The diameter of a set $S$ in a metric space $(M,d)$ is defined to be {$\sup d(x,y)|x,y\in S$}. In Euclidean space the diameter of a triangle is the length of the largest side.
In the hyperbolic plane, is the diameter of a triangle the length of the bigger side (edge)? If yes, how to prove it?
The answer is yes. It can be proved e.g. from Hilbert's axioms for neutral geometry.
Given a triangle $\triangle abc$ and points $p,q$ inside this triangle (including a boundary), $|pq|\leq \max\{|ab|,|bc|,|ac|\}$.
Sketch of the proof. We can assume $p\neq q$. There are few cases to consider depending on whether $p$ and $q$ lie in the interior of the triangle, on the sides or coincide with vertices.
First assume $p=a$ and $q$ lies on side $bc$. Since the angles $\angle aqb,\angle aqc$ are complementary, one of them is not acute. Say $\angle aqb$ is right or obtuse. Then angle $\angle abq$ must be acute and therefore smaller than $\angle aqb$. Then $|pq|=|aq|<|ab|$ (the side opposite smaller angle is shorter than the side opposite greater angle). Analogously $|pq|<|ac|$ in case $\angle aqc$ is right or obtuse.
Assume now that $p$ lies on side $ac$ and $q$ on side $bc$. By the same reasoning as before (applied to triangle $\triangle pbc$) we conclude that $|pq|<\max\{|pb|,|pc|\}$. But $|pb|<|ab|$ and $|pc|<\max\{|bc|,|ac|\}$ (again the same reasons).
Finally, when both $p$ and $q$ lie in the interior, then line connecting these points must intersect the boundary in two points, say $u,v$ such that $p,q$ lie inside segment $uv$. So $|pq|<|uv|<\max\{|ab|,|bc|,|ac|\}$ by previous cases.
Remaining cases are easy to consider or symmetrical to ones above.