Join the diagonals AC and BD of the given square, intersecting
at E. In the right triangles △ABE and △ADE,
SSS (I.8) gives ∠EAB=∠EAD, so AE bisects the right
angle at A; similarly at every vertex. The four triangles at E
are then isosceles with equal vertex angles (I.6), so EA=EB=EC=ED. The circle with centre E and radius EA passes through all
four vertices.