What manifold is produced if all similar triangles in the plane are identified? Consider Euclidean space $\mathbb{R}^6$, we can consider this as the $x$ and $y$ coordinates of the $3$ points of a triangle in $\mathbb{R}^2$. If we identify all points in $\mathbb{R}^6$ whose triangles have the same 3 angles, what manifold do we get under the quotient topology?
 A: We can exploit the scale invariance of homogeneous coordinates to capture efficiently the similarity condition and produce a description for which the action of the relevant symmetry groups are especially apparent.
For any oriented triangle $\Delta ABC$, mark one of its vertices, say, $A$, and denote the marked triangle by $(\Delta ABC, A)$. Then, if the lengths of the sides opposite $A, B, C$ are respectively, $a, b, c$, consider the triple $$(t_a, t_b, t_c) := (-a + b + c, -b + c + a, -c + a + b) .$$ We can recover the relative lengths from the triple via $$(a, b, c) = \left(\tfrac{1}{2} (t_b + t_c), \tfrac{1}{2} (t_b + t_c), \tfrac{1}{2} (t_b + t_c)\right) .$$
Geometric construction of $(t_a, t_b, t_c)$ (see this GeoGebra activity for an interactive version)

Now, observe the following:


*

*The sides $a, b, c$ form a triangle iff the inequality $a + b > c$ and its cyclic permutations hold, but these conditions are equivalent to $t_a, t_b, t_c > 0$. Conversely, every triple of positive numbers determine a triangle.

*If we scale $(\Delta ABC, A)$ by $\lambda$, the resulting triple scales by the same factor---we get $\lambda t_a, \lambda t_b, \lambda t_c$.

*Similar marked triangles give rise to proportional triples.


Thus, by regarding triples as homogeneous coordinates $[t_a, t_b, t_c]$ on a reference triangle $T$, we get an identification between oriented similarity classes of marked triangles and the interior $T^{\circ}$ of $T$.
The reference triangle $T$

Some elementary algebra shows that the equilateral class corresponds to the red point, $[1, 1, 1]$, and the isosceles classes correspond to the green lines, the loci of the points of the form $[t, t, \ast]$ and its cyclic permutations.
Now, we want to disregard the markings, that is, identify $[(\triangle ABC, A)] \sim [(\triangle ABC, B)] \sim [(\triangle ABC, C)]$; we write this equivalence class as $[\triangle ABC]$. These classes are exactly the orbits of the natural $\Bbb Z_3$-action whose generator acts by, say, $[(\triangle ABC, A)] \mapsto [(\triangle ABC, B)]$. Unwinding definitions, via the above identification $\sigma$ just acts on $T^{\circ}$ by cyclic permutation: $$\sigma: [t_a, t_b, t_c] \to [t_b, t_c, t_a] ,$$ or equivalently clockwise rotation by $1/3$ of a turn. Of course, viewed this way $\Bbb Z_3$ is just the oriented symmetry group of $T^{\circ}$. Thus:

We may naturally identify the space of triangles modulo oriented similarity with $T^{\circ} / \Bbb Z_3$.

By the argument in my other answer, this is an orbifold).
If instead we want to consider the space of unoriented triangles modulo similarity, we must also identify $[\triangle ABC, A]$ with $[\triangle ACB, A]$, but this corresponds to the involution $$\tau: (t_a, t_b, t_c) \mapsto (t_a, t_c, t_b) ,$$ that is, reflection across the line through $[1, 0, 0]$ and $[1, 1, 1]$. This reflection and $\sigma$ together generate the full group of symmetries of $T^{\circ}$ and is isomorphic to $S_3$. Thus:

We may naturally identify the space of unoriented triangles modulo similarity with the orbifold $T^{\circ} / S_3$.

Remark In fact, for any triangle $\Delta ABC$, regarded as homogeneous coordinates on $\Delta ABC$, $[t_a, t_b, t_c]$ is the Nagel point of $\Delta ABC$; this is the triangle center complementary to the incenter.
A: Here we realize the plane as $\Bbb C$.
Given an oriented triangle $\triangle ABC$, there is unique translation that maps a selected vertex $A$ to $0$, a unique rotation that then maps the next vertex anticlockwise $B$ to the positive real axis (so that the third vertex $C$ is in the upper half-plane), and a unique dilation that then maps $B$ to $1$. By construction, this procedure gives an identification of the space of similarity classes of marked triangles with the upper half-plane $\Bbb H$: $$\{[(\triangle ABC, A)]\} \leftrightarrow \Bbb H .$$
The standard coordinate $z$ on $\Bbb H$ is essentially the parameter $z$ Somos identifies in his answer.
Now we forget about markings, which is the same as identifying $[(\triangle ABC, A)] \sim [(\triangle ABC, B)] \sim [(\triangle ABC, C)]$, and we just write this equivalence class as $[\triangle ABC]$. By construction, these classes are exactly the orbits of the natural action of $\Bbb Z_3$ on $\Bbb H$, where a choice of generator $\sigma$ just rotates the choice of basepoint, say, anticlockwise: $\sigma : [(\triangle ABC, A)] \mapsto [(\triangle BCA, B)]$, so we can identify the space of similarity classes of oriented triangles: $$\{[\triangle ABC]\} \leftrightarrow \Bbb H / \Bbb Z_3 .$$ Some elementary geometry shows that in terms of the standard coordinate $z$ of $\Bbb H$, $\sigma : z \mapsto \frac{1}{1 - z}$.
Since $\Bbb Z_3$ is finite and $\Bbb H$ is Hausdorff, this action is properly discontinuous, and so $\Bbb H / \Bbb Z_3$ is an orbifold, a space modeled on $\Bbb R^n$ modulo a finite group action. All manifolds are orbifolds (taking a trivial action), but the converse is not true.
If we want instead to treat unoriented similarity classes of triangles, then we also need  to identify $[\triangle ABC]$ with its mirror image $[\triangle ACB]$. If we denote the order $2$ map on $\Bbb H$ that exchanges these by $\tau$, then (1) $\tau : z \mapsto \frac{1}{\bar z}$, and (2) $\sigma \tau = \tau \sigma^2$, so $\sigma$ and $\tau$ generate a group isomorphic to $S_3$ that acts on $\Bbb H$. We can thus identify the space of unoriented similarity classes with $\Bbb H / S_3$.
A: Up to similarity, triangles are determined by their three angles which must add up to 180 degrees, hence two degrees of freedom which determines a surface which is a variation of a kaleidoscope with three mirrors forming an equilateral triangle. The equilateral triangle is the fundamental domain and the sides are reflected back into the triangle by the mirrors which represent the edge cases of the similarity class of isosceles triangles. An even more exceptional case is the similarity class of equilateral triangles which corresponds to the vertices of the fundamental domain which are identified with each other. 
You can get a more conformal model using complex numbers. If  complex numbers $\,A,B,C\,$ are the vertices of a triangle, the division ratio $\,z:=(A-C)/(A-B)\,$ is invariant under linear transformations and hence is a similarity invariant. Using a linear transformation, we can let $A=0, B=1$ and $C$ determines the triangle up to similarity. However, $A,B,C$ can be permuted in six ways which transform $\,z\,$ into
$$\{\,1-z,\,1/z,\,1/(1-z),\,(z-1)/z,\,z/(z-1)\}.$$
A fundamental domain is given by
 $\, \textrm{Re}(z)\le 1/2, \, |z-1|\le 1.\, $
If orientation reversing is allowed, the fundamental domain is now restricted to $\, \textrm{Im}(z)>0.\,$ Note that
$\, \textrm{Im}(z)=0\,$ corresponds to degenerate triangles.
