Area method in Lobachevskian geometry There are many proofs in Euclidean geometry using the area method; for example, Ceva's theorem or the proof of Pythagorean theorem shown below.


Do you know such proofs in hyperbolic geometry? 

I mean something where area helps, but does not appear in the statement.
I suspect that the answer is no, but I hope to be surprised.
 A: Ok, if your question is "Are there proofs by dissection in hyperbolic geometry?" then the answer is very much yes.  Probably the first one is due to Gauss.  
Let $T(\alpha, \beta, \gamma)$ be the hyperbolic triangle with angles $\alpha$, $\beta$, and $\gamma$.  Gauss computes the area of $T(\alpha, \beta, \gamma)$ using the "windmill" figure. That is, Gauss decomposes $T(0, 0, 0)$ into four triangles - the central one is $T(\alpha, \beta, \gamma)$ and the others are $T(\pi - \alpha, 0, 0)$ and so on. 

There are also many, many dissection theorems in higher dimensions.  There are important conjectures here - see Conjecture 1.3 of Walter Neumann's article. 

However, if you are instead asking about "hinged dissections" then I am a stuck.  Here is the best I can do.

Liebmann [Nichteuklidische Geometrie, page 43] fixes a gap in Gauss' proof that the area of $T(\pi - \alpha, 0, 0)$ is $\alpha$.  I learned this from Coxeter's "Introduction to Geometry" [Figure 16.4a, page 295]. See also Coxeter's article "Angles and arcs in the hyperbolic plane" [last paragraph on page 19]. 
Here is the figure: 

Of course, this is a bit complicated.  Here is a somewhat simplified version.  We work in the upper half plane model of $\mathbb{H}^2$.  We must show that the area of an ideal triangle is finite.  It suffices to show that the following region has finite area:
$C = \{ z = x + iy \in \mathbb{H}^2 \mid x \in [1, 2], y \geq 1 \}$
We'll give a dissection of $C$ and reassemble it into $D$:
$D = \{ z \in \mathbb{H}^2 \mid x \in (0, 2], y \in [1, 2] \}$
Since the closure of $D$ is compact, it has finite area and we will win. 
We now define 
$C_n = \{ z \in C \mid x \in [1, 2], y \in [2^n, 2^{n+1}] \}$
and
$D_n = \{ z \in D \mid x \in [2^{-n}, 2^{-n+1}], y \in [1, 2] \}$
So, $C = \cup C_n$ and $D = \cup D_n$.  Also, $C_n$ is isometric to $D_n$ using a power of the map $z \mapsto z/2$, and we are done. 

Remarks: 


*

*Coxeter uses two families of geodesics to dissect his figure; I instead use one family of geodesics and one family of horocycles.  

*This example is also helpful for understanding "spiralling geodesics in incomplete finite area hyperbolic surfaces", but that is a story for a different time. 
A: Well, you may get hyperbolic Ceva's theorem using area in the manner similar to Euclidean's one. Namely, using the formula for the area $E$ of a triangle with sides $a,b,c$ and angles $A,B,C$: $\sin \frac{E}2 \cosh \frac{c}2=\sinh \frac{a}2\sinh\frac{b}2\sin C$ (analogue of Euclidean $E=\frac12ab\sin C$) we get the following interpretation of cevian: a line $\ell$ through a vertex $A$ of triangle $\triangle ABC$ is a locus of points $P$ for which the ratio
$$\sin \frac{E(PAB)}2\cdot\cosh\frac{BP}2:\sin \frac{E(PAC)}2\cdot\cosh\frac{PC}2$$
is a constant which I denote $\varkappa(\ell;BA,CA)$. The above areas should be oriented but let's ignore this. When are three lines $\ell_a$ through $A$, $\ell_b$ through $B$, $\ell_c$ through $C$ concurrent? 
When $\varkappa(\ell_a;BA,CA)\cdot \varkappa(\ell_b;CB,AB)\cdot \varkappa(\ell_c;AC,BC)=1$. 
(Again, I ignore the question of concurrent lines without common point which should be treated differently in hyperbolic geometry.) 
Now if $\ell_a\cap BC=A_1$, we have 
$$
\varkappa(\ell_a;BA,CA)=
\sin \frac{E(A_1AB)}2\cdot\cosh\frac{BA_1}2:\sin \frac{E(A_1AC)}2\cdot\cosh\frac{CA_1}2=\\
\frac{\sinh\frac{BA_1}2\cosh\frac{BA_1}2}{\cosh \frac c2}:
\frac{\sinh\frac{CA_1}2\cosh\frac{CA_1}2}{\cosh \frac b2}=
\frac{\sinh BA_1}{\cosh \frac c2}:
\frac{\sinh CA_1}{\cosh \frac b2}
$$
that yields hyperbolic Ceva.
