Let ABCD be the given circle with centre E. Draw two diameters
AC and BD at right angles (I.11). Join AB, BC, CD, DA.
The four right triangles at E are congruent by I.4 (two radii and
the common right angle), so AB=BC=CD=DA. The inscribed
angles standing on the diameters are right (III.31), so all four
angles of ABCD are right. Therefore ABCD is a square.