In a given circle to inscribe a square
be the given circle; thus it is required to inscribe a square in the circle ABCD
Let two diameters AC
of the circle ABCD
be drawn at right angles to one another, and let AB
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
is also equal to each of the straight lines AB
; 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
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.