previous next

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.

PROPOSITION 31.

In right-angled triangles the figure on the side subtending the right angle is equal to the similar and similarly described figures on the sides containing the right angle.

Let ABC be a right-angled triangle having the angle BAC right; I say that the figure on BC is equal to the similar and similarly described figures on BA, AC.

Let AD be drawn perpendicular.

Then since, in the right-angled triangle ABC, AD has

Euclid. Euclid's Elements. Sir Thomas Little Heath. New York. Dover. 1956.

The National Science Foundation provided support for entering this text.

Purchase a copy of this text (not necessarily the same edition) from Amazon.com

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.

hide Display Preferences
Greek Display:
Arabic Display:
View by Default:
Browse Bar: