# Proof 4.11

Prove that the perpendicular bisectors of a quadrilateral are concurrent if and only if the quadrilateral is cyclic.
[] Let ABCD be a quadrilateral with concurrent perpendicular bisectors (Proposition 10). Let the point of concurrency be I. Let the midpoint of  be E, so . By construction,  because they are perpendicular. So by SAS (Proposition 4), . This means that . Similarly, all of the vertices of the quadrilateral are equal length from point I. Therefore by the definition of a circle, we know that the quadrilateral ABCD is cyclic based on the definition of a cyclic quadrilateral.
Let MNOP be a cyclic quadrilateral and let A be the center of the circle. Based on the definition of a circle, we know that . Let B me the midpoint of  (Proposition 10), so that . We also know that  by Common Notion 1. Therefore, we know that  by SSS (Proposition 8). This indicates that the angles within the triangles are also congruent. Therefore, we know  and Proposition 13 tells us that these angles are supplementary, so we can conclude that they are 90 degrees. Therefore,  is the perpendicular bisector of . Similarly, we can create all of the perpendicular bisectors of MNOP with the center A or concurrent point.
In conclusion based on the two arguments above, we can conclude that the perpendicular bisectors of a quadrilateral are concurrent if and only if the quadrilateral is cyclic.