Proposition 2

Euclid prefers to prove a pair of converses in two stages, but in some propositions, as this one, the proofs in the two stages are almost inverses of each other, so both could be proved at once.

In this proposition we have a given triangle ABC and a line DE joining a point D on the side BC to a point E on the side AC. The claim is that

BD:AD = CE:AE if and only if DE || BC.

By the previous proposition VI.1 we know in any case that

CE:AE = triangle CDE : triangle ADE.

Hence,

By propositions V.7 and V.9 the latter condition is equivalent to BDE = CDE, and that, in turn, by propositions I.37 and I.39 is equivalent to DE || BC.

Note

It should be noted that a proportion such as BD:AD = AE:CE is not intended. In that case the sides are cut proportionally, but the correspondence is not the intended one.

Use of this theorem

This proposition is frequently used in the rest of Book VI starting with the next proposition. It is also used in Books XI and XII.

Next proposition: VI.3

Previous: VI.1

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