Consider [math]\triangle ABC[/math] in the hyperbolic plane. Let the midpoint of [math]\overline{AB}[/math] be point [math]D[/math], let the midpoint of [math]\overline{BC}[/math] be point [math]E[/math], and let the midpoint of [math]\overline{AC}[/math] be point [math]F[/math]. Assume that the perpendicular bisectors of each side of [math]\triangle ABC[/math] intersect at point [math]O[/math]. [br][br]We can conclude that [math]\triangle ADO\cong\triangle BDO[/math] by SAS congruence because [math]\overline{AD}\cong\overline{BD}[/math], [math]\overline{DO}\cong\overline{DO}[/math], and all right angles are congruent thus, [math]\angle ADO\cong\angle BDO[/math]. Thus, we can conclude that [math]\overline{AO}\cong\overline{BO}[/math]. [br][br]Similarly, we can conclude that [math]\triangle BEO\cong\triangle CEO[/math] by SAS congruence because [math]\overline{BE}\cong\overline{CE}[/math], [math]\overline{EO}\cong\overline{EO}[/math], and [math]\angle BEO\cong\angle CEO[/math]. Thus, we can conclude that [math]\overline{BO}\cong\overline{CO}[/math]. [br][br]Then, by the transitivity of equality, [math]\overline{AO}\cong\overline{BO}\cong\overline{CO}[/math]. Thus, a circle center at [math]O[/math] with radius [math]\overline{OA}[/math], must also contain the points B and C. [br][br]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 [math]O[/math] and passing through the points [math]A[/math], [math]B[/math], and [math]C[/math]. This implies that the radius of one circumcircle does not equal that of another circumcircle. But the segments [math]\overline{OA}[/math], [math]\overline{OB}[/math], and [math]\overline{OC}[/math] are also radii and are the same for each circumcircle. Thus, the two circumcircles must be congruent. That is, if the circumcircle of [math]\triangle ABC[/math] in the hyperbolic plane exists, then it is unique.