AoPS Wiki talk:Problem of the Day/September 12, 2011

Dropping an altitude, we see that it has length $3\sqrt{2}$ by the 45-45-90 triangle. Thus, the midsegment must have length $5\sqrt{2}$ . Dropping altitudes from B and C, we see that the sum of the bases is $2BC+6\sqrt{2}$ , and thus $BC+3\sqrt{2}=5\sqrt{2}$ and $\boxed{BC=2\sqrt{2}}$ .