About a given circle to circumscribe a square
be the given circle; thus it is required to circumscribe a square about the circle ABCD
Let two diameters AC
of the circle ABCD
be drawn at right angles to one another, and through the points A
be drawn touching the circle ABCD
. [III. 16, Por.
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]
For the same reason the angles at the points B, C, D are also right.
Now, since the angle AEB
is right, and the angle EBG
is also right, therefore GH is parailel to AC. [I. 28]
For the same reason AC is also parallel to FK, so that GH is also parallel to FK. [I. 30]
Similarly we can prove that each of the straight lines GF, HK is parallel to BED.
are parallelograms; therefore GF
is equal to HK
, and GH
. [I. 34
And, since AC
is equal to BD
, and AC
is also equal to each of the straight lines GH
, while BD is equal to each of the straight lines GF, HK, [I. 34] therefore the quadrilateral FGHK is equilateral.
I say next that it is also right-angled.
For, since GBEA
is a parallelogram, and the angle AEB
is right, therefore the angle AGB
is also right. [I. 34
Similarly we can prove that the angles at H, K, F are also right.
But it was also proved equilateral; therefore it is a square;
and it has been circumscribed about the circle ABCD
Therefore about the given circle a square has been circumscribed. Q. E. F.