[Folio 0098 | p. 92]

PROPOSITION 7.

About a given circle to circumscribe a square.

Let ABCD be the given circle; thus it is required to circumscribe a square about the circle ABCD.

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.]

Then, since FG touches the circle ABCD, and EA has been joined from the centre E to the point of contact at A,

[III. 18]therefore the angles at A are right.

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,

[I. 28]therefore GH is parallel to AC.

[Folio 0099 | p. 93]

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.

Therefore GK, GC, AK, FB, BK are parallelograms; therefore GF is equal to HK, and GH to FK. [I. 34]

And, since AC is equal to BD, and AC is also equal to each of the straight lines GH, FK,

[I. 34]while BD is equal to each of the straight lines GF, HK,

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.

Therefore FGHK is right-angled.

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.

Dependency Graph

Required for

XII.10, XII.11, XII.12, XII.13, XII.14, XII.15

---

See Commentary on Proposition IV.7.