1. In trapezoid $A B C D$, it is known that $A D / / B C(B C>$ $A D), \angle D=90^{\circ}, B C=C D=12, \angle A B E=45^{\circ}$. If $A E=10$, then the length of $C E$ is $\qquad$
(2004, "Xinli Cup" National Junior High School Mathematics Competition)
(提示: Extend $D A$ to point $F$, complete the trapezoid into a square $C B F D$, then rotate $\triangle A B F$ $90^{\circ}$ around point $B$ to the position of $\triangle B C G$. It is easy to see that $\triangle A B E \cong \triangle G B E$. Therefore, $A E$ $=E G=E C+A F=10$. Let $E C=x$, then $A F=10-$ $x, D E=12-x, A D=2+x$. By the Pythagorean theorem, $(12-x)^{2}+(2+x)^{2}=10^{2}$. Solving this, we get $x=4$ or 6. )