I say also that from the point F only two equal straight lines will fall on the circle ABCD, one on each side of the least FD.
For on the straight line EF, and at the point E on it, let the angle FEH be constructed equal to the angle GEF [I. 23], and let FH be joined.
Then, since GE is equal to EH, and EF is common,
the two sides GE, EF are equal to the two sides HE, EF;
and the angle GEF is equal to the angle HEF;
therefore the base FG is equal to the base FH. [I. 4]
I say again that another straight line equal to FG will not fall on the circle from the point F.
For, if possible, let FK so fall.
Then, since FK is equal to FG, and FH to FG,
FK is also equal to FH,
the nearer to the straight line through the centre being thus equal to the more remote: which is impossible.
Therefore another straight line equal to GF will not fall from the point F upon the circle;
therefore only one straight line will so fall.