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Proposition 6

To inscribe a square in a given circle.
Let ABCD be the given circle. It is required to inscribe a square in the circle ABCD.
Draw two diameters AC and BD of the circle ABCD at right angles to one another, and join AB, BC, CD, and DA. | III.1 I.11
| Then, since BE equals ED, for E is the center, and EA is common and at right angles, therefore the base AB equals the base AD. | I.4
For the same reason each of the straight lines BC and CD also equals each of the straight lines AB and 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, and CDA is also right. Therefore the quadrilateral ABCD is right-angled.
But it was also proved equilateral, therefore it is a square, and it has been inscribed in the circle ABCD.
Therefore the square ABCD has been inscribed in the given circle.
Q.E.F.

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This construction is used in a few propositions of Book XII, the first being XII.2.

Book IV Introduction - Proposition IV.5 - Proposition IV.7.

© 1996
D.E.Joyce
Clark University