[b][u]Theorem 4.1:[/u][/b] A quadrilateral is cyclic if and only if the perpendicular bisectors of the sides are concurrent.[br][br]Proof:[br][math]\Longrightarrow[/math]Let ABCD be a quadrilateral with concurrent perpendicular bisectors which intersect at point E. Let X be the midpoint of AB. From this, we know that AX = XB. By construction, we know that [math]\angle[/math]AXE = [math]\angle[/math]BXE because XE is perpendicular to AB since we have assumed each perpendicular bisector share point E and X is the midpoint of AB. By SAS, we have that [math]\Delta[/math]AXE = [math]\Delta[/math]BXE. From this, we can conclude that AE = BE. By an analogous argument for each side of ABCD, we would have that all line segments from a vertex of ABCD to point E are equal. By the definition of a circle, it follows that points A,B,C,D all lie on some circle, O. Therefore, by the definition of cyclic, we know that our quadrilateral is a cyclic quadrilateral.

[math]\Longleftarrow[/math] Let ABCD be a cyclic quadrilateral inscribed in a circle centered at X. By the definition of circle, we know that AX = DX = BX = CX. Let S be the midpoint of AD. Thus, we know that AS = DS. We also know that SX = SX. By SSS criterion, we have that [math]\Delta[/math]ASX = [math]\Delta[/math]DSX. Therefore, we know that [math]\angle[/math]ASX = [math]\angle[/math]DSX. By Proposition 13, we know that these angles must be supplementary. Thus, we can conclude [math]\angle[/math]ASX and [math]\angle[/math]DSX are right angles. Therefore, SX is the perpendicular bisector of AD. Analogously, we find the midpoint of each side and using the same argument to show that all perpendicular bisectors of ABCD are concurrent at X.