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13.6: Teorema de Pitágoras
https://espanol.libretexts.org/Matematicas/Geometria/Avi%C3%B3n_Euclideano_y_sus_Familiares_(Petrunin)/13%3A_Geometr%C3%ADa_del_plano_h/13.06%3A_Teorema_de_Pit%C3%A1goras
\[\begin{array} {rcl} {\cosh c} & = & {\dfrac{AB \cdot A'B' + AB' \cdot A'B}{AA' \cdot BB'} = } \\ {} & = & {\dfrac{\sqrt{x^2 + y^2} \cdot \sqrt{\dfrac{1}{x^2} + \dfrac{1}{y^2}} + \sqrt{\dfrac{1}{x^2}... $\begin{array} {rcl} {\cosh c} & = & {\dfrac{AB \cdot A'B' + AB' \cdot A'B}{AA' \cdot BB'} = } \\ {} & = & {\dfrac{\sqrt{x^2 + y^2} \cdot \sqrt{\dfrac{1}{x^2} + \dfrac{1}{y^2}} + \sqrt{\dfrac{1}{x^2} + y^2} \cdot \sqrt{x^2 + \dfrac{1}{y^2}}}{(\dfrac{1}{y} - y) \cdot (\dfrac{1}{x} - x)} =} \\ {} & = & {\dfrac{x^2 + y^2 + 1 + x^2 \cdot y^2}{(1 - y^2) \cdot (1 - x^2)} =} \\ {} & = & {\dfrac{1 + x^2}{1 - x^2} \cdot \dfrac{1 + y^2}{1 - y^2}.} \end{array}$

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