \times B'P \times \frac{1}{4}AB &= \frac{3}{4}BC \times AP \times \frac{3}{4}AB \\
\dfrac{B'P}{AP} &= \dfrac{9}{4}\\
\dfrac{AB' - AP}{AP} &= \dfrac{9}{4}\\
\dfrac{AB'}{AP} &= \dfrac{13}{4} \implies \dfrac{AP}{AB'} = \dfrac{4}{13}\\ 
\end{align*}$$
Thus, as desired $\dfrac{AP}{4AB'} = \dfrac{1}{13}$. Thus, the area of the central triangle is $$\triangle ABC(1 - 3\times \frac{1}{4} + 3 \times \frac{1}{13}\times \frac{1}{4}) = \triangle ABC \times \frac{4}{13}$$

This is my first solution using trigonometry and the particular case of an equiangular triangle. According to Blue, due to affine equivalence of all triangles, this proof suffices for all cases.

Let $AB = BC = AC = 4k$, and hence $AA'=BB'=CC'=k$ and $\angle{ABC} = \angle{BCA} = \angle{BAC} = 60$. 

I claim the central triangle is equiangular. Examine one of the three
  small triangles at the corners of $\triangle ABC$. For convenience,
  I'll focus on the top one. Call it $AA'P$, where $P$ is the
  intersection of $CA'$ and $AB'$. Note the following $$ \angle{A'AP} = \angle{A'AB}\text{, and } \\ \angle{AA'P} = \angle{A'AC} = \angle{BB'A}$$ Furthermore, we know that $\angle{A'AB} + \angle{BB'A} = 120$, since the sum of the angles of any triangle equals 180. By this same principle, $\angle{A'PA} = 60$. This is true for all the
  small triangles, and by the vertical angles theorem, the central
  triangle is equiangular.
Notice that triangle $PA'A$ is similar to triangle $BB'A$. Thus we can
  find the lengths of $PA$ and $PA'$. But first we must compute the
  length of $B'A$. Via the law of cosines, $B'A = \sqrt{(4k)^2 +k^2 - 2(k)(4k)\cos(60)} = k\sqrt{13}$. Now using properties of similar
  triangles, we have the following equality
$$\dfrac{PA}{AB}=\dfrac{A'A}{B'A} \implies PA=\dfrac{AB\times A'A}{B'A} = \dfrac{4k}{\sqrt{13}}$$
By a similar argument, we can calculate the length $PA' =
> \dfrac{k}{\sqrt{13}}$. So, the side lengths of the central triangle
  all equal $k\sqrt{13} - \dfrac{5k}{\sqrt{13}} = \dfrac{8k}{\sqrt{13}}$. Finally, the ratio of the areas of two
  equilateral triangles equals the ratio of the squares of their sides,
  hence
$$\dfrac{\triangle Outer}{\triangle Inner} = \dfrac{16k^2}{\frac{64k^2}{13}} = \boxed{\dfrac{13}{4}}$$

A: 
Evaluate the area ratio
$$\frac{[ABB'']}{[ABC']} = \frac{[ABB']}{[ABB'C']} 
= \frac{\frac14[ABC]}{[ABC]-[B'C'C]}= \frac{\frac14[ABC]}{[ABC]-\frac14\frac34[ABC]}
=\frac4{13} $$
which leads to
$\frac{[ABB'']}{[ABC]} =\frac{[ABB'']}{\frac43[ABC']}  =\frac3{13}$.
Likewise, $\frac{[BCC'']}{[ABC]} =\frac{[CAA'']}{[ABC]} =\frac3{13}$, Then
$$\frac{[A''B''C'']}{[ABC]} = \frac{[ABC]-[ABB'']-[BCC'']-[CAA'']}{[ABC]} = 1-3\cdot\frac3{13}=\frac4{13}
$$
