**Corollary:** A circle is symmetrical about any of its diameter.

By *symmetrical*, we mean that the circle can be *divided into two congruent parts *by any of its diameter. Consider a circle with center O and AB as a diameter:

We need to show that this circle can be divided into two congruent parts by AB.

**Proof:** Take any point X on the circumference of the circle, and draw XY perpendicular to AB, such that it intersects AB at Z:

In our theorem, we proved that the perpendicular bisector of any chord of a circle passes through its center. Here, we have a chord XY, and a perpendicular AB to XY passing through O. This must mean that AB bisects XY as well, or XZ = YZ.

Thus, for any point X on the circumference of the circle, its *mirror image* Y in the *mirror* AB also lies on the circle (reflect on this for a minute). This means that the circle is symmetrical about AB.