Triangle \(ABC\) has \(AB=13, BC=14\) and \(AC=15\). Let \(P\) be the point on \(\overline{AC}\) such that \(PC=10\). There are exactly two points \(D\) and \(E\) on line \(BP\) such that quadrilaterals \(ABCD\) and \(ABCE\) are trapezoids. The distance \(DE\) can be expressed in the form \(k\sqrt{m}\). Please find the value of \(k + m\).

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