Goursat's Lemma and ill-defined sequence of triangles I'm working on the proof of the Goursat's lemma, but I think that the sequence of triangle is "ill-defined" in the proof (due to the choice of subtriangles).
Given a open subset of $\mathbb C$ containing a triangle $T$ and $f$ an holomorphic function on this open set, we can split $T$ in four similar triangles $T_i$, $i=1,2,3,4$ and there is a $j$ such that the following inequality holds : $$\left\vert\int_{\partial T_j} f\right\vert \geqslant \frac{1}{4}\left\vert\int_{\partial T} f\right\vert\tag{$\star$}$$However, such a $j$ is clearly not unique (since Goursat's lemma will provide a zero integral for every triangle) so how to choose such a triangle ?
I thought that the vertices of the triangle $T$ could be ordered by this way : if we denote $\arg$ the main argument in the $[0,2\pi[$ range, the vertices of $T$ are denoted by $v_1,v_2,v_3$ with : $\arg(v_1 - b)<\arg(v_2-b)<\arg(v_3-b)$ where $b$ denotes the barycenter of $T$. 
Now split $T$ in four subtriangles as follow :


*

*$T_1 = \text{the triangle with vertices $v_1$, $(v_1+v_2)/2$ and $(v_1+v_3)/2$}$

*$T_2 = \text{the triangle with vertices $v_2$, $(v_1+v_2)/2$ and $(v_2+v_3)/2$}$

*$T_3 = \text{the triangle with vertices $v_3$, $(v_2+v_3)/2$ and $(v_1+v_3)/2$}$

*$T_4 = \text{the triangle with vertices $(v_1+v_2)/2$, $(v_2+v_3)/2$ and $(v_1+v_3)/2$}$


We choose $T_1$ if $(\star)$ holds with it, else if it holds for $T_2$ we select this one, else if it works for $T_3$ we pick out this guy, otherwise we must take $T_4$. By this way the next triangle in the sequence is uniquely selected and satisfies $(\star)$ by construction. Then we perform the rest of the proof normally... get the required relations at each step, apply Cantor theorem, etc.
Does this "selecting process" work to define recursively the sequence ?
 A: It seems you are trying to make this too complicated and invoking Goursat's lemma in your argument is irrelevant since that is what you are trying to prove.
Furthermore, the proof rests only on providing one sequence of triangles with the desired property.
Given triangle $T$, form four congruent triangles $T_1, \,$ $T_2,\,$$T_3,\,$ and $T_4$ using the vertices and the mid-points of the sides of triangle $T$.  Each pair of triangles $T_i$ and $T_j$ share one common side.
If we sum the integrals over the boundaries of the smaller triangles, each taken in the positive (counterclockwise) sense, we get cancellation of the contributions over the common sides.
Hence,
$$\int_{\partial T} f = \sum_{k=1}^4\int_{\partial T_k} f$$
Taking the modulus and applying the triangle inequality we get
$$\left|\int_{\partial T} f\right| \leqslant \sum_{k=1}^4\left|\int_{\partial T_k} f\right|$$
Given this inequality, there must be one triangle $T_j$ such that
$$\left|\int_{\partial T_j}f \right| \geqslant \frac{1}{4} \left|\int_{\partial T} f \right|$$
