Let \(T(a,b,c)\subset\mathbb H\) be a hyperbolic triangle with interior angles \(a,b,c\). Prove that \[ \operatorname{area}T(a,b,c)=\pi-a-b-c \] as follows.
Note that \(T(0,0,0)\) is an ideal triangle with all vertices at infinity. Show that all ideal triangles have area \(\pi\).
Set \[ A(a)=\operatorname{area}T(\pi-a,0,0). \] Show geometrically that \[ A(a+a')=A(a)+A(a'), \] and conclude that \(A(a)=a\).
Extend the sides of \(T(a,b,c)\) to rays and use the corresponding ideal vertices to relate the area of \(T(a,b,c)\) to the areas of \(T(a,0,0)\), \(T(b,0,0)\), and \(T(c,0,0)\).
Solution
The area is \[ \boxed{\operatorname{area}T(a,b,c)=\pi-a-b-c}. \]
1. Every ideal triangle has area \(\pi\).
Any ordered triple of distinct boundary points of \(\mathbb H\) can be sent to \(0,1,\infty\) by an automorphism of \(\mathbb H\). Since such automorphisms are hyperbolic isometries, it is enough to compute the area of the ideal triangle with vertices \(0,1,\infty\).
Its sides from \(0\) and \(1\) to \(\infty\) are the vertical lines \(x=0\) and \(x=1\), while the third side is the semicircle \[ \left(x-\frac12\right)^2+y^2=\frac14, \qquad y=\sqrt{x(1-x)}. \] For the metric \(|dz|/y\), the area element is \(dx\,dy/y^2\). Therefore \[ \begin{aligned} \operatorname{area}T(0,0,0) &=\int_0^1\int_{\sqrt{x(1-x)}}^\infty\frac{dy\,dx}{y^2}\\ &=\int_0^1\frac{dx}{\sqrt{x(1-x)}}. \end{aligned} \] With \(x=\sin^2 t\), \(0\le t\le\pi/2\), this becomes \[ 2\int_0^{\pi/2}dt=\pi. \]
2. If\[ A(t)=\operatorname{area}T(\pi-t,0,0), \qquad 0\le t\le\pi, \]then\[ A(t+t')=A(t)+A(t') \]whenever \(t,t',t+t'\in[0,\pi]\).
The parameter \(t\) is the angular defect at the unique finite vertex: its interior angle is \(\pi-t\). Extend a geodesic side through the finite vertex to the ideal boundary, and insert the geodesic from the finite vertex to an intermediate ideal endpoint chosen so that the two pieces have the prescribed defects. The two resulting one-finite-vertex triangles have angular defects \(t\) and \(t'\), while their union has defect \(t+t'\). Their interiors are disjoint, so hyperbolic area is additive and gives the displayed identity.
3. One has \(A(t)=t\) for every \(0\le t\le\pi\).
First, \[ A(0)=0 \] because \(T(\pi,0,0)\) is degenerate, while by 1, \[ A(\pi)=\operatorname{area}T(0,0,0)=\pi. \] The area varies continuously with the finite angle: after fixing the two ideal vertices by an isometry, the remaining geodesic side depends continuously on that angle, and the corresponding regions vary monotonically on compact subranges. Thus \(A\) is continuous.
By 2, for every positive integer \(m\), \[ mA(\pi/m)=A(\pi)=\pi, \] so \(A(\pi/m)=\pi/m\). Additivity then gives \[ A(k\pi/m)=k\pi/m \] for \(0\le k\le m\). These rational multiples of \(\pi\) are dense in \([0,\pi]\); continuity therefore yields \[ A(t)=t \] throughout the interval.
4. An ideal triangle decomposes into \(T(a,b,c)\) and threeone-finite-vertex triangles of areas \(A(a)\), \(A(b)\), and \(A(c)\).
Extend each side of \(T(a,b,c)\) beyond one of its endpoints, cyclically, to a geodesic ray reaching the ideal boundary. The three resulting ideal vertices bound an ideal triangle containing the original triangle. At the vertex whose interior angle in the original triangle is \(a\), the adjacent complementary piece has two ideal vertices and finite angle \(\pi-a\); hence it is a triangle of type \(T(\pi-a,0,0)\) and has area \(A(a)\). The same angle chase gives complementary pieces of types \(T(\pi-b,0,0)\) and \(T(\pi-c,0,0)\) at the other two vertices.
These four regions have disjoint interiors and fill the ideal triangle. Consequently \[ \pi =\operatorname{area}T(a,b,c)+A(a)+A(b)+A(c). \]
5. Therefore\[ \operatorname{area}T(a,b,c)=\pi-a-b-c. \]
Substitute \(A(t)=t\) from 3 into the decomposition in 4: \[ \pi =\operatorname{area}T(a,b,c)+a+b+c. \] Rearranging gives the claimed formula.