If $\bar{AD}=r\bar{AB}$ and $\bar{AE}=r\bar{AC}$, is $\bar{DE}\parallel\bar{BC}$?

Question

Suppose we start with triangle $ABC$ and then scale sides $AB$ and $AC$ by the same factor $r$ so as to produce a new triangle $ADE$, as on the figure. I'm convinced that the bases of these two triangles will be parallel, but how can I show that without using the triangle similarity postulates (AAA, SAS)? (The reason for this is because I'm trying to prove those postulates using the basic proportionality theorem, but I need this result to complete the proof). Thanks.

Answer 1

That's Euclid's Elements - Book VI - Prop. 2
"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. "
Refer for instance to this site.

Answer 2

Yes, of course!

Let $C_1\in AE$ such that $BC_1||DE$.

Thus, $AE=rAC_1$, which says $C\equiv C_1$ and we are done!