# Proof that in hyperbolic geometry the length of the mid-segment is never half the length of the segment.

Prove that in hyperbolic geometry there is no triangle $ABC$ such that the length of a mid-segment is half the length of the corresponding segment.

In other words: If there exist a triangle $ABC$, such that the length of a mid-segment is half the length of the corresponding segment, then it is only possible in Euclid geometry.

My attempt:

I think the structure of the solution must be the following: We should suppose that in hyperbolic geometry there exists triangle with mentioned property and then we should conclude that axiom about parallel lines in hyperbolic geometry fails : so there exists the line $L$ and the point outside the line $P$, such that there are 0 or 1 lines through $P$ that are parallel to $L$ and no more, thus contradicting the definition of hyperbolic geometry. I also think that I should find $L$ and $P$ by looking at $ABC$. But I am not able to move forward.

Update:

There is another possible approach: it is enough to prove that in hyperbolic geometry the length of mid segment is always less than the length of the segment.

• Here's a trigonometric solution: In $\triangle ABC$, writing $a_2$ and $b_2$ for $a/2$ and $b/2$, the hyperbolic Law of Cosines gives $$\cosh c = \cosh 2 a_2 \cosh 2b_2 - \sinh 2 a_2 \sinh 2 b_2 \cos C$$ $$\cosh m = \cosh a_2 \cosh b_2 - \sinh a_2 \sinh b_2 \cos C$$ where $m$ is the mid-segment corresponding to $c$. With a bit of work, we find $$\cosh c - \cosh 2m = \cosh c - ( 2 \cosh^2 m - 1 ) = 2\sinh^2 a_2 \sinh^2 b_2 \sin^2 C \geq 0$$ with "$=0$" in degenerate cases. We conclude that $c > 2 m$ for non-degenerate $\triangle ABC$. – Blue May 29 '17 at 16:23
• I think it's more-appropriate for me to post a full solution here (which I will do shortly), and for you to delete your other question. – Blue May 29 '17 at 16:27
• Ok, I've deleted another question, please post your answer. As long as I understand it and work thought it I will be happy to accept. – Hedgehog May 29 '17 at 16:38

Converting a comment into an answer, as requested:

Here's a trigonometric solution: In $\triangle ABC$, writing $a_2$ and $b_2$ for $a/2$ and $b/2$, the hyperbolic Law of Cosines gives $$\cosh c = \cosh 2 a_2 \cosh 2b_2 - \sinh 2 a_2 \sinh 2 b_2 \cos C$$ $$\cosh m = \cosh a_2 \cosh b_2 - \sinh a_2 \sinh b_2 \cos C$$ where $m$ is the mid-segment corresponding to $c$. With a bit of work (see below), we find $$\cosh c - \cosh 2m = \cosh c - ( 2 \cosh^2 m - 1 ) = 2\sinh^2 a_2 \sinh^2 b_2 \sin^2 C \geq 0 \tag{\star}$$ with "$=0$" in degenerate cases. We conclude that $c > 2 m$ for non-degenerate $\triangle ABC$. $\square$

Getting $(\star)$ is a matter of slogging-through some symbol manipulation. To save some space and typing effort, I'll abbreviate "$\cosh u$" as simply "$u$" and "$\sinh u$" as "$\widehat{u}$".

The double-angle formulas for hyperbolic sine and cosine give $$\cosh 2u = 2 \cosh^2 u - 1 \qquad\qquad \sinh 2u = 2 \sinh u \cosh u$$

So, from the above, we have \begin{align} \cosh c &= ( 2 a_2^2 - 1 )( 2 b_2^2 - 1 )- 4 a_2 b_2 \widehat{a_2}\widehat{b_2} \cos C \\ &= 4 a_2^2 b_2^2 - 2 a_2^2 - 2 b_2^2 + 1 - 4 a_2 b_2 \widehat{a_2}\widehat{b_2} \cos C \end{align} Also, \begin{align} \cosh 2m &= 2 \cosh^2 m - 1 \\ &= 2 ( a_2 b_2 - \widehat{a_2}\widehat{b_2} \cos C)^2 - 1 \\ &= 2 a_2^2 b_2^2 + 2 \widehat{a_2}^2 \widehat{b_2}^2 \cos^2 C - 4 a_2 b_2 \widehat{a_2}\widehat{b_2} \cos C - 1 \end{align} Therefore, \begin{align} \cosh c - \cosh 2m &= 2 a_2^2 b_2^2 - 2 a_2^2 -2 b_2^2 + 2 - 2 \widehat{a_2}^2 \widehat{b_2}^2\cos^2 C \\ &= 2 ( a_2^2 - 1 )( b_2^2 - 1 ) - 2 \widehat{a_2}^2 \widehat{b_2}^2\cos^2 C \\ &= 2 \widehat{a_2}^2 \widehat{b_2}^2 - 2 \widehat{a_2}^2 \widehat{b_2}^2\cos^2 C \\ &= 2 \widehat{a_2}^2 \widehat{b_2}^2 \left( 1- \cos^2 C \right) \\ &= 2 \sinh^2 a_2 \sinh^2 b_2 \sin^2 C \quad \square \end{align}

• Terrific, thanks, accepted even without understanding it, but I will understand soon. I have second hyperbolic geometry unanswered question, it is also half-solved by me, can you please check? Thanks a lot again. – Hedgehog May 29 '17 at 16:50
• I may not have time to get to your other question today. We'll see. – Blue May 29 '17 at 16:54