<s>[INST] Geometry question with three triangles Any help on this question would be great, I'm not sure how to solve it.
Thanks in advance!
Given $\triangle ABC$, let $A'$ be the point $\frac{1}{3}$ of the way from $B$ to $C$, as shown. Similarly, $B'$ is the point $\frac{1}{3}$ of the way from $C$ to $A$, and $C'$ is the point $\frac{1}{3}$ of the way from $A$ to $B$. In this way, we have constructed a new triangle starting with an arbitrary triangle. Now apply the same procedure to $\triangle A'B'C'$, thereby creating $\triangle A''B''C'' $. Show that the sides of $\triangle A''B''C'' $ are parallel to the appropriate sides of $\triangle ABC$. What fraction of the area of $\triangle ABC$ is the area of $\triangle A''B''C'' $?

 [/INST] This is nicely done in barycentric coordinates. We have:
$$ A' = \frac{2B+C}{3},\quad B'=\frac{2C+A}{3},\quad C'=\frac{2A+B}{3} $$
hence:
$$ A''=\frac{2B'+C'}{3} = \frac{4A+B+4C}{9},\quad B''=\frac{4A+4B+C}{9},\quad C''=\frac{A+4B+4C}{9} $$
The last relation implies:
$$ A''-B'' = \frac{1}{3}(C-B),\quad B''-C''=\frac{1}{3}(A-C),\quad C''-A''=\frac{1}{3}(B-A) $$
so $B''A''\parallel BC,C''B''\parallel CA,A''C''\parallel AB$ and the ratio between the area of $ABC$ and the area of $A''B''C''$ is $9$.
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