Triangle $ABC$ is a right triangle with $\angle ACB$ as its right angle, $m\angle ABC = 60^\circ$, and $AB = 10$. Let $P$ be randomly chosen inside $ABC$, and extend $\overline{BP}$ to meet $\overline{AC}$ at $D$. Find the probability that $BD > 5\sqrt{2}$. The original answer is in the form \(\frac{k-\sqrt{m}}{n}\), where the fraction is in simplest form. Please provide the value of \(k + m + n\).

Let's think step by step and output the final answer within \boxed{}.
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