9. As shown in the figure, in the isosceles right triangle $A B C$, $\angle B A C=90^{\circ}$, points $D, E$ are on $B C$, and $C D=B E, C N \perp A E$ at point $N, C N$ intersects $A D$ at point $M$. If $\triangle A M N$ is an isosceles right triangle, $A N=\sqrt{2-\sqrt{2}}$, then the area of $\triangle A B C$ is $\qquad$
$2$