therefore the quadrilateral figure FGHK is rectangular.
And it has been shewn to be equilateral; therefore it is a square.
Wherefore a square has been described about the given circle. q.e.f.
PROPOSITION 8. PROBLEM.
To inscribe a circle in a given square.
Let ABCD be the given square: it is required to inscribe a circle in ABCD.
Bisect each of the sides AB, AD at the points F, E; [I. 10.
through E draw EH parallel to AB or DC, and through F draw FK parallel to AD or BC. [I. 31.
Then each of the figures AK, KB, AH, HD, AG, GC, BG, GD is a right-angled parallelogram;
and their opposite sides are equal. [I. 34.
And because AD is equal to AB, [I. Definition 30.
and that AE is half of AD, and AF half of AB, [Constr. therefore AE is, equal to AF. [Axiom 7.
Therefore the sides opposite to these are equal, namely, FG equal to GE. [I. 34.
In the same manner it may be shewn that the straight lines GH, GK are each of them equal to FG or GE.
Therefore the four straight lines GE, GF, GH, GK are equal to one another, and the circle described from the centre G, at the distance of any one of them, will pass through the extremities of the other three;
and it will touch the straight lines AB, BC, CD, DA, because the angles at the points E, F, H, K are right angles, and the straight line which is drawn from the extremity of a diameter, at right angles to it, touches the circle. [III. 16. Corollary.
Therefore the straight lines AB, BC, CD, DA do each of them touch the circle.
Wherefore a circle has been inscribed in the given square. q.e.f.
