Kitap 3
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Then, since the point G is the centre of the circle ABCD, BG is equal to GD; therefore BG is greater than HD; therefore BH is much greater than HD.
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Again, since the point H is the centre of the circle EBFD, BH is equal to HD; but it was also proved much greater than it: which is impossible.
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Therefore a circle does not touch a circle internally at more points than one.
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I say further that neither does it so touch it externally.
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For, if possible, let the circle ACK touch the circle ABDC at more points than one, namely A, C, and let AC be joined.
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Then, since on the circumference of each of the circles ABDC, ACK two points A, C have been taken at random, the straight line joining the points will fall within each circle; [III. 2] but it fell within the circle ABCD and outside ACK [III. Def. 3]: which is absurd.
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Therefore a circle does not touch a circle externally at more points than one.
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And it was proved that neither does it so touch it internally.
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Therefore etc. Q. E. D. ABDC. Euclid writes ABCD (here and in the next proposition), notwithstanding the order in which the points are placed in the figure. does it so touch it. It is necessary to supply these words which the Greek (ὅτι οὐδὲ ἐκτός and ὅτι οὐδὲ ἐντός) leaves to be understood.
PROPOSITION 14.
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In a circle equal straight lines are equally distant from the centre, and those which are equally distant from the centre are equal to one another.
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Let ABDC be a circle, and let AB, CD be equal straight lines in it; I say that AB, CD are equally distant from the centre.
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For let the centre of the circle ABDC be taken [III. 1], and let it be E; from E let EF, EG be drawn perpendicular to AB, CD, and let AE, EC be joined.
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Then, since a straight line EF through the centre cuts a straight line AB not through the centre at right angles, it also bisects it. [III. 3] Therefore AF is equal to FB; therefore AB is double of AF.
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For the same reason CD is also double of CG; and AB is equal to CD; therefore AF is also equal to CG.
