Problem - 2998
In isosceles right triangle $\triangle{ABC}$, $\angle{C}$ is the right angle. Points $D$ and $E$ are on side $AB$ such that $\angle{DCE}=45^\circ$, $AD=25$, and $EB=16$. Let the length of $DE$ be $x$. Find $x^2$.
Rotate $\triangle{ACD}$ around point $C$ for $90^\circ$ counterclockwise, and let the new location of $D$ be $D'$. Because $AC=BC$ and $\angle{ACB}=90^\circ$, point $A$ will move to $B$ after rotation.
Thus $$\angle{BCD'}=\angle{ACD} \implies \angle{ECD'} = \angle{ECB} + \angle{BCD'} = \angle{ECB}+\angle{ACD}= 45^\circ = \angle{DCE}$$
By SAS, we find $\triangle{CDE}\cong\triangle{CD'E}$, or $DE=D'E$. Furthermore, it is easy to see that $$\angle{D'BA} = \angle{D'BC} + \angle{CBA} = \angle{DAC}+\angle{CBA}=90^\circ$$ Therefore $\triangle{D'BE}$ is a right triangle. Hence $$x^2={EB^2 + D'E^2} = 25^2 + 16^2=\boxed{881}$$