falls without the triangle, the angle opposite to the side beyond which it is, being in a segment less than a semicircle, is greater than a right angle. [III. 31.
Therefore, conversely, if the given triangle be acuteangled, the centre of the circle falls within it; if it be a right-angled triangle, the centre is in the side opposite to the right angle; and if it be an obtuse-angled triangle, the centre falls without the triangle, beyond the side opposite to the obtuse angle.
PROPOSITION 6. PROBLEM.
To inscribe a square in a given circle.
Let ABCD be the given circle: it is required to inscribe a square in ABCD.
Draw two diameters AC, BD of the circle ABCD, at right angles to one another; [III. 1, I. 11.
and join AB, BC, CD, DA. The figure ABCD shall be the square required.
Because BE is equal to DE, for E is the centre;
and that EA is common, and at right angles to BD;
therefore the base BA is equal to the base DA. [I. 4.
And for the same reason BC, DC are each of them equal to BA, or DA.
Therefore the quadrilateral figure ABCD is equilateral.
It is also rectangular.
For the straight line BD being a diameter of the circle ABCD, BAD is a semicircle; [Construction.
therefore the angle BAD is a right angle. [III. 31.
For the same reason each of the angles ABC, BCD, CDA is a right angle;
therefore the quadrilateral figure ABCD is rectangular.
And it has been shewn to be equilateral; therefore it is a square.
Wherefore a square has been inscribed in the given circle, q.e.f.
