Prove that if the three perpendicular bisectors of the sides of [math]\Delta[/math]ABC in the hyperbolic plane are concurrent at a point O, then the circle with center O and radius OA also contains the points B and C. Prove that this circumcircle is unique.[br][br][b][u]Proof:[/u][/b][br][br]([math]\Longrightarrow[/math])[br][br]Let [math]\Delta[/math]ABC be a triangle in the hyperbolic plane such that the three perpendicular bisectors of the sides of [math]\Delta[/math]ABC are concurrent at point O. Let [math]l_1[/math] be the perpendicular bisector of AC which intersects AC at point X, [math]l_2[/math] be the perpendicular bisector of BC which intersects BC at point Y, and [math]l_3[/math] be the perpendicular bisector of BC which intersects BC at point Z. We wish to show that points A, B, and C lie on a common circle, E, centered at O with radius OA. Thus, we need to show that [br]OA = OB = OC.[br][br]First, construct line segments OA, OB, and OC. Note that X is the midpoint of AC because [math]l_1[/math] is the perpendicular bisector of AC. From this, we know that [math]\angle[/math]AXO = 90 = [math]\angle[/math]CXO. Furthermore, we know that OX = OX and AX = CX. Therefore, by SAS congruence, we know that [math]\Delta[/math]AXO [math]\cong[/math] [math]\Delta[/math]CXO. Thus, we know that OA = OC.[br][br]Similarly, note that Y is the midpoint of BC because [math]l_2[/math] is the perpendicular bisector of BC. From this, we know that [math]\angle[/math]BYO = 90 = [math]\angle[/math]CYO. Furthermore, we know that OY = OY and BY = CY. Therefore, by SAS congruence, we know that [math]\Delta[/math]BYO = [math]\Delta[/math]CYO. Thus, we know that OC = OB.[br][br]By transitivity, we know that because OC = OB and OA = OC, we have that OB = OA.[br][br]Since OA = OB = OC, by the definition of circle, we have that A, B, and C are all common points on a circle, E, centered at O with radius OA.[br][br]Also note, as a consequence of our result that OA = OB = OC, we know that for every point on the perpendicular bisector of each side of our triangle, [math]\Delta[/math]ABC, they are all equidistant from the endpoints of the respective side.[br][br]([math]\Longleftarrow[/math])[br][br]Let [math]\Delta[/math]ABC be a triangle in the hyperbolic plane. For any circle that contains A, B, and C, the center of the circle is equidistant from A, B, and C. Using our consequence from the previous section, we know that the center must lie on each of the perpendicular bisectors of [math]\Delta[/math]ABC. However, there is only one intersection, if it exists. Therefore, the intersection, if it exists, is the center of our circle which contains A, B, and C.[br][br]Note: The picture below is glitchy because of the custom tools, but as long the the perpendicular bisectors are concurrent, the proof stands. (Discussed with Dr. Cochran)[br]