I am wondering what ways there are of constructing similar triangles from a right triangle by non-trivial means.
By non-trivial, I mean not:
- The triangle itself or any translations or rotations of it
- A 'shrinking' of the right triangle; that is, for $\triangle ABC$ with $\angle ABC = 90^ \circ$, given points $D$ on $\overline{AC}$ and $E$ on $\overline{BC}$ such that $\angle DEC = 90 ^ \circ$ we have $\triangle DEC \sim \triangle ABC$, along with any arbitrary renaming of points. Or, more generally, any scaling of the triangle.
One of the most well-known ways to construct a similar triangle is by dropping a cevian from the hypotenuse to the vertex which is perpendicular to the hypotenuse. In other words, an altitude.
I can only think of one more case, and I'm not sure if it is true (verification would be appreciated); given right triangle $\triangle ABC$, let $D$ be the midpoint of $\overline{AC}$ and $E$ be on $\overline{BC}$ such that $\angle BED = 90^ \circ$. Then, $\triangle BED \sim \triangle ABC$. In fact, I think this triangle is congruent to the one mentioned in the last paragraph. Also not sure on that, though.
The last one came about as a surprise while I was working on a proof, but I could not prove it nor could I think of any other ways to construct a similar right triangle to one given. So, my question: given a right triangle, what are some ways to construct a similar triangle (preferably ones that you found clever/interesting)?
Let $M$, $N$, $P$ be the middle of $AB$, $BC$, and $CA$. You have now 4 triangles similar to the initial one.