Proving an intuitive property for families of non-overlapping convex polygons on the plane Let $\mathscr{F}$ be a finite family of non-overlapping convex polygons on the plane. I'd like to prove that

$\exists F \in \mathscr{F}\exists F'\subset \mathbb{R}^2 \mathscr{F}\cup \{F'\}$ is a family of non-overlapping convex polygons, $F'$ is congruent to $F$ and shares one side with it.

For now, I'm trying to prove the result for triangles, but with no success. This is quite an intuitive proposition, especially for triangles. Indeed one expects to always find a triangle in the "external region" with enough space around to construct a congruent triangle on one of its sides.
Observation If $P\in\mathscr{F}, |P\cap ConvHull(\mathscr{F})|\geq 2$ then $\mathscr{F}$ respects the thesis.

After Hagen von Eitzen answered I went back to thinking about my original problem, i.e. what happens if the polygons are all congruent triangles. I was not able to adapt his answer, which I very much liked, to this special case.
 A: Complete this image to a circle.

The big triangles cannot be used, except perhaps with their base edge.
The smaller triangles filling the gaps can obviously not be used.
The out-pointing small triangles cannot be used either because we can obviously not use their inner edge, and by choosing the height smaller than the height of the gaps below them, their other two edges cannot be used either.
So what about the base edges of the big triangles? All we need to ensure is that the height of the triangles exceeds the diameter of the $n$-gon formed by theses edges (so make them significantly taller than in the sketch) and we are done!

After fine-tuning the details, the following seems to be a minimal working example reachable with this method (21 triangles):

The blue triangles are wlog. equilateral. Hence we need $\gamma<60^\circ$, so $\beta>60^\circ$ and $\alpha<60^\circ$. Moreover, $n\cdot(\alpha-\epsilon)=360^\circ$. Thus we certainly need $n>6$. With $n=7$ as in the image, we can pick $\alpha$ anywhere strictly between $51\frac37^\circ$ and $60^\circ$ (which make $\epsilon<8\frac 57^\circ$).
