smc

Toss on the Cartesian plane with $A=(5,12),B=(0,0),$ and $C=(14,0)$. Then $\overline{AD}\parallel \overline{BC},\overline{AB}\parallel \overline{CE}$ by the trapezoid condition, where $D,E\in \overline{BP}$. Since $PC=10$, point $P$ is $\frac{10}{15}=\frac{2}{3}$ of the way from $C$ to $A$ and is located at $(8,8)$. Thus line $BP$ has equation $y=x$. Since $\overline{AD}\parallel \overline{BC}$ and $\overline{BC}$ is parallel to the ground, we know $D$ has the same $y$-coordinate as $A$, except it'll also lie on the line $y=x$. Therefore, $D=(12,12).■$ To find the location of point $E$, we need to find the intersection of $y=x$ with a line parallel to $\overline{AB}$ passing through $C$. The slope of this line is the same as the slope of $\overline{AB}$, or $\frac{12}{5}$, and has equation $y=\frac{12}{5}x-\frac{168}{5}$. The intersection of this line with $y=x$ is $(24,24)$. Therefore point $E$ is located at $(24,24).■$ The distance $DE$ is equal to the distance between $(12,12)$ and $(24,24)$, which is $\boxed{12\sqrt{2}}$.