Let $$ABCD$$ be a rectangle and let $$E$$ be a point on $$BD$$ such that $$AE$$ is perpendicular to $$BD$$. If $$AE = 12$$ and $$CE = \sqrt{193}$$, compute the area of the rectangle $$ABCD$$.

Draw $$CF \perp BD$$. A half-turn about the centre of the rectangle takes $$A$$ to $$C$$ and $$E$$ to $$F$$, so $$CF = AE = 12$$ and $$EF^2 = CE^2 - CF^2 = 193 - 144 = 49$$, giving $$EF = 7$$. Putting $$BE = p$$ and $$ED = q$$, the symmetry gives $$|p - q| = 7$$ while the altitude relation in right triangle $$ABD$$ gives $$pq = AE^2 = 144$$, so $$BD^2 = (p - q)^2 + 4pq = 49 + 576 = 625$$ and $$BD = 25$$. The area is $$2 \times \tfrac{1}{2} \times BD \times AE = 25 \times 12 = 300$$.