Reflection and Angle Bisector

Since point $E$ is point $D$ reflected across $\overline{AB}$ , $BE=BD$ . Therefore, triangle $BED$ is isoscles, implying that $\overline{AB}$ bisects $\angle DBE$ . Let $\frac{1}{2}\measuredangle DBE=x$ .

$\overline{AB}$ is a transversal of the two parallel lines $\overline{BE}$ and $\overline{AC}$ , so by alternate interior angles $\measuredangle BAC = x$ . Therefore, triangle $ABC$ is isosceles with $AC=BC=6$ . Use angle bisector theorem to find that $2BD = CD$ , so $BD=2$ and $CD=4$ . Since $BE=BD$ , $BE=2$ .

Drawing the altitude of triangle $ABC$ that passes through $\overline{AB}$ creates a right triangle with leg $1.5$ and hypotenuse $6$ . Therefore, $\cos{x} = \frac{1}{4} \implies \cos{2x} = 2(\frac{1}{4})^2-1 = -\frac{7}{8}$ . Now use law of cosines on triangle $BCE$ to find $CE^2$ :

$CE^2=2^2+6^2-2(2)(6)\cos{2x}=40-24(-\frac{7}{8})=\boxed{61}$ .