If a straight line is drawn parallel to one of the sides of a triangle, then it cuts the sides of the triangle proportionally; and, if the sides of the triangle are cut proportionally, then the line joining the points of section is parallel to the remaining side of the triangle.

Let *DE* be drawn parallel to *BC,* one of the sides of the triangle ABC.

I say that *BD* is to *AD* as *CE* is to *AE.*

Join *BE* and *CD.*

Therefore the triangle *BDE* equals the triangle *CDE,* for they are on the same base *DE* and in the same parallels *DE* and *BC.*

And *ADE* is another triangle.

But equals have the same ratio to the same, therefore the triangle *BDE* is to the triangle *ADE* as the triangle *CDE* is to the triangle *ADE.*

But the triangle *BDE* is to *ADE* as *BD* is to *AD,* for, being under the same height, the perpendicular drawn from *E* to *AB,* they are to one another as their bases.

For the same reason, the triangle *CDE* is to *ADE* as *CE* is to *AE.*

Therefore *BD* is to *AD* also as *CE* is to *AE.*

Next, let the sides *AB* and *AC* of the triangle *ABC* be cut proportionally, so that *BD* is to *AD* as *CE* is to *AE.* Join *DE.*

I say that *DE* is parallel to *BC.*

With the same construction, since *BD* is to *AD* as *CE* is to *AE,* but *BD* is to *AD* as the triangle *BDE* is to the triangle *ADE,* and *CE* is to *AE* as the triangle *CDE* is to the triangle *ADE,* therefore the triangle *BDE* is to the triangle *ADE* as the triangle *CDE* is to the triangle *ADE.*

Therefore each of the triangles *BDE* and *CDE* has the same ratio to *ADE.*

Therefore the triangle *BDE* equals the triangle *CDE,* and they are on the same base *DE.*

But equal triangles which are on the same base are also in the same parallels.

Therefore *DE* is parallel to *BC.*

Therefore, *if a straight line is drawn parallel to one of the sides of a triangle, then it cuts the sides of the triangle proportionally; and, if the sides of the triangle are cut proportionally, then the line joining the points of section is parallel to the remaining side of the triangle.*

Q.E.D.

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

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

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.*