第4卷命题 6 · 圆内接正方形作图

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.

设ABCD为给定圆，作互相垂直的两条直径AC和BD。

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.

连接AB、BC、CD、DA。因E为圆心，BE=ED，EA公共且垂直，故AB=AD。

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.

同理，BC=CD=AB=AD，故四边形ABCD等边。

[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.

因BD为直径，∠BAD为直角；同理∠ABC、∠BCD、∠CDA均为直角，故ABCD为正方形，且内接于圆。