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For $g(z)=(az+b)/(cz+d)$ with real coefficients and $ad-bc=1$, direct subtraction from its conjugate givesThe imaginary part remains positive, and the inverse transformation has the same form, so $g$ is a bijection of the upper half-plane. MoreoverThus every path has the same hyperbolic length as its image. Taking the infimum over connecting paths, and using the inverse for the reverse inequality, proves every $g\in \Gamma$ is an isometry.

LikewiseIts squared modulus and the two imaginary-part factors cancel in the quotient, proving

We can normalize any ordered distinct pair to $(i,iy)$ with $y>1$. First send $z=x_0+i{y}_0$ to $i$ by $(\zeta -x_0)/y_0$, a real determinant-one Möbius transformation after rescaling its representing matrix. The stabilizer transformationsfix $i$. Under the Cayley map $C(\zeta )=(\zeta -i)/(\zeta +i)$, direct substitution gives $C(k_{\tau }\zeta )=e^{2i\tau }C(\zeta )$. Choose $\tau$ to rotate the second point's nonzero disc coordinate onto the positive real axis. Its inverse image is $iy$ with $y=(1+|C|)/(1-|C|)>1$.

The vertical path from $i$ to $iy$ has length $\log y$. Any other path has length at least $\int |dy|/y\ge |\int dy/y|=\log y$, so $\rho (i,iy)=\log y$. ConsequentlyInvariance of both quantities transfers this equality back to every pair; coincident points satisfy it as well. Therefore

For the right-triangle identity, normalize its right vertex to $i$ and one leg to the vertical geodesic, with endpoint $iy$, $y>1$. The other leg lies on the unit semicircle, with endpoint $s+it$, $s^2+t^2=1$, $t>0$. These geodesics meet orthogonally at $i$, since the metric is conformal to the Euclidean metric. The semicircle is a geodesic because a real Möbius transformation sends it to a vertical line; vertical lines are minimizing by the same length bound. The stabilizer rotations above can align any given tangent direction, giving this normal form for every right triangle.

If the legs have lengths $a,b$ and the opposite side has length $c$, the distance formula givesandThus the hyperbolic Pythagorean theorem is