1. **Problem statement:** Prove that the diameter of a circle which bisects a chord also bisects the angle subtended by the chord at the center of the circle. 2. **Given:** A circle with center $O$, a chord $AB$, and a diameter $CD$ that bisects the chord $AB$ at point $M$. 3. **To prove:** The diameter $CD$ bisects the angle $AOB$ subtended by chord $AB$ at the center $O$, i.e., $\angle AOM = \angle MOB$. 4. **Key properties:** - The diameter passes through the center $O$. - $M$ is the midpoint of chord $AB$, so $AM = MB$. - Radii $OA$ and $OB$ are equal. 5. **Proof steps:** - Since $M$ is midpoint of $AB$, $AM = MB$. - Triangles $OAM$ and $OBM$ share side $OM$. - Radii $OA = OB$. 6. **By the Side-Side-Side (SSS) congruence criterion:** $$\triangle OAM \cong \triangle OBM$$ 7. **Therefore, corresponding angles are equal:** $$\angle AOM = \angle MOB$$ 8. **Conclusion:** The diameter $CD$ bisects the angle $AOB$ subtended by chord $AB$ at the center $O$. This completes the proof.