Let DE be drawn parallel to BC, one of the sides of the triangle ABC: BD shall be to DA as CE is to EA.
Join BE, CD.
Then the triangle BDE is equal to the triangle CDE, because they are on the same base DE and between the same parallels DE, BC. [I. 37.
And ADE is another triangle;
and equal magnitudes have the same ratio to the same magnitude; [V. 7.
therefore the triangle BDE is to the triangle ADE as the triangle CDE is to the triangle ADE.
But the triangle BDE is to the triangle ADE as BD is to DA;
because the triangles have the same altitude, namely, the perpendicular drawn from E to AB, and therefore they are to one another as their bases. [VI. 1,
For the same reason the triangle CDE is to the triangle ADE as CE is to EA
Therefore BD is to DA as CE is to EA. [V. 11.
Next, let BD be to DA as CE is to EA, and join DE: DE shall be parallel to BC.
For, the same construction being made, because BD is to DA as CE is to EA, [Hypothesis.
and as BD is to DA, so is the triangle BDE to the triangle ADE, [VI. 1.
and as CE is to EA so is the triangle CDE to the triangle ADE; [VI. 1.
