1. Problem (i): Prove that a point O equidistant from points A and B lies on the perpendicular bisector of segment AB. 2. Recall the definition: The perpendicular bisector of a segment is the line that is perpendicular to the segment at its midpoint. 3. Let M be the midpoint of AB. Since O is equidistant from A and B, we have $OA = OB$. 4. By the definition of midpoint, $AM = MB$. 5. Consider triangles $\triangle OAM$ and $\triangle OBM$. 6. They share side $OM$. 7. We have $OA = OB$ (given) and $AM = MB$ (midpoint). 8. By the Side-Side-Side (SSS) congruence criterion, $\triangle OAM \cong \triangle OBM$. 9. Therefore, angles $\angle OMA$ and $\angle OMB$ are equal. 10. Since these two angles are adjacent and form a straight line, each must be $90^\circ$. 11. Hence, $OM$ is perpendicular to $AB$ at $M$, so O lies on the perpendicular bisector of $AB$. 12. Problem (ii): Construct a diameter of a circle with unknown center using a straightedge and compass, then locate the center. 13. Step 1: Use the straightedge to draw any chord $AB$ on the circle. 14. Step 2: Use the compass to find the midpoint $M$ of chord $AB$ by constructing the perpendicular bisector of $AB$. 15. Step 3: Draw the perpendicular bisector line of $AB$ through $M$. 16. Step 4: Repeat steps 13-15 with another chord $CD$ to get its perpendicular bisector. 17. Step 5: The intersection point $O$ of the two perpendicular bisectors is the center of the circle. 18. Step 6: Draw the diameter by connecting $O$ to any point on the circle, for example $A$ and its opposite point on the circle along the line through $O$. This completes the construction and location of the center.