Let ABCD be the given square. Bisect the sides at E, F, G,
H (I.10). Join EG and FH, meeting at K. By I.34 and the
construction, KE=KF=KG=KH. Drop perpendiculars from K to
each side; each is equal to KE. Thus the circle with centre K
and radius KE touches each side at its midpoint (III.16).