Kitap 4
5
And it is manifest that, when the centre of the circle falls within the triangle, the angle BAC, being in a segment greater than the semicircle, is less than a right angle; when the centre falls on the straight line BC, the angle BAC, being in a semicircle, is right; and when the centre of the circle falls outside the triangle, the angle BAC, being in a segment less than the semicircle, is greater than a right angle. [III. 31]
PROPOSITION 6.
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In a given circle to inscribe a square.
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Let ABCD be the given circle; thus it is required to inscribe a square in the circle ABCD.
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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.
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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]
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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.
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I say next that it is also right-angled.
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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]
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For the same reason each of the angles ABC, BCD, CDA is also right; therefore the quadrilateral ABCD is right-angled.
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But it was also proved equilateral; therefore it is a square; [I. Def. 22] and it has been inscribed in the circle ABCD.
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Therefore in the given circle the square ABCD has been inscribed. Q. E. F.
PROPOSITION 7.
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About a given circle to circumscribe a square.
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Let ABCD be the given circle; thus it is required to circumscribe a square about the circle ABCD.
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Let two diameters AC, BD of the circle ABCD be drawn at right angles to one another, and through the points A, B, C, D let FG, GH, HK, KF be drawn touching the circle ABCD. [III. 16, Por.]
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Then, since FG touches the circle ABCD, and EA has been joined from the centre E to the point of contact at A, therefore the angles at A are right. [III. 18]
