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.
Dem.—Because AOBE is a parallelogram, its opposite sides are equal; therefore AO is equal to EB; but EB is half the side of the given square; therefore AO is equal to half the side of the given square; and so in like manner is each of the lines OB, OC, OD; therefore the four lines OA, OB, OC, OD are all equal; and since they are perpendicular to the sides of the given square, the circle described with O as centre, and OA as radius, will be inscribed in the square.