1. In trapezoid $A B C D, A D$ is parallel to $B C . \angle A=\angle D=45^{\circ}$, while $\angle B=\angle C=135^{\circ}$. If $A B=6$ and the area of $A B C D$ is 30 , find $B C$.
Solution: $2 \sqrt{2}$
Draw altitudes from $B$ and $C$ to $A D$ and label the points of intersection $X$ and $Y$, respectively. Then $A B X$ and $C D Y$ are $45^{\circ}-45^{\circ}-90^{\circ}$ triangles with $B X=C Y=3 \sqrt{2}$. So, the area of $A B X$ and the area of $C D Y$ are each 9 , meaning that the area of rectangle $B C Y X$ is 12. Since $B X=3 \sqrt{2}, B C=12 /(3 \sqrt{2})=2 \sqrt{2}$.