Rectangle $ABCD$ has $AB=5$ and $BC=12$. Point $E$ is the foot of the perpendicular from $B$ to diagonal $\overline{AC}$. What is the area of $\triangle AED$?

1. Place $A=(0,0)$, $B=(5,0)$, and $C=(5,12)$. Then the diagonal vector is $\overrightarrow{AC}=(5,12)$, with $AC^2=169$. Since $E$ is the projection of $B$ onto $AC$, its position along $AC$ is
   $$\frac{\overrightarrow{AB}\cdot\overrightarrow{AC}}{AC^2}=\frac{25}{169},$$
   so $E=\frac{25}{169}(5,12)=(125/169,300/169)$.
2. The base $AD$ has length $12$, and the perpendicular distance from $E$ to the line $AD$ is its $x$-coordinate, $125/169$. Therefore,
   $$[AED]=\frac12\cdot12\cdot\frac{125}{169}=\frac{750}{169}.$$