In Parallelogram ABCD, \(AB=13\), \(BC=11\), and \(AC=20\). Two congruent circles are inscribed in \(\triangle ACD\) and \(\triangle ACB\) as shown in the figure.

The shortest distance between the centers of the two inscribed circles can be expressed in the form \(a\sqrt{b}\). Where \(b\) is a square-free integer. Find the value of \(a+b\).

- Found this somewhere.

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