To circumscribe a square about a given circle.

Let *ABCD* be the given circle.

It is required to circumscribe a square about the circle *ABCD.*

Draw two diameters *AC* and *BD* of the circle *ABCD* at right angles to one another.

Draw *FG, GH, HK,* and *KF* through the points *A, B, C,* and *D* touching the circle *ABCD.*

Then, since *FG* touches the circle *ABCD,* and *EA* has been joined from the center *E* to the point of contact at *A,* therefore the angles at A are right. For the same reason the angles at the points *B, C,* and *D* are also right.

Now, since the angle *AEB* is right, and the angle *EBG* is also right, therefore *GH* is parallel to *AC.*

For the same reason *AC* is also parallel to *FK,* so that *GH* is also parallel to *FK.*

Similarly we can prove that each of the straight lines *GF* and *HK* is parallel to *BED.*

Therefore *GK, GC, AK, FB,* and *BK* are parallelograms, therefore *GF* equals *HK,* and *GH* equals *FK.*

And, since *AC* equals *BD,* and *AC* also equals each of the straight lines *GH* and *FK,* and *BD* equals each of the straight lines *GF* and *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.

Similarly we can prove that the angles at *H, K,* and *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 a square has been circumscribed about the given circle.

Q.E.F.