Let ABCD be the given square.
It is required to circumscribe a circle about the square ABCD.
Join AC and BD, and let them cut one another at E.
Then, since DA equals AB, and AC is common, therefore the two sides DA and AC equal the two sides BA and AC, and the base DC equals the base BC, therefore the angle DAC equals the angle BAC.
Therefore the angle DAB is bisected by AC.
Similarly we can prove that each of the angles ABC, BCD, and CDA is bisected by the straight lines AC and DB.
Now, since the angle DAB equals the angle ABC, and the angle EAB is half of the angle DAB, and the angle EBA half of the angle ABC, therefore the angle EAB also equals the angle EBA, so that the side EA also equals EB.
Similarly we can prove that each of the straight lines EA and EB equals each of the straight lines EC and ED.
Therefore the four straight lines EA, EB, EC, and ED equal one another.
Therefore the circle described with center E and radius one of the straight lines EA, EB, EC, or ED also passes through the remaining points, and it is circumscribed about the square ABCD.
Let it be circumscribed, as ABCD.
Therefore a circle has been circumscribed about the given square.