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](elem.3.18)]

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](elem.1.28)]

For the same reason AC is also parallel to FK , so that GH is also parallel to FK . [[I. 30](elem.1.30)]

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](elem.1.34)]

And, since AC is equal to BD , and AC is also equal to each of the straight lines GH , FK , while BD is equal to each of the straight lines GF , HK , [[I. 34](elem.1.34)] 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](elem.1.34)]

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.