Sol.—Draw any two diameters AC, BD at right angles to each other. Join AB, BC, CD, DA. ABCD is a square.
Dem.—Let O be the centre. Then the four angles at O, being right angles, are equal. Hence the arcs on which they stand are equal [III. xxvi.], and hence the four chords are equal [III. xxix.]. Therefore the figure ABCD is equilateral.
Again, because AC is a diameter, the angle ABC is right [III. xxxi.]. In like manner the remaining angles are right. Hence ABCD is a square.
PROP. VII.—Problem.
About a given circle (ABCD) to describe a square.
Sol.—Through the centre O draw any two diameters at right angles to each other, and draw at the points A, B, C, D the lines HE, EF, FG, GH touching the circle. EFGH is a square.
Dem.—Because AE touches the circle at A, the angle EAO is right [III. xviii.], and therefore equal to BOC, which is right (const.). Hence AE is parallel to OB. In like manner EB is parallel to AO; and since AO is equal to OB, the figure AOBE is a lozenge, and the angle AOB is right; hence AOBE is a square. In like manner each of the figures BC, CD, DA is a square. Hence the whole figure is a square.
Cor.—The circumscribed square is double of the inscribed square.
PROP. VIII.—Problem.
In a given square (ABCD) to inscribe a circle.
Sol.—Bisect (see last diagram) two adjacent sides EH, EF in the points A, B, and through A, B draw the lines AC, BD, respectively parallel to EF, EH; then O, the point of intersection of these parallels, is the centre of the required circle.