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Two corresponding side lengths and an equal nonincluded angle need not determine a hyperbolic triangle. Set $A=\pi /6$, $\cosh a=3/2$, $\cosh b=2$. The hyperbolic law of cosines has two distinct positive solutions $c=\ln (3\pm \sqrt{2})$. The two triangles share $a,b,A$ but have different third sides. An included angle instead determines the third side by the cosine rule.

Three corresponding side lengths determine a hyperbolic triangle up to an isometry of the hyperbolic plane. First match an ordered pair of vertices by an isometry. The third vertex lies at the intersection of two prescribed hyperbolic distance circles. There are at most two choices, interchanged by reflection in the geodesic through the matched pair; either choice can therefore be matched.

In the hyperbolic law of cosines, the angle opposite $a$ being at least $\pi /2$ makes its cosine nonpositive. ThusSince $\cosh$ is increasing on $[0,\infty )$, the opposite side is longest: $\boxed{a\ge b,a\ge c}$. For a nondegenerate hyperbolic triangle both comparisons are strict.

At an interior point $w$ of the geodesic segment from $z_1$ to $z_2$, the two angles made with the segment toward $w^{\prime }$ add to $\pi$. At least one is at least $\pi /2$. In the corresponding hyperbolic triangle, part (a) says that its opposite side $w^{\prime }{z}_i$ is at least the adjacent side $w^{\prime }w$. ThereforeIf $w$ is an endpoint the assertion is immediate. The same inequality also holds when $w^{\prime }$ lies on the complete geodesic through $z_1,z_2$: parameterize that geodesic by arclength and use the elementary inequality $|t-s|\le \max (|t-a|,|t-b|)$ for $a\le s\le b$. This extension will allow repeated use of the result in part (c).

Consider the filled hyperbolic triangle with vertices $A,B,C$. Every point $P$ is on some geodesic segment $Aw$ with $w\in BC$: for instance, in the Beltrami-Klein model these are the ordinary straight segments filling a convex triangle. For any other point $Q$ of the triangle, apply part (b) first on $Aw$ and then on $BC$:Applying the same argument to $Q$, with each vertex held fixed, bounds every term on the right by the maximum of the three side lengths. The endpoints of a longest side attain that bound. Thus the diameter is exactly the longest side length:

In the curvature $-1$ normalization, the hyperbolic triangle area is its angle defect:Join one vertex of a convex geodesic polygon to its nonadjacent vertices. This gives $n-2$ hyperbolic triangles with disjoint interiors. Their angles at every polygon vertex sum to that vertex's interior angle. Adding the hyperbolic triangle areas therefore gives the hyperbolic polygon areaFor a regular hyperbolic polygon with prescribed area, place $n$ equally spaced vertices on a hyperbolic circle of radius $R>0$ and join consecutive vertices by geodesics. Rotations through $2\pi /n$ and reflections in radial lines show that this is a convex regular polygon. The triangle from its centre to a vertex and the midpoint of an adjacent side is right angled, with angles $\pi /n$, $\alpha (R)/2$, and $\pi /2$, and hypotenuse $R$. The angle form of the hyperbolic law of cosines givesConsequently $A(R)=(n-2)\pi -n\alpha (R)$ is continuous and strictly increasing. As $R\downarrow 0$, $\alpha (R)\to \pi -2\pi /n$, so $A(R)\to 0$; as $R\to \infty$, $\alpha (R)\to 0$, so $A(R)\to (n-2)\pi$. The intermediate value theorem proves existence for every permitted $A$. Indeed the required radius isThe argument also proves uniqueness of the radius in this construction; neither endpoint area is attained by a nondegenerate finite-radius polygon.

Because the half-plane metric is a conformal rescaling of the Euclidean metric, its angles are Euclidean angles. For each $y>0$, the hyperbolic lines meeting the imaginary axis at a fixed angle $\alpha$ at $iy$ are the Euclidean circleswhere $Y$ denotes the vertical coordinate. These circles have centres on the real axis and are therefore hyperbolic lines; varying $y$ gives the required collection.

To construct a triangle, put one vertex at $i$ and a second at $it$ on ${\ell }_{+}$, with $t>1$. Draw from these vertices the lines making interior angles $\alpha$ and $\beta$ with ${\ell }_{+}$. Their Euclidean centres and radii areIf their other intersection has angle $\gamma$, the law of cosines in the Euclidean triangle formed by the two centres and that intersection givesEquivalently,The right side is greater than one exactly when $\alpha +\beta +\gamma <\pi$. Since $(t+t^{-1})/2$ ranges continuously and strictly increasingly from $1$ to infinity for $t>1$, such a $t$ exists. The three lines therefore bound a hyperbolic triangle with the prescribed angles.

A hyperbolic triangle is bounded by three hyperbolic geodesic segments or rays. Its vertices may lie in the hyperbolic plane; an ideal vertex is their endpoint on the boundary at infinity. Every ideal angle is zero. Gauss-Bonnet theorem with curvature $-1$ givesso an all-ideal triangle has area $\pi$.

For fixed admissible angles, hyperbolic trigonometry determines all three side lengths from the angles, for exampleThus two such triangles are congruent. An orientation-preserving isometry can send one chosen vertex and oriented tangent to the corresponding data of the other, and the determined side lengths and angles then send the entire triangle to it. Since the orientation-preserving isometry group is $PS{L}_2(\mathbb{R})$, represented by $S{L}_2(\mathbb{R})$, the stated action is transitive.