previous next


PROPOSITION 30.

To cut a given finite straight line in extreme and mean ratio.

Let AB be the given finite straight line;

thus it is required to cut AB in extreme and mean ratio.

On AB let the square BC be described; and let there be applied to AC the parallelogram CD equal to BC and exceeding by the figure AD similar to BC. [VI. 29]

Now BC is a square;

therefore AD is also a square.

And, since BC is equal to CD, let CE be subtracted from each;

therefore the remainder BF is equal to the remainder AD.

But it is also equiangular with it; therefore in BF, AD the sides about the equal angles are reciprocally proportional; [VI. 14]

therefore, as FE is to ED, so is AE to EB.

But FE is equal to AB, and ED to AE.

Therefore, as BA is to AE, so is AE to EB.

And AB is greater than AE;

therefore AE is also greater than EB.

Therefore the straight line AB has been cut in extreme and mean ratio at E, and the greater segment of it is AE. Q. E. F.

Creative Commons License
This work is licensed under a Creative Commons Attribution-ShareAlike 3.0 United States License.

An XML version of this text is available for download, with the additional restriction that you offer Perseus any modifications you make. Perseus provides credit for all accepted changes, storing new additions in a versioning system.

load focus Greek (J. L. Heiberg, 1883)
hide Display Preferences
Greek Display:
Arabic Display:
View by Default:
Browse Bar: