If the two large circles are congruent, what is the radius of the smallest circle?

The answer is 1.0.

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We note that $\triangle CDE$ and $\triangle CDF$ are congruent. This means that $CD$ bisects $\angle C$ and by angle bisector theorem ,

$\frac {DB}{AB} = \frac {BC}{CA} \implies \frac {DB}{14-DB} = \frac {13}{15} \implies DB = \frac {13 \cdot 14}{15+13} = 6.5$

We also note that a $\text{13-14-15}$ triangle is made up of a $\text{3-4-5}$ right triangle and a $\text{5-12-13}$ right triangle sharing an altitude of $12$ ; and that $\triangle BDF$ is a $\text{5-12-13}$ right triangle. Since $DB = 6.5$ , $BF = \frac 5{13} \cdot 6.5 = 2.5$ and $DF = \frac {12}{13} \cdot 6.5 = 6$ . And the radius of the incircle of $\triangle BDF$ ,

$r = \frac {\frac 12 BF \cdot DF}{\frac 12(BF+DF+DB)} = \frac {2.5 \cdot 6}{2.5+6+6.5} = \boxed 1$