1. **State the problem:** We need to prove that the perpendicular bisector of a chord in a circle bisects the central angle subtended by the chord. 2. **Set up the scenario:** Consider a circle with center $O$ and a chord $AM$. Let $OT$ be the perpendicular bisector of $AM$, where $T$ is the midpoint of $AM$. 3. **Recall definitions:** The perpendicular bisector means $OT \perp AM$ and $T$ is the midpoint of $AM$, so $AT = TM$. 4. **Consider triangles:** Look at triangles $\triangle OAT$ and $\triangle OMT$. 5. **Equal sides:** Both triangles share side $OT$, and by definition $AT = TM$ since $T$ is the midpoint. 6. **Right angles:** Since $OT$ is perpendicular to $AM$, $\angle OTA = \angle OTM = 90^\circ$. 7. **Use RHS (Right angle-Hypotenuse-Side) congruence:** Triangles $\triangle OAT$ and $\triangle OMT$ have: - Right angle at $T$, - Hypotenuse $OA = OM$ (both are radii of the circle), - Side $AT = TM$ (since $T$ bisects $AM$). Therefore, $\triangle OAT \cong \triangle OMT$. 8. **Corresponding angles:** From congruent triangles, angles $\angle AOT$ and $\angle MOT$ are equal. 9. **Conclusion:** Line $OT$ bisects $\angle AOM$ because it divides it into two equal angles. **Final answer:** The perpendicular bisector $OT$ of chord $AM$ bisects the central angle $\angle AOM$ subtended by $AM$.