Answer: 810

Small Hint:

Put $B = (0, 0),$ $C = (s, 0),$ $A = (0, s),$ $D = (s, s),$ and write $E = (a, s),$ $F = (b, 0)$

Big Hint:

$BE = FD$ forces $b = s - a,$ and then $EF = BE$ forces $a = \frac{s}{3};$ finish with $a^2 + s^2 = 900$

Solution:

Let the side length be $s,$ and place $B = (0, 0),$ $C = (s, 0),$ $A = (0, s),$ $D = (s, s).$ Write $E = (a, s)$ and $F = (b, 0).$ Then $BE^2 = a^2 + s^2,$ $FD^2 = (s - b)^2 + s^2,$ and $EF^2 = (a - b)^2 + s^2.$

From $BE = FD$ we get $a = s - b,$ so $a - b = 2a - s.$ Then $EF = BE$ gives $(2a - s)^2 = a^2,$ whose solutions are $a = \frac{s}{3}$ and $a = s$ (the latter identifies the separately named points $E$ and $F$ with corners $D$ and $B,$ making all three segments the same diagonal). So $a = \frac{s}{3}.$

Now $900 = BE^2 = \frac{s^2}{9} + s^2 = \frac{10s^2}{9},$ so the area is $s^2 = \frac{9}{10} \cdot 900 = 810.$