Kitap 3
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.
8
For, if possible, let a straight line so fall, and let it be DN. Then, since DK is equal to DN,
8
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.
8
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.
8
Therefore etc.
PROPOSITION 9.
9
If a point be taken within a circle, and more than two equal straight lines fall from the point on the circle, the point taken is the centre of the circle.
9
Let ABC be a circle and D a point within it, and from D let more than two equal straight lines, namely DA, DB, DC, fall on the circle ABC; I say that the point D is the centre of the circle ABC.
9
For let AB, BC be joined and bisected at the points E, F, and let ED, FD be joined and drawn through to the points G, K, H, L.
9
Then, since AE is equal to EB, and ED is common, the two sides AE, ED are equal to the two sides BE, ED; and the base DA is equal to the base DB; therefore the angle AED is equal to the angle BED. [I. 8]
9
Therefore each of the angles AED, BED is right; [I. Def. 10] therefore GK cuts AB into two equal parts and at right angles.
9
And since, if in a circle a straight line cut a straight line into two equal parts and at right angles, the centre of the circle is on the cutting straight line, [III. 1, Por.] the centre of the circle is on GK.
9
For the same reason the centre of the circle ABC is also on HL.
9
And the straight lines GK, HL have no other point common but the point D; therefore the point D is the centre of the circle ABC.
9
Therefore etc. Q. E. D.
PROPOSITION 10.
