Book 4
7
For the same reason the angles at the points B, C, D are also right.
7
Now, since the angle AEB is right, and the angle EBG is also right, therefore GH is parailel to AC. [I. 28]
7
For the same reason AC is also parallel to FK, so that GH is also parallel to FK. [I. 30]
7
Similarly we can prove that each of the straight lines GF, HK is parallel to BED.
7
Therefore GK, GC, AK, FB, BK are parallelograms; therefore GF is equal to HK, and GH to FK. [I. 34]
7
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] therefore the quadrilateral FGHK is equilateral.
7
I say next that it is also right-angled.
7
For, since GBEA is a parallelogram, and the angle AEB is right, therefore the angle AGB is also right. [I. 34]
7
Similarly we can prove that the angles at H, K, F are also right.
7
Therefore FGHK is right-angled.
7
But it was also proved equilateral; therefore it is a square; and it has been circumscribed about the circle ABCD.
7
Therefore about the given circle a square has been circumscribed. Q. E. F.
PROPOSITION 8.
8
In a given square to inscribe a circle.
8
Let ABCD be the given square; thus it is required to inscribe a circle in the given square ABCD.
8
Let the straight lines AD, AB be bisected at the points E, F respectively [I. 10], through E let EH be drawn parallel to either AB or CD, and through F let FK be drawn parallel to either AD or BC; [I. 31] therefore each of the figures AK, KB, AH, HD, AG, GC, BG, GD is a parallelogram, and their opposite sides are evidently equal. [I. 34]
8
Now, since AD is equal to AB, and AE is half of AD, and AF half of AB, therefore AE is equal to AF, so that the opposite sides are also equal; therefore FG is equal to GE.
8
Similarly we can prove that each of the straight lines GH, GK is equal to each of the straight lines FG, GE; therefore the four straight lines GE, GF, GH, GK are equal to one another.
