What are some non-trivial ways to construct a similar triangle from a right triangle?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 △ A B C with ∠ A B C = 90 ∘ , given points D on A C ¯ and E on B C ¯ such that ∠ D E C = 90 ∘ we have △ D E C ∼ △ A B C, 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 △ A B C, let D be the midpoint of A C ¯ and E be on B C ¯ such that ∠ B E D = 90 ∘ . Then, △ B E D ∼ △ A B C. 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)?

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What are some non-trivial ways to construct a similar triangle from a right triangle?By non-trivial, I mean not:- The triangle itself or any translations or rotations of it- A &#039;shrinking&#039; of the right triangle; that is, for   △  A  B  C with   ∠  A  B  C  =      90    ∘  , given points   D on             A      C        ¯   and   E on             B      C        ¯   such that   ∠  D  E  C  =      90    ∘   we have   △  D  E  C  ∼  △  A  B  C, 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&#039;m not sure if it is true (verification would be appreciated); given right triangle   △  A  B  C, let   D be the midpoint of             A      C        ¯   and   E be on             B      C        ¯   such that   ∠  B  E  D  =      90    ∘  . Then,   △  B  E  D  ∼  △  A  B  C. 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)?

Nancy Sandoval · Accepted Answer

Let   M,   N,   P be the middle of   A  B,   B  C, and   C  A. You have now 4 triangles similar to the initial one.

What are some non-trivial ways to construct a similar triangle from a right triangle? By non-trivia

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