15. As shown in the figure, in the right-angled $\triangle \mathbf{ABC}$, $\angle C=90^{\circ}, AC=2, BC=1, D$ is on $\mathbf{AC}$. After folding $\triangle \mathbf{ADB}$ along the line $\mathbf{BD}$, point $\mathrm{A}$ lands at point $\mathrm{E}$. If $\mathrm{AD} \perp \mathrm{ED}$, then the area of $\triangle \mathrm{ABE}$ is $\qquad$ _.
【Analysis】 $\angle A D B=\left(360^{\circ}-90^{\circ}\right) \div 2=135^{\circ}, \angle B D C=45^{\circ}$
Therefore, $D C=B C=1, D E=A D=2-1=1$ $S_{\triangle A B E}=2 \times \frac{1}{2} \times A D \times B C+\frac{1}{2} \times A D \times D E=1.5$