Theorem (Hyperbolic Pythagoras).
Let $\triangle ABC$ be a hyperbolic triangle with angles $\alpha$, $\beta$ and $\gamma=\frac\pi2$.
Then the lengths of the edges $a$, $b$ and $c$ satisfy
\[\cosh c=\cosh a\cdot\cosh b\,.\]

Proof.

We calculate the length in the upper half-plane model after normalizing the triangle to the following:

We use the formula derived for the hyperbolic distance in the upper half-plane using complex variables:

\[
\cosh d(z_1,z_2)-1
=\frac{\lvert z_1-z_2\rvert^2}{2\operatorname{Im}z_1\operatorname{Im}z_2}\,,
\]

So for the length of the edges of the triangle we obtain:
\begin{align*}
\cosh a&=1+\frac{\lvert B-C\rvert^2}{2\operatorname{Im}B\operatorname{Im}C}\\
&=1+\frac{(k-1)^2}{2k}=\frac{k^2+1}{2k}\\
\cosh c&=1+\frac{\lvert\mathrm e^{\mathrm i\phi}-k\mathrm i\rvert^2}{2k\sin\phi}\\
&=1+\frac{\left(\mathrm e^{\mathrm i\phi}-k\mathrm i\right)\left(\mathrm e^{-\mathrm i\phi}+k\mathrm i\right)}{2k\sin\phi}\\
&=1+\frac{k^2+1-2k\sin\phi}{2k\sin\phi}\\
&=\frac{k^2+1}{2k\sin\phi}\\
\cosh b&=\frac{1^2+1}{2\cdot1\cdot\sin\phi}\\
&=\frac1{\sin\phi}\,.
\end{align*}

This calculation implies the required equality.

What is the relation of hyperbolic and Euclidean triangles?

Looking at “small” hyperbolic triangles (i.e. triangles with small edge lengths and area) hyperbolic triangles behave similar to Euclidean triangles.
If the area is very small, then the sum of its interior angles is almost $\pi$.
The hyperbolic Pathagoras’ Theorem becomes the Euclidean Pythagoras’ Theorem for small edge lengths:
The Taylor series for $\cosh$ is $\cosh x=\sum\frac{f^{(k)}(0)}{k!}x^k=1+\frac12x^2+\dotsb$.
Thus $\cosh c=\cosh a\cdot\cosh b$ becomes
\[1+\frac{c^2}2+\dotsb=\left(1+\frac{a^2}2+\dotsb\right)\left(1+\frac{b^2}2+\dotsb\right)\,,\]
hence
\[1+\frac{c^2}2=1+\frac{a^2}2+\frac{b^2}2+\underbrace{\frac{a^2b^2}4+\dotsb}_{\approx0}\]
which gets $c^2=a^2+b^2$ for small $a$, $b$ and $c$.

Hyperbolic motions in $H_+^2$

Consider the following maps:
\[g\colon\hat\C\to\hat\C\,,\ z\mapsto\frac{az+b}{cz+d}\,,\]
where $ad-bc>0$ and $a,b,c,d\in\R$.
Here, $g$ maps $\infty$ to $\frac ac$ and $-\frac dc$ to $\infty$.
Also,
\[\tilde g\colon\hat\C\to\hat\C\,,\ z\mapsto\frac{-a\bar z+b}{-c\bar z+d}\,.\]

• The maps $g$ and $\tilde g$ are bijective (the inverse is just the transformation corresponding to $\begin{pmatrix}a&b\\c&d\end{pmatrix}^{-1}$).
• These maps preserve the upper half-plane:
  We have
  \begin{align*}
  \operatorname{Im}g(z)
  &=\frac1{2\mathrm i}\left(\frac{az+b}{cz+d}-\frac{a\bar z+b}{c\bar z+d}\right)\\
  &=\frac1{2\mathrm i\lvert cz+d\rvert^2}\left(ac\lvert z\rvert^2+adz-ac\lvert z\rvert^2-ad\bar z+bc\bar z+bd-bcz-bd\right)\\
  &=\frac{ad-bc}{2\mathrm i\lvert cz+d\rvert^2}(z-\bar z)\\
  &=\frac{ad-bc}{\lvert cz+d\rvert^2}\operatorname{Im}z>0\,.
  \end{align*}
  Similarly for $\tilde g$.
• These maps are isometries:
  \begin{align*}
  \frac{\lvert g(z_1)-g(z_2)\rvert^2}{2\operatorname{Im}g(z_1)\operatorname{Im}g(z_2)}
  &=\frac{\lvert cz_1+d\rvert^2\lvert cz_2+d\rvert^2}{(ad-bc)^2}\frac1{\operatorname{Im}z_1\operatorname{Im}z_2}\frac{(ad-bc)^2\lvert z_2-z_1\rvert^2}{2\lvert cz_1+d\rvert^2\lvert cz_2+d\rvert^2}\\
  &=\frac{\lvert z_1-z_2\rvert^2}{2\operatorname{Im} z_1\operatorname{Im}z_2}\,.
  \end{align*}
  Again, a similar calculation for $\tilde g$ yields that $\tilde g$ is also an isometry.
• Hyperbolic lines are mapped to hyperbolic lines.

Theorem.
An arbitrary hyperbolic isometry of $H_+^2$ is of the form
\[z\mapsto\frac{az+b}{cz+d}\quad\text{or}\quad z\mapsto\frac{-a\bar z+b}{-c\bar z+d}\,,\]
where, again, $ad-bc>0$ and $a,b,c,d\in\R$.

Proof.
Let $\phi\colon H_+^2\to H_+^2$ be an isometry. We can construct a map $g$ which is one of the following, such that $g\circ\phi$ maps the imaginary axis onto itself and $g\circ\phi(\mathrm i)=\mathrm i$ and $g\circ\phi((0,\mathrm i))=(0,\mathrm i)$:

• $z\mapsto\frac{z-u}v$ with $\phi(\mathrm i)=u+\mathrm i v$,
• $z\mapsto\frac v{\bar z-u}$,
• $z\mapsto\alpha\frac{z-x}{z-y}$ where $\phi(0)=x$, $\phi(\infty)=y$.

Then $g\circ\phi$ is an isometry, since $g$ and $\phi$ are. Further $g\circ\phi$ is the identity on the imaginary axis, as it is isometric and maps the imaginary axis onto itself preserving the intervals $(0,\mathrm i)$ and $(\mathrm i,\infty)$.
Now consider $x+\mathrm i y$ with $g\circ\phi(x+\mathrm i y)=u+\mathrm i v$.
We have $d(k\mathrm i,x+\mathrm i y)=d(g\circ\phi(k\mathrm i),u+\mathrm i v)=d(k\mathrm i,u+\mathrm i v)$ for all $k\in\R^+$.
Thus
\[\frac{\lvert k\mathrm i-(x+\mathrm i y)\rvert^2}{2ky}=\frac{\lvert k\mathrm i-(u+\mathrm i v)\rvert^2}{2kv}\]
and hence
\[(x^2+(k-y)^2)v=(u^2+(v-k)^2)y\]
for all $k$.
This implies $y=v$ and $u=\pm x$.
So $g\circ\phi$ is either the identity or it the reflection with respect to the imaginary axis, $z\mapsto-\bar z$.
As $\phi=g^{-1}\circ g\circ\phi$, $\phi$ is of the given type.

Lengths/distances in $H_+^2$.

Two points $P_1$ and $P_2$ in the upper half-plane model either have the same $x$ coordinate or not. If they share the $x$ coordinate the geodesic through two points is a straight line. Otherwise it is a semi-circle. Let us calculate the hyperbolic distance between two points depending on the two cases.

Case 1.

\[
\gamma\colon[t_1,t_2]\to H_+^2\,,\ \gamma(t)=\begin{pmatrix}0\\t\end{pmatrix}\,.
\]
\begin{align*}
d(p_1,p_2)&=\int_{t_1}^{t_2}\sqrt{g_{\gamma(t)}\left(\begin{pmatrix}0\\1\end{pmatrix},\begin{pmatrix}0\\1\end{pmatrix}\right)}\mathrm dt \\
&=\int_{t_1}^{t_2}\frac1t\sqrt{\left(\begin{pmatrix}0\\1\end{pmatrix},\begin{pmatrix}0\\1\end{pmatrix}\right)}\mathrm dt=\log\frac{t_2}{t_1}\,.
\end{align*}
So the length/distance only depends on the ratio $\frac{t_2}{t_1}$, not on the values.

Case 2.

\begin{align*}
P_1&=r\begin{pmatrix}\cos\alpha_1\\\sin\alpha_1\end{pmatrix}&
P_2&=r\begin{pmatrix}\cos\alpha_2\\\sin\alpha_2\end{pmatrix}\,.
\end{align*}
Here,
\[\gamma\colon[\alpha_1,\alpha_2]\to H_+^2\,,\ \gamma(\alpha)=r\begin{pmatrix}\cos\alpha\\\sin\alpha\end{pmatrix}\,.\]
By this we get
\[d(p_1,p_2)=\int_{\alpha_2}^{\alpha_1}\frac1{\sin t}\mathrm dt=\left.\log\tan\frac t2\right\vert_{\alpha_1}^{\alpha_1}=\log\frac{\tan\frac{\alpha_1}2}{\tan\frac{\alpha_2}2}\,,\]
i.e. the distance depends on the angle, not on the radius.

Remrk.
The following maps are isometries of $H_+^2$:

• Horizontal translations, $\begin{pmatrix} x\\y\end{pmatrix}\mapsto\begin{pmatrix} x+c\\y\end{pmatrix}$,
• scalings, $\begin{pmatrix} x\\y\end{pmatrix}\mapsto\lambda\begin{pmatrix} x\\y\end{pmatrix}$ for $\lambda>0$.

Both transformations leave the Riemmanian metric invariant.

Proposition.
The distance in the Poincaré halfplane model $H_+^2\subset\C$ between two points $z_1$, $z_2$ is given by
\[\cosh d(z_1,z_2)-1=\frac{\lvert z_1-z_2\rvert^2}{2\operatorname{Im} z_1\operatorname{Im} z_2}\,.\]

Proof. Calculation using some trigonometric identities.

Hyperbolic triangles

Let $A$, $B$, $C$ be three points in the hyperbolic plane $H^2$ not all on one line.
Then the hyperbolic triangle is given by
\[\triangle ABC=\{\lambda A+\mu B+\nu C:\lambda,\mu,\nu>0\}\cap H^2\,.\]

The triangle defined by $A$, $B$, $C$ in the Klein model is
\[\triangle ABC=\{\lambda A+\mu B+\nu C:\lambda,\mu,\nu\geq0,\lambda+\mu+\nu=1\}\,.\]
We label the vertices, angles and sides of a triangle as shown in the above picture.

Observation: $\alpha+\beta+\gamma<\pi$ for any hyperbolic triangle.
Consider a triangle in the Poincaré disc. We may use a hyperbolic motion to move $A$ to the center of the disc.

The angles between the hyperbolic lines at $B$ and $C$ are smaller than the angles of the Euclidean triangle, hence $\alpha+\beta+\gamma<\pi$.

Proposition.
For $\alpha,\beta,\gamma>0$ with $\alpha+\beta+\gamma < \pi$ there exists a hyperbolic triangle. The triangle is unique up to hyperbolic motions.

Proof. We proof the proposition by construction: Let $\delta > 0$ such that $\alpha+\beta+\gamma+\delta = \pi$. Construct a Euclidean triangle with angles $\alpha$, $\tilde\beta = \beta+\delta/2$ and $\tilde\gamma=\gamma+\delta/2$. Then construct a circle $S$ through $B$ and $C$ such that $\angle OBC=\delta$ and $O$ on the other side of $a$ than $A$. The angles between $S$ and $b$, $c$ are $\beta$ and $\gamma$, resp. Now, construct a circle $S_0$ centered in $A$ which intersects $S$ orthogonally. Scaling this construction such that the second circle has radius $1$ yields a triangle with the prescribed angles in the Poincaré model.

Theorem.
The area of a hyperbolic triangle with angles $\alpha$, $\beta$, $\gamma$ is $\pi-(\alpha+\beta+\gamma)$.

Proof.
Consider a triangle with one point at infinity, i.e. on the boundary of the disc in the Poincaré model.

We may normalize the picture and transform it to the upper half-plane model, such that we obtain the following picture:

The area of the triangle is
\begin{align*}
T(\alpha,\beta,0)&=\int_{-\cos\alpha}^{\cos\beta}\int_{\sqrt{1-x^2}}^\infty\frac1{y^2}\mathrm dy\mathrm dx \\
&=\int_{-\cos\alpha}^{\cos\beta}-\left.\frac1y\right\vert_{\sqrt{1-x^2}}^\infty\mathrm dx \\
&=\int_{\cos(\pi-\alpha)}^{\cos\beta}\frac1{\sqrt{1-x^2}}\mathrm dx \\
&=\pi-(\alpha+\beta)\,.
\end{align*}

Now,
\begin{align*}
T(\alpha,\beta,\gamma)&=T(\alpha,\gamma+\gamma_1,0)-T(\gamma_1,\pi-\beta,0) \\
&=\pi-\alpha-\gamma-\gamma_1-(\pi-\gamma_1-\pi+\beta)\\
&=\pi-(\alpha+\beta+\gamma)\,.
\end{align*}