Point $$P$$ is inside the square $$ABCD$$ such that $$PA=PB$$ = Distance of $$P$$ from $$CD$$. The ratio of the area of triangle $$PAB$$ to the area of square $$ABCD$$ is $$\frac{m}{n}$$, where $$m$$ and $$n$$ are relatively prime integers. Then the value of $$m+n$$ is

Let the side of the square be $$s$$ and the distance from $$P$$ to $$CD$$ be $$y$$. Since $$PA=PB$$, point $$P$$ lies on the perpendicular bisector of $$AB$$, and $$PA^2=(s-y)^2+\left(\frac{s}{2}\right)^2=y^2$$. This gives $$5s=8y$$. Therefore the area ratio is $$\frac{\frac{1}{2}s(s-y)}{s^2}=\frac{s-y}{2s}=\frac{3}{16}$$, so $$m+n=3+16=19$$.