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Based on the above diagram, the theorem can be written as follows.
⊙A≅⊙B⇔r1=r2
The biconditional statement will be proved separately.
Consider two congruent circles ⊙A and ⊙B and a point on each one.
Because ⊙A≅⊙B, the distance from the center to a point on the circle is the same for both circles. Therefore, AC≅BD, which implies that both circles have the same radius. That is, r1=r2.
It has been proved that if two circles are congruent, then they have the same radius.
Consider now two circles with the same radius.
Next, ⊙B can be translated so that point B is mapped onto point A. The image of ⊙B is ⊙B′, which is a circle centered at A. Since the circles have the same radius, this translation maps ⊙B onto ⊙A.