Converse Perpendicular Chord Bisector Theorem

It is given that $\overline{QS}$ is a perpendicular bisector of $\overline{TR}.$ Let $L$ be the center of the circle. Consider $△LPT$ and $△LPR.$

Segments $\overline{LT}$ and $\overline{LR}$ are radii of $\odot L.$ Since all radii of a circle are congruent, $\overline{LT}$ and $\overline{LR}$ are congruent. Additionally, by the definition of a perpendicular bisector, it can be stated $\overline{TP}$ and $\overline{RP}$ are congruent segments. Furthermore, by the Reflexive Property of Congruence, it can be said that $\overline{LP}$ is congruent to itself. This information can be seen in the diagram. Here, three sides of $△LPT$ are congruent to three sides of $△LPR.$ Therefore, by the Side-Side-Side Congruence Theorem, $△LPT$ and $△LPR$ are congruent triangles. Since corresponding parts of congruent triangles are congruent, $\angle LPT$ and $\angle LPR$ are congruent angles. By the definition of a chord, $\angle TPR$ is a straight angle. By dividing $180^{\circ }$ by $2,$ it can be calculated that $\angle LPT$ and $\angle LPR$ measure $90^{\circ }$ each. This means that they are right angles and that $\overline{LP}$ is perpendicular to $\overline{TR}.$ Therefore, $\overline{LP}$ is the perpendicular bisector of $\overline{RT}.$ Since there is only one perpendicular bisector that can be drawn to a segment, $L$ must lie on $\overline{QS}.$

Finally, the fact that $\overline{QS}$ contains the center of the circle leads to the conclusion that $\overline{QS}$ is a diameter of $\odot L.$