Prove that a line that divides two sides of a triangle proportionally is parallel to the third side. Be sure to create and name the appropriate geometric figures. I've gotten as far as saying this: If we said we had triangle ABC, like drawn here: http://prntscr.com/94jhd5 If we drew the line that would proportionally divide the triangle, like this: http://prntscr.com/94ji00 We can name that line DE, like I did in the example figure. I've read up on this a lot and I figured it would be the converse of the Triangle Proportionality Theorem, but I'm lost as to how to show that.
@phi
If the corresponding legs of two triangles are "in proportion", then the two triangles are similar. I think there is a theorem that says that , or we have to prove that. once we know that, then we have "corresponding angles" are congruent and then we use that to show corresponding angles of a transversal are equal which means the lines crossed by the transversal are parallel
I saw someone proving this by saying angles B and D are corresponding, but I wasn't sure how that proved proportionality because I only know that to prove similarity.
You are given a line that divides two sides of a triangle proportionally so we know the sides are divided proportionally we use that fact to show the triangles are similar.
And how do we do that?
I think it is a definition of similar: if corresponding sides of two shapes are "in proportion", then the shapes are similar.
How would I write a proof for all of this though? That's what I need help with.
Given triangle ABC and line DE that divides the sides AB and BC in proportion. 1) triangle ABC is similar to triangle ADE; reason: sides are proportion (given) 2) angle D is congruent to angle B; reason: corresponding angles of similar triangles 3) corresponding angles of transversal AB crossing lines AC and DE are congruent; reason: step 2 4) lines AC and DE are parallel; reason: corresponding angles are congruent by step 3
here is the picture I am assuming |dw:1447949309778:dw|
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