第3卷命题 13 · 圆与圆相切至多一个点

For, if possible, let the circle ABDC touch the circle EBFD, first internally, at more points than one, namely D, B. Let the centre G of the circle ABDC, and the centre H of EBFD, be taken. Therefore the straight line joined from G to H will fall on B, D.

假设圆ABDC与圆EBFD内切于多于一点，比如D和B。取圆ABDC的圆心G和圆EBFD的圆心H，则连接G与H的直线必经过B和D。

[III. 11] Let it so fall, as BGHD. 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. 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.

由于G是圆ABDC的圆心，BG等于GD，因此BG大于HD，从而BH远大于HD。但H是圆EBFD的圆心，BH等于HD，矛盾。故内切不能多于一点。

Therefore a circle does not touch a circle internally at more points than one. I say further that neither does it so touch it externally. For, if possible, let the circle ACK touch the circle ABDC at more points than one, namely A, C, and let AC be joined.

再假设圆ACK与圆ABDC外切于多于一点，比如A和C。连接AC。

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. Therefore a circle does not touch a circle externally at more points than one. And it was proved that neither does it so touch it internally.

由于A和C在两个圆的圆周上，线段AC应落在每个圆内部，但它在圆ABDC内部而在圆ACK外部，矛盾。故外切也不能多于一点。