2. As shown in Figure 15, in trapezoid $A B C D$, $A D / / B C (B C > A D), \angle D=90^{\circ}, B C=$ $C D=12, \angle A B E=45^{\circ}$. If $A E=10$, then the length of $C E$ is $\qquad$ .
(2004, National Junior High School Mathematics Competition)
(Hint: First, complete the trapezoid $ABCD$ into a square $CBFD$, then rotate $\triangle ABF$ $90^{\circ}$ around point $B$ to the position of $\triangle BCG$. It is easy to see that $\triangle ABE \cong \triangle GBE$. Therefore, $AE = EG = EC + AF = 10$. Let $EC = x$. Then $AF = 10 - x$, $DE = 12 - x$, $AD = 2 + x$. By the Pythagorean theorem, $(12 - x)^2 + (2 + x)^2 = 10^2$. Solving this, we get $x = 4$ or 6.)