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Proposition 15

Of straight lines in a circle the diameter is greatest, and of the rest the nearer to the center is always greater than the more remote.
Let ABCD be a circle, AD its diameter, and E its center. Let BC be nearer to the center AD, and FG more remote. I say that AD is greatest and BC greater than FG.
| Draw EH and EK from the center E perpendicular to BC and FG. | I.12
Then, since BC is nearer to the center and FG more remote, EK is greater than EH. | III.Def.5
Make EL equal to EH. Draw LM through L at right angles to EK, and carry it through to N. Join ME, EN, FE, and EG. | I.3 I.11
Then, since EH equals EL, BC also equals MN. | III.14
Again, since AE equals EM, and ED equals EN, AD equals the sum of ME and EN.
But the sum of ME and EN is greater than MN, and MN equals BC, therefore AD is greater than BC. | I.20
And, since the two sides ME and EN equal the two sides FE and EG, and the angle MEN greater than the angle FEG, therefore the base MN is greater than the base FG. | I.24
But MN was proved equal to BC. Therefore the diameter AD is greatest and BC greater than FG.
Therefore of straight lines in a circle the diameter is greatest, and of the rest the nearer to the center is always greater than the more remote.
Q.E.D.

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This proposition is not used in the rest of the Elements.

Book III Introduction - Proposition III.14 - Proposition III.16.

© 1996
D.E.Joyce
Clark University