Right Angled Triangle Park Paths

Let right-angled vertex be $B$, hypotenuse $AC = 80$. $a^2 + b^2 = 80^2 = 6400$. Points $D, E, F$ divide $AC$ into 4 equal segments of 20 m each. $BE$ is the median to hypotenuse $\implies BE = 40 \implies BE^2 = 1600$. By Apollonius theorem on $\triangle ABE$ and $\triangle EBC$: $BD^2 + BF^2 + 20^2 + 20^2 = \frac{1}{2}(a^2 + b^2 + 40^2 + 40^2) = \frac{1}{2}(6400 + 3200) = 4800$. $BD^2 + BF^2 = 4800 - 800 = 4000$. Sum of squares = $BD^2 + BE^2 + BF^2 = 4000 + 1600 = 5600\text{ m}^2$.