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Rule

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.

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 $⊙A.$
Because a rigid motion maps one circle onto the other, it is concluded that both circles are congruent.

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