AoPS - Problem

Let a = BC be the square side. Area $|ABCDE| = \frac{5a^2}{4}.$ Let x be the leg of the isosceles right triangle with the area $\frac{x^2}{2} = \frac 5 4 a^2.$ Then $x = a\ \sqrt{\frac 5 2.}$ Since $AC = AD = a\ \sqrt{\frac 1 4 + \frac 9 4} = a\ \sqrt{\frac 5 2}$, one of these is the 1st cut, say AD. Rotate the triangle $\triangle ADE$ by $90^\circ$ around A, so that AE coincides with AB and D goes into $D' \in BE$. Let M be the midpoint of BC. Since the triangles $\triangle MCD \cong \triangle MBD'$ are congruent, DM is the 2nd cut.