Final answer:
The perpendicular bisector of a chord in a circle contains the center of the circle.
Step-by-step explanation:
To prove that the perpendicular bisector of a chord in a circle contains the center of the circle, we can use the fact that the perpendicular bisector of a chord passes through the center of a circle.
Let ABC be a chord in a circle with center O. We want to show that the perpendicular bisector of AB passes through O.
Since AB is a chord, it intersects the center O. Let P be the midpoint of AB. The line segment OP is a radius of the circle, and the line segment AB is perpendicular to OP at P. Therefore, the perpendicular bisector of AB passes through O, the center of the circle.