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.
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.
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]
I say that no other straight line equal to the straight line DK will fall on the circle from the point D.
For, if possible, let a straight line so fall, and let it be DN.
Then, since DK is equal to DN,
while DK is equal to DB,
DB is also equal to DN,
that is, the nearer to the least DG equal to the more remote: which was proved impossible.
Therefore no more than two equal straight lines will fall on the circle ABC from the point D, one on each side of DG the least.