19. 如图所示, $\triangle A C D$ 是等边三角形, $\triangle A B C$ 是等腰直角三角形, $\angle A C B=90^{\circ}$, $B D$ 交 $A C$ 于 $E, A B=2$.

(1)求 $\cos \angle C B E$ 的值;

<ImageHere>

(2)求 $A E$.
(1) $\because \angle B C D=90^{\circ}+60^{\circ}=150^{\circ}, C B=A C=C D$,

$\therefore \angle C B E=15^{\circ}$.

$\therefore \cos \angle C B E=\cos \left(45^{\circ}-30^{\circ}\right)=\frac{\sqrt{6}+\sqrt{2}}{4}$.

(2)在 $\triangle A B E$ 中, $A B=2$,

由正弦定理得 $\frac{A E}{\sin \angle A B E}=\frac{A B}{\sin \angle A E B}$,

即 $\frac{A E}{\sin \left(45^{\circ}-15^{\circ}\right)}=\frac{2}{\sin \left(90^{\circ}+15^{\circ}\right)}$,

故 $A E=\frac{2 \sin 30^{\circ}}{\cos 15^{\circ}}=\frac{2 \times \frac{1}{2}}{\frac{\sqrt{6}+\sqrt{2}}{4}}=\sqrt{6}-\sqrt{2}$.