Intouch and Extouch triangles - An Interesting Result 
Consider a ∆ABC. Let the triangle formed by the points of contact of the sides of a given triangle with the excircles corresponding to these sides be called the extouch triangle; and let the triangle formed by joining, the points of contact of the three sides with the inscribed circle be called the intouch triangle. Prove that these two triangles have equal areas, and that the midpoint of their centroids is the centroid of ∆ABC.

Attached is a proof I found: Check Page 3 of this PDF
I've been trying really hard for an analytical proof, one using only coordinate geometry - but haven't been able to get anywhere. 
Could someone please help me with a proof using coordinate geometry only?
Assuming coordinates for ∆ABC and then finding the vertices of the intouch and extouch triangles will definitely solve the problem, but it's getting really lengthy and cumbersome. (The areas can be proved equal using the determinant, and calculation of centroid is easy) 
Is there anything I'm missing? 
Could someone help me find the vertices of the intouch and extouch triangles? (the result should be interesting, for the midpoint of the two centroids coincides with the centroid of ∆ABC)
A detailed solution, or guidance in the right direction would help! (The former, preferably) Thanks! 
P.S. You can find a neat diagram on Page 4 of the PDF, I'll add it here to the question if need be. 
 A: Elaborating on my comment: A coordinate derivation isn't terribly difficult if the intouch triangle's vertices are placed conveniently.


Let the origin-centered circle of radius $r$ be the incircle of $\triangle ABC$, and locate intouch vertices $D$, $E$, $F$ thusly:
$$D := r(\cos 2\beta,-\sin 2\beta) \qquad E := r( \cos 2\alpha, \sin 2\alpha) \qquad F := r(1,0)$$
Since $\overline{OF}\perp\overline{AB}$, we see immediately that
$$A = r(1,\tan\alpha) \qquad B = r(1,-\tan\beta)$$
For $C$, we easily write the normal forms of the equations of the tangent lines at $D$ and $E$ ...
$$\begin{align}
\overleftrightarrow{BC}:&\quad x \cos 2\beta - y \sin 2\beta = r \\
\overleftrightarrow{CA}:&\quad x \cos 2\alpha + y \sin 2 \alpha = r
\end{align}$$
... and calculate the intersection:
$$C = \frac{r(\cos(\alpha-\beta), \sin(\alpha-\beta))}{\cos(\alpha+\beta)}$$
Paying attention to similar triangles $\triangle OAF \sim \triangle AC^\prime F^\prime$ and $\triangle OBF \sim \triangle B C^\prime F^\prime$, we have
$$\frac{|\overline{AF^\prime}|}{|\overline{BF^\prime}|} = \frac{|\overline{C^\prime F^\prime}| \cot\alpha}{|\overline{C^\prime F^\prime}|\cot\beta} = \frac{|\overline{OF}| \tan\beta}{|\overline{OF}| \tan\alpha} = \frac{|\overline{BF}|}{|\overline{AF}|}\;\to\;
|\overline{AF}|=|\overline{BF^\prime}|\;\text{and}\;
|\overline{BF}|=|\overline{AF^\prime}|$$
This is exactly the "isotomic" property discussed in Dalcín's article. Since, as a consequence, $\overline{AB}$ and $\overline{FF^\prime}$ have a common midpoint (likewise for other sides), we have
$$A + B = F + F^\prime \qquad B + C = D + D^\prime \qquad C + A = E + E^\prime \tag{1}$$
Therefore,

$$\frac12\left(\;\frac13(D+E+F)+\frac13(D^\prime+E^\prime+F^\prime)\;\right) = \frac13(A+B+C) \tag{$\star$}$$

which is the centroid property. Confirming $|\triangle DEF| = |\triangle D^\prime E^\prime F^\prime|$ is a little messier, but can be done using $(1)$ to get explicit coordinates for $D^\prime$, $E^\prime$, $F^\prime$; for now, I'll leave that as an exercise to the reader.
A: Once you know that $BX=s-b$, $CY=s-c$, $AZ=s-a$ and so on, it is easy to find the coordinates of the points of contact of the three sides with the inscribed circle:
$$
X=B+{s-b\over a}(C-B)={s-c\over a}B+{s-b\over a}C,
$$ 
and the analogous expressions for $Y$ and $Z$:
$$
Y={s-a\over b}C+{s-c\over b}A,
\quad
Z={s-b\over c}A+{s-a\over c}B.
$$ 
In the same way you can find the coordinates of the other contact points $A_b$, $A_c$, etc.

