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Proposition 39

Fri, 2006-11-10 17:27 — ashley

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 $, [Prop. 1.31]

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. [Prop. 1.37]

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.