# Why are perpendicular bisectors 'lines'?

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.

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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.
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: en.wikipedia.org/wiki/Hyperbolic_geometry – Christian Blatter Feb 23 '12 at 10:26