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.
Complete text
Book I