Kitap 3
8
But EM, MD are greater than ED; [I. 20] therefore AD is also greater than ED.
8
Again, since ME is equal to MF, and MD is common, therefore EM, MD are equal to FM, MD; and the angle EMD is greater than the angle FMD; therefore the base ED is greater than the base FD. [I. 24]
8
Similarly we can prove that FD is greater than CD; therefore DA is greatest, while DE is greater than DF, and DF than DC.
8
Next, since MK, KD are greater than MD, [I. 20] and MG is equal to MK, therefore the remainder KD is greater than the remainder GD, so that GD is less than KD.
8
And, since on MD, one of the sides of the triangle MLD, two straight lines MK, KD were constructed meeting within the triangle, therefore MK, KD are less than ML, LD; [I. 21] and MK is equal to ML; therefore the remainder DK is less than the remainder DL.
8
Similarly we can prove that DL is also less than DH; therefore DG is least, while DK is less than DL, and DL than DH.
8
I say also that only two equal straight lines will fall from the point D on the circle, one on each side of the least DG.
8
On the straight line MD, and at the point M on it, let the angle DMB be constructed equal to the angle KMD, and let DB be joined.
8
Then, since MK is equal to MB, and MD is common, the two sides KM, MD are equal to the two sides BM, MD respectively; and the angle KMD is equal to the angle BMD; therefore the base DK is equal to the base DB. [I. 4]
8
I say that no other straight line equal to the straight line DK will fall on the circle from the point D.
