Triangle $ABC$ has $AB=13,BC=14$ and $AC=15$. Let $P$ be the point on $\overline{AC}$ such that $PC=6$ (so $AP=9$). There are exactly two points $D$ and $E$ on line $BP$ such that quadrilaterals $ABCD$ and $ABCE$ are trapezoids. What is the distance $DE?$

1. Place $A=(0,0)$, $C=(15,0)$, and $B=(33/5,56/5)$. Since $PC=6$, $P=(9,0)$ and $\overrightarrow{PB}=(-12/5,56/5)$, so $PB=4\sqrt{205}/5$.
2. Write $D=P+s\overrightarrow{PB}=(9-\frac{12}5s,\frac{56}5s)$. The condition $AD\parallel BC$, with $\overrightarrow{BC}=(\frac{42}5,-\frac{56}5)$, gives $-(9-\frac{12}5s)-\frac{42}5s=0$, so $s=-\frac32$. Similarly $E=P+s\overrightarrow{PB}$ with $CE\parallel AB$, $\overrightarrow{AB}=(\frac{33}5,\frac{56}5)$, gives $-6-\frac{12}5s-\frac{33}5s=0$, so $s=-\frac23$. Thus $D=P-\frac32\overrightarrow{PB}$ and $E=P-\frac23\overrightarrow{PB}$.
3. Therefore, $DE=\left(\frac32-\frac23\right)PB=\frac56\cdot\frac{4\sqrt{205}}5=\frac{2\sqrt{205}}3$.