I was looking at geometric constructions of similar triangles, and at one point I came across the statement in the question.
A triangle delimited by vertices which are intersections of lines parallel to the sides of some other triangle is similar to said triangle.
I drew an example of the statement as follows:
Where the small green triangle's vertices are given by the intersections of some lines which are parallel to the sides of the blue triangle. The statement seems intuitively obvious, but I can't seem to find a way to justify it. I tried using the intercept theorem which is what I normally see is used to prove statements about similar triangles, but since I'm not sharing a side in both triangles I didn't think I could use it. Does anyone know a simple argument to justify this? Thank you!
Best Answer
An easy way to make the connection:
Given triangle $ABC$, and through external points $D$, $E$, $F$ parallels drawn to $AB$, $BC$, $CA$, respectively, forming triangle $GHJ$.
For explicit proof that $\triangle ABC\sim\triangle GHJ$, extend $AB$ to meet the parallels to $CA$ and $BC$ at $K$ and $L$. Since by parallels$$\angle CAB=\angle JKL$$and$$\angle JKL=\angle JGH$$therefore$$\angle CAB=\angle JGH$$For the same reason$$\angle CBA=\angle JLK=\angle JHG$$Therefore remaining angles$$\angle ACB=\angle GJH$$ and$$\triangle ABC\sim\triangle GHJ$$