Proof 11.5.12 (circumcircle proof)
Prove that if the three perpendicular bisectors of the sides of triangle ABC in the hyperbolic plane are concurrent at a point O, then the circle with center O and radius OA also contain the points B and C. Prove that this circumcircle is unique.
Consider in the hyperbolic plane. Let the midpoint of be point , let the midpoint of be point , and let the midpoint of be point . Assume that the perpendicular bisectors of each side of intersect at point .
We can conclude that by SAS congruence because , , and all right angles are congruent thus, . Thus, we can conclude that .
Similarly, we can conclude that by SAS congruence because , , and . Thus, we can conclude that .
Then, by the transitivity of equality, . Thus, a circle center at with radius , must also contain the points B and C.
This circumcircle must be unique. To prove this, assume it is not unique. That is, assume that there is more than one circumcircle centered at and passing through the points , , and . This implies that the radius of one circumcircle does not equal that of another circumcircle. But the segments , , and are also radii and are the same for each circumcircle. Thus, the two circumcircles must be congruent. That is, if the circumcircle of in the hyperbolic plane exists, then it is unique.