In trapezoid $A B C D, A B=4, C D=6, \angle D A B=90^{\circ}, \angle B C D=45^{\circ}$, and $A B$ is parallel to $C D$. What is the length of $B D$ ?

![](https://cdn.mathpix.com/cropped/2024_04_17_587bc2a8915214d42f9eg-2.jpg?height=233&width=680&top_left_y=290&top_left_x=777)
Let $E$ be on $C D$ such that $B E$ is perpendicular to $C D$ as shown.

![](https://cdn.mathpix.com/cropped/2024_04_17_c3c8a275c8ba62c19ff3g-02.jpg?height=239&width=677&top_left_y=1813&top_left_x=781)

Since $A B$ and $D E$ are parallel and $\angle D A B$ and $\angle D E B$ are right angles, $\angle A D E$ and $\angle A B E$ are right angles as well, and so $A B E D$ is a rectangle.

Using $A B=4$ and the fact that $A B E D$ is a rectangle, we get $D E=4$.

Using $C D=C E+D E, D E=4$, and $C D=6$, we get $C E=6-4=2$.

In $\triangle B E C, \angle B E C=90^{\circ}$ and $\angle B C E=\angle B C D=45^{\circ}$, and so $\angle E B C=180^{\circ}-90^{\circ}-45^{\circ}=45^{\circ}$.

This means $\triangle B E C$ is isosceles, and so $B E=C E=2$.

By the Pythagorean theorem applied to right-angled $\triangle B E D$, we have $D E^{2}+B E^{2}=B D^{2}$, and so $B D^{2}=4^{2}+2^{2}=20$.

Since $B D>0, B D=\sqrt{20}=2 \sqrt{5}$.