Proof 4.3.11
Cyclic quadrilateral ABCD.
Conjecture: The perpendicular bisectors of a quadrilateral are concurrent if and only if the quadrilateral is cyclic.
Forwards implication: If the perpendicular bisectors of a quadrilateral are concurrent, then the quadrilateral is cyclic.
Proof of forwards implication (by direct proof): Consider the quadrilateral ABCD. Assume that the perpendicular bisectors of ABCD are concurrent. That is, the perpendicular bisectors of ABCD all intersect at some point, P.
Consider the side AB of the quadrilateral. Because any point on a perpendicular bisector of a line segment must be an equal distance from each endpoint (proved in class), we can assume that .
Similarly, when considering side BC, we can assume that .
When considering side CD, we can assume that .
Finally, when considering side DA, we can assume that .
That is, . Thus, the points, A, B, C, D lie on a circle centered at P. Therefore, the quadrilateral ABCD is cyclic.
Backwards implication: If a quadrilateral is cyclic, then the perpendicular bisectors of the quadrilateral are concurrent.
Proof of backwards implication (by direct proof): Consider the cyclic quadrilateral ABCD.By definition, we know that the points A, B, C, D must lie on a circle such that A, B, C, D are all equal lengths from the center of the circle, P. That is, .
Given that a point lies on a perpendicular bisector of a segment if it is an equal distance from both endpoints (proved in class) and , we can conclude that P is on the perpendicular bisector of side AB.
Similarly, because , we can conclude that P is on the perpendicular bisector of side BC. Because , we can conclude that P is on the perpendicular bisector of side CD. Finally, because , we can conclude that P is on the perpendicular bisector of side DA.
Because P lies on every perpendicular bisector of quadrilateral ABCD, we know that the perpendicular bisectors are concurrent at point P.