[Folio 0018 | p. 12]

PROPOSITION 5.

If two circles cut one another, they will not have the same centre.

For let the circles ABC, CDG cut one another at the points B, C;

I say that they will not have the same centre.

For, if possible, let it be E; let EC be joined, and let EFG be drawn through at random.

Then, since the point E is the centre of the circle ABC,

[I. Def. 15]EC is equal to EF.

Again, since the point E is the centre of the circle CDG,

EC is equal to EG.

But EC was proved equal to EF also; therefore EF is also equal to EG, the less to the greater: which is impossible.

Therefore the point E is not the centre of the circles ABC, CDG.

Therefore etc. Q. E. D.

Dependency Graph

Required for

III.10, III.24, III.26, III.27, III.28, III.29, III.30, IV.11, IV.12, IV.15, IV.16, VI.33, XII.1, XII.2, XII.5, XII.6, XII.7, XII.8, XII.10, XII.11, XII.12, XII.13, XII.14, XII.15, XII.17, XII.18, XIII.8, XIII.9, XIII.10, XIII.11, XIII.12, XIII.13, XIII.16, XIII.18

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See Commentary on Proposition III.5.