2. $A B C D$ is a square. $B D E F$ is a rhombus with $A$, $E$ and $F$ collinear. Find $\angle A D E$.
SolUtion
Let $X$ be the foot of the perpendicular from $A$ to $B D$ and let $Y$ be the foot of the perpendicular from $E$ to $B D$.

Let $B D$ have length 2 .
The length of $A X$ is 1 because it is half the diagonal $A C$, which is the same as $B D$. This is also the same length of $E Y$ because $A X Y E$ is a rectangle.

Triangle $E Y D$ is half of an equilateral triangle (since $D E=D B=2$ ), so $\angle E D Y=30^{\circ}$.
Since $\angle B D A=45^{\circ}, \angle A D E=45^{\circ}-30^{\circ}=15^{\circ}$.