Because $\angle BCA$ is common to both $\triangle ABC$ and $\triangle DBE$, $AB = 4$, and $BE = 3$, we have that $\triangle ABC\sim \triangle DBE$ by the $\textit{SAS}$ similarity criterion. Since the ratios of the corresponding sides of two similar isosceles right triangles are equal, $\triangle ABC \sim \triangle DBE$ means We also note that $CE = CA + AE$, so $CA = CE - AE = 8 - DE$.
$\frac {DE}{DB} = \frac {4}{3}$. Multiplying both sides of this equation by $4$ yields
\[DE = \frac {4}{3}DB\]
Finding $DB$ is the key step to solving the problem. Drop an altitude from $E$ to $\overline{AD}$, and let the point of intersection be $G$. Note that $GE = EG = 4 - 3 = 1$, {By the Pythagorean Theorem} $AG = \sqrt {3^2 - 1^2} = \sqrt {8}$. $\triangle DGB$ and $\triangle AGB$ are similar right triangles, so
\[\frac {BG}{GB} = \frac {AG}{DB}\]
\[DB = \frac {AG\cdot BG}{GB}\]
Because $\triangle GBE$ is an isosceles right triangle, $BG = GE = 1$. We see that $GB + GE = 8 - DE = 1 + 1 = 2$ (by segments on a triangle), so $GB = 2 - DE$. Substituting gives
\[DB = \frac {\sqrt {8}\cdot (2 - DE)}{2 - DE} = \sqrt {8}\]
Thus
\[\frac {4}{3} = \frac {\sqrt {8}}{DE}\]
\[DE = \frac {\sqrt {8\cdot 9}\cdot 2}{4} = \boxed{3\sqrt {2}}.\]
Let $A$, $B$, $C$, $D$, and $E$ be points in the plane such that:
* $D$ is on line $AE$ with $\overline{CD} \perp \overline{AE}$.
* $B$ is on line $CE$ with $\overline{AB} \perp \overlin{CE}$.
* $AB = 4$, $BE = 3$, and $CE = 8$.
Find the length $DE$.
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