Formula for length of diagonal in a Lambert quadrilateral

Question

Given a Lambert quadrilateral $AOBF$ where the angles $ \angle FAO , \angle AOB , \angle OBF $ are right, and $F$ is opposite $O , \angle AFB$ is the acute angle , and the Gaussian curvature = -1 (so all is standard)

Then a whole lot of equations hold see https://en.wikipedia.org/wiki/Lambert_quadrilateral#Lambert_quadrilateral_in_hyperbolic_geometry

(these are the relations I found in different sources)

There is also another relation (follows from the Klein Disk model)

$$ \tanh^2 OF =\tanh^2 OA +\tanh^2 OB $$

But I was not able to deduce this formula from the others.

How can $ \tanh^2 OF =\tanh^2 OA +\tanh^2 OB $ be deduced from the other formula's?

Answer 1

Maybe take a look at the paper "The Right Right Triangle on the Sphere" by Dickinson and Salmassi. In spherical (rather than hyperbolic) geometry, they derive and champion the similar-looking $\sin^2 OF = \sin^2 FA + \sin^2 FB$. I'm sure that in the hyperbolic case one would have $\sinh^2 OF = \sinh^2 FA + \sinh^2 FB$, although this is still using the 'wrong' (from your perspective) legs FA and FB. But perhaps with further application of the 'usual' Pythagorean Theorem you can deduce what you want.

As a remark, I worked out the relation $\sin^2 OF = \sin^2 FA + \sin^2 FB$ myself from the same Lambert quadrilateral Wikipedia formulas you cited (assuming they held also on the sphere), and since it's so simple and pleasant-looking I assumed it was a "well known" fact. But it doesn't seem to be; it took me a long time to find the Dickinson--Salmassi paper, and it's the only source I managed to find.

Answer 2

Defining $a:=|OA|$, $b:=|OB|$, $f:=|OF|$, $a':=|AF|$, $b':=|BF|$, we have

$$\begin{align} \cosh f &= \cosh a\cosh a' &&\text{(Hyperbolic Pythagoras)}\\[4pt] \operatorname{sech}^2f &= \operatorname{sech}^2a \operatorname{sech}^2a' \\[4pt] \operatorname{sech}^2f&=\operatorname{sech}^2a(1-\tanh^2a') \\[4pt] \operatorname{sech}^2f&=\operatorname{sech}^2a(1-\color{red}{\cosh^2 a\tanh^2 b}) &&\text{(Lambert quad property)}\\[4pt] \operatorname{sech}^2f&=\operatorname{sech}^2a-\tanh^2 b \\[4pt] 1-\tanh^2f &= (1-\tanh^2a)-\tanh^2b \\[4pt] \therefore \tanh^2f &= \tanh^2a+\tanh^2b \end{align}$$