All Three Concurrency

Geometry Level 5

Triangle ABCABC is drawn with B=90\angle B=90^{\circ} and AC=1AC=1.

An angle bisector is drawn from AA hitting BCBC at DD. An altitude is drawn from BB hitting ACAC at EE. Finally, a median is drawn from CC hitting ABAB at FF.

Given that ADAD, BEBE, and CFCF are concurrent, then the area of triangle DEFDEF can be represented by 1a+bc\dfrac{1}{\sqrt{a+b\sqrt{c}}} where a,b,ca,b,c are integers with cc square-free. Find a+b+ca+b+c.

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