#11 Prove that the perpendicular bisectors of a cyclic quadrilateral are concurrent if and only if the quadrilateral is cyclic.
#11 Prove that the perpendicular bisectors of a cyclic quadrilateral are concurrent if and only iff the quadrilateral is cyclic.
A cyclic quadrilateral has all its points on a common circle. Given any two points we construct the line segment between them this is a chord of the circle. Construct perpendicular bisector of the chord. Claim the center lies on the bisector. Form segments CA, AG, DA. Since CA and DA are radii they are congruent. Since GA is a bisector CG and DG are congruent and AG is self congruent. Thus by SSS triangle congruence gives us the triangle ACG and triangle ADG are congruent. Since angle AGC and AGD are congruent 90 degree angles by construction we can conclude that the center A is on the perpendicular bisector of any chord. Thus given any two chords the perpendicular bisectors will be concurrent at the center. Thus the perpendicular bisectors of a cyclic quadrilateral are concurrent.
Conversely, given 2, 3 or 4 lines concurrent at the center of a circle then the quadrilaterals that have those concurrent lines as perpendicular bisectors of their sides is not unique and many different quadrilaterals can be constructed.