Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given two points $p$ and $q$ their bisector is defined to be $l(p,q)=\{z:d(p,z)=d(q,z)\}$.

Due to the construction in Euclidean geometry, we know that $l(p,q)$ is a line, that is, for $x,y,z\in l(p,q)$, we have $d(x,y)+d(y,z)=d(x,z)$, which charactorizes lines.

I wonder whether this is true for other geometries. That is, does the bisector always satisfy the above charactorization?

I think about this problem when trying to prove bisectors are 'lines' in hyperbolic geometry (upper half plane) where the metric is different from Euclidean, only to notice even the Euclidean case is not so easy.

Any advice would be helpful!

share|cite|improve this question
up vote 1 down vote accepted

Let $A$ and $B$ be the two given points and let $M$ be the midpoint of $AB$, i.e., $M\in A\vee B$ and $d(M,A)=d(M,B)$. Let $X\ne M$ be an arbitrary point with $d(X,A)=d(X,B)$. Then the triangles $\Delta(X,A,M)$ and $\Delta(X,B,M)$ are congruent as corresponding sides have equal length. It follows that $\angle(XMA)=\angle(XMB)={\pi\over2}$ which implies that the line $m:=X\vee M$ is the unique normal to $A\vee B$ through $M$. Conversely, if $Y$ is an arbitrary point on this line, then $d(Y,M)=d(Y,M)$, $d(M,A)=d(M,B)$ and $\angle(Y,M,A)=\angle(Y,M,B)={\pi\over2}$. Therefore the triangles $\Delta(Y,M,A)$ and $\Delta(Y,M,B)$ are congruent, and we conclude that $d(Y,A)=d(Y,B)$.

The above argument is valid in euclidean geometry as well as in spherical and hyperbolic geometry. Note that a spherical triangle is completely determined (up to a motion or reflection on $S^2$) by the lengths of its three sides or by the lengths of two sides and the enclosed angle, and the same is true concerning hyperbolic triangles.

share|cite|improve this answer
Well, basically you are saying triangles are rigid in hyperbolic geometry if I got it right. But I am not sure that this is true. Could you point to some reference? Thanks! – Hui Yu Feb 23 '12 at 7:52
In hyperbolic trigonometry you have pretty much the same theorems and formulas as in spherical trigonometry, only the trigonometric functions $\sin$, etc., are replaced by the corresponding hyperbolic functions $\sinh$, etc. See here: – Christian Blatter Feb 23 '12 at 10:26
Actually, what confuses me is how you proved two triangles are congruent. It seems you used SSS condition and SAS, which are both OK for Euclidean. I am not sure they are true in hyperbolic case. Sorry but I am not really an expert in this area. Thanks! – Hui Yu Feb 23 '12 at 10:53
@Hui Yu: This might interest you: – Christian Blatter Feb 23 '12 at 10:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.