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'' $?

 A: 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 $9$.
A: I'll use complex coordinates. Without loss of generality, I'll assume $A,B,C$ have coordinates $9z$, $0$, $9$ respectively. (We only care about the relative areas, so assuming $|BC|=1$ loses nothing; the $9's$ are for convenience.) If we have two points with coordinates $\alpha,\beta$ then the point which is 1/3 of the way from $\alpha$ to $\beta$ is given by $\frac{2}{3}\alpha+\frac{1}{3}\beta$. Thus the coordinates of $A',B',C'$ are $$\frac{1}{3}(9)=3,\quad \frac{2}{3}(9)+\frac{1}{3}(9z)=6+3z,\quad \frac{2}{3}(9z)=6z$$ respectively, and those of $A'',B'',C''$ are $$\frac{2}{3}(6+3z)+\frac{1}{3}(6z)=4z+4,\quad \frac{2}{3}(6z)+\frac{1}{3}(3)=4z+1,\quad \frac{2}{3}(3)+\frac{1}{3}(6+3z)=z+4$$ respectively. This gives the respective side-lengths of $\Delta A'' B''C''$ as $$|(4z+1)-(4z+4)|=3,\quad |(z+4)-(4z+1)|=3|z-1|,\quad |(4z+4)-(z+4)|=3|z|.$$ But these represent a dilation by a factor of $3$ the side-lengths $9,9|z-1|,9|z|$ as $\Delta ABC$. Consequently the triangles are similar, and their areas will differ by a factor of $3^2=27$.
