Cut-the-Knot's SSS proof page has a number of solutions, including Euclid's. As the author indicates, however, only Hadamard's proof "goes through without a hitch", with the important aside: "assuming of course that isosceles triangles have been fairly treated previously".
I'll give a full development of Hadamard's argument, including the necessary bits needed about isosceles triangles. I include a couple of "obvious" sub-proofs just to make clear which axioms are in play.
Preliminaries:
- SAS triangle congruence is an axiom.
- (1) implies one direction of the Isosceles Triangle Theorem, namely: If two sides of a triangle are congruent, then the angles opposite those sides are congruent. $[\star]$
- (2) implies that A point equidistant from distinct points $P$ and $Q$ lies on the perpendicular bisector of the $\overline{PQ}$. $[\star\star]$
Now, we'll cut to the part of Hadamard's argument where we have $\triangle ABC$ and $\triangle A^\prime BC$ with corresponding edges congruent, constructed with $A$ and $A^\prime$ on the same side of $\overleftrightarrow{BC}$. (That last part is key.) We need only show that $A$ and $A^\prime$ coincide to prove SSS. Hadamard makes this argument by contradiction ...
- Assume $A \neq A^\prime$.
- $B$ is equidistant from $A$ and $A^\prime$, and therefore, $B$ lies on the perpendicular bisector of $\overline{AA^\prime}$ by (3) above. The same is true of $C$.
- Therefore, $\overleftrightarrow{BC}$ is the perpendicular bisector of $\overline{AA^\prime}$.
- Therefore, $\overleftrightarrow{BC}$ separates $A$ and $A^\prime$.
- This contradicts the fact that we constructed $A$ and $A^\prime$ to be on the same side of $\overleftrightarrow{BC^\prime}$. $[\star\star\star]$
- Therefore, the assumption that $A \neq A^\prime$ must be invalid, so that triangles $\triangle ABC$ and $\triangle A^\prime BC$ coincide; this gives us SSS. $\square$
$[\star]$ Proof: Isosceles $\triangle ABC$ with base $\overline{BC}$ is congruent to $\triangle ACB$ by SAS, and those opposite angles are corresponding angles of the congruent triangles.
$[\star\star]$ The midpoint, $M$, of $\overline{PQ}$ is certainly on the perpendicular bisector. A point $R \neq M$ equidistant from $P$ and $Q$ creates isosceles $\triangle RPQ$ with congruent angles at $P$ and $Q$ by (2). Thus, $\triangle RPM \cong \triangle RQM$ by SAS. Corresponding angles $\angle RMP$ and $\angle RMQ$ must be congruent; as supplements, they must be right. $\overleftrightarrow{MR}$, then, is the perpendicular bisector of $\overline{PQ}$. (I'll note that an easier proof could assert that, from the get-go, $\triangle RPM \cong RQM$ by SSS ... but we can't use that argument in a proof of SSS itself.)
$[\star\star\star]$ The contradiction here is one of Euclid's oversights later made explicit by Hilbert as the Plane Separation Axiom. The PSA has slightly-different formulations in different treatments of geometric foundations, but effectively it's the "Because I said so!" assertion that lines break planes into two disjoint "sides" and gives us our contradiction.
Your $H$ is a linear transformation, not an affine one. Write $H(x) = Ax + b$ with $A\in\mathbb R^{2\times 2}, b\in\mathbb R^2$ instead. Using this directly shouldn't be too hard, noting that the things you describe are given by
$$p_1-p_2 = \lambda (p_3-p_4)$$
and you have to show that
$$H(p_1)-H(p_2) = \lambda (H(p_3)-H(p_4))$$
Best Answer
For convenience with the unit circle, let's place the first (right isosceles) triangle with its base on the positive horizontal axis, so it has an acute angle at the origin and its right angle at $(1, 0)$. If each successive triangle is rotated $2\pi/n$ clockwise, the right angle of the $n$th triangle is at $(\cos(2\pi/n), \sin(2\pi/n))$. Consequently, we have $x = \sin(2\pi/n)$, $y = 1 - \cos(2\pi/n)$, and $$ \frac{x}{y} = \frac{\sin(2\pi/n)}{1 - \cos(2\pi/n)} = \frac{2\sin(\pi/n)\cos(\pi/n)}{1 - \cos^{2}(\pi/n) + \sin^{2}(\pi/n)} = \cot\frac{\pi}{n}. $$ The question is, what choice of $n > 1$ makes $x/y$ as close as possible to $\gamma = \frac{1}{2}(1 + \sqrt{5}) \approx 1.618034$.
The diagram shows the unit circle (centered at the origin) and the "final positions" of the right angle if we take $n = 2, \dots, 24$. The points $n = 5$ and $n = 6$ bracket the blue line (whose slope is $-\gamma$). The ratio $x/y$ is monotone in $n$, so one of these is optimal. We have $$ \frac{\sin(2\pi/5)}{1 - \cos(2\pi/5)} = \frac{\sqrt{2}\sqrt{5 + \sqrt{5}}}{\sqrt{5} - 1} \approx 1.37638,\qquad \frac{\sin(2\pi/6)}{1 - \cos(2\pi/6)} = \sqrt{3} \approx 1.73205. $$ Of these, $n = 6$ is closer.
The aspect ratio of the right triangles is immaterial, because the second acute angle (not at the origin) has no effect on the position of the right angle.