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Based on the above diagram, the theorem can be written as follows.
⊙ A ≅ ⊙ B ⇔ r_1=r_2
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, r_1=r_2.
It has been proved that if two circles are congruent, then they have the same radius.
Consider now two circles with the same radius.