#### PROPOSITION 6.

In a given circle to inscribe a square.

Let ABCD be the given circle; thus it is required to inscribe a square in the circle ABCD.

Let two diameters AC, BD of the circle ABCD be drawn at right angles to one another, and let AB, BC, CD, DA be joined.

Then, since BE is equal to ED, for E is the centre, and EA is common and at right angles, therefore the base AB is equal to the base AD. [I. 4]

For the same reason each of the straight lines BC, CD is also equal to each of the straight lines AB, AD;

therefore the quadrilateral ABCD is equilateral.

I say next that it is also right-angled.

For, since the straight line BD is a diameter of the circle ABCD, therefore BAD is a semicircle;

therefore the angle BAD is right. [III. 31]

For the same reason each of the angles ABC, BCD, CDA is also right;

therefore the quadrilateral ABCD is right-angled.

But it was also proved equilateral; therefore it is a square; [I. Def. 22] and it has been inscribed in the circle ABCD.

Therefore in the given circle the square ABCD has been inscribed. Q. E. F.