#### Proposition 39.

Equal triangles which are on the same base and on the same side are also in the same parallels.

Let ABC, DBC be equal triangles which are on the same base BC and on the same side of it;
[I say that they are also in the same parallels.]

And [For] let AD be joined; I say that AD is parallel to BC.

For, if not, let AE be drawn through the point A parallel to the straight line
BC, [I. 31] and let EC be joined.

Therefore the triangle ABC is equal to the triangle EBC; for it is on the same base BC with it and in the same
parallels. [I. 37]

But ABC is equal to DBC;

therefore DBC is also equal to EBC, [C.N. 1] the greater to the less: which is impossible.

Therefore AE is not parallel to BC.

Similarly we can prove that neither is any other straight line except AD;

therefore AD is parallel to BC.

Therefore etc.

Q. E. D.

1 [I say that they are also in the same parallels.] Heiberg has proved (Hermes, XXXVIII., 1903, p. 50) from a recently discovered papyrus-fragment (Fayūm towns and their papyri, p. 96, No. IX.) that these words are an interpolation by some one who did not observe that the words “And let AD be joined” are part of the setting-out (ἔκθεσις), but took them as belonging to the construction (κατασκευή) and consequently thought that a διορισμός or “definition” (of the thing to be proved) should precede. The interpolator then altered “And” into “For” in the next sentence.

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: Unicode (precombined) Unicode (combining diacriticals) Beta Code SPIonic SGreek GreekKeys Latin transliteration Arabic Display: Unicode Buckwalter transliteration View by Default: Original Language Translation Browse Bar: Show by default Hide by default