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In a general homothetic transformation, if two triangles have corresponding sides parallel then the lines joining respective vertices are concurrent at the homothetic center. I was wondering if the converse is true. Given two similar triangles in which the lines joining corresponding vertices are concurrent, then are the corresponding sides of the two triangles necessarily parallel?

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edit: Er, oops, here's a counterexample. Let $\triangle ABC$ and $\triangle DEF$ be equilateral triangles, $\triangle ABC\sim\triangle DEF$, with $A$ on $\overline{DF}$ and $B$ on $\overline{EF}$. This means that $\overleftrightarrow{AD}$ contains $F$ and $\overleftrightarrow{BE}$ contains $F$, so $\overleftrightarrow{AD}$, $\overleftrightarrow{BE}$, and $\overleftrightarrow{CF}$ are concurrent at $F$, but the sides of the triangles need not be parallel.

possible diagram of the described counterexample


original wrong answer: I believe (though I do not yet have a proof) that, given two similar triangles in which the lines joining corresponding vertices are concurrent, there exists a dilation (possibly with negative scale factor) centered at the point of concurrency that maps one triangle onto the other. If this is the case, then corresponding sides of the triangles are parallel because dilations map a line to a parallel line.

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