$\begingroup$

I know that if two triangles are mapped to one another by a homothety at a point as center, they are similar. But is the converse true?

If two triangles are similar then they always possess an insimilicenter and exsimilicenter?

More specifically, are two similar triangles always perspective from a point?

I tried to use Menelaos and Desargues, but cannot arrive at a conclusion, but i think this is not a correct statement.

If it is wrong, does any extra condition with similar triangles criteria make them perspective from a point?(except being perspective from a line)

Thanks

$\endgroup$

1 Answer

$\begingroup$

Given two similar triangles ($\rm ABC$ and $\rm A'B'C'$), draw three straight lines through the corresponding vertices of both triangles ($\rm AA'$, $\rm BB'$, and $\rm CC'$). Either these three lines intersect at one point, or they don't. In the former case, that's your homothety center. In the latter case, you don't have one.

If the triangles are symmetric (that is, at least isosceles), you might even have two centers, but that's another story. I was wrong, you can't have two centers at once. You have either the insimilicenter, or the exsimilicenter, or no center at all. Why? Well, because triangles don't have the central symmetry, while circles do.

$\endgroup$ 1

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy