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Is it a property that if $\triangle ABC$ is a right angled triangle, all right angled triangles inscribed within $\triangle ABC$ are similar to $\triangle ABC$?

If so, is there a name for this property?

http://gogeometry.blogspot.ca/2009/07/problem-317-right-triangle-inscribed.html

Math1
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    First of all, that's not true for all right triangles inscribed in a right triangle, because the slope of the hypotenuse could be different, making them dissimilar. – kamoroso94 Nov 06 '16 at 02:15
  • Thank-you! That's all I needed to know. I will attach the website where I read it --- I wasn't sure if it was something I'd just never read before. I'll update the post with the website. – Math1 Nov 06 '16 at 02:18
  • In that example, there is a way you can prove all those interior right triangles are similar to the outer right triangle. – kamoroso94 Nov 06 '16 at 02:22
  • Oh, yes - I am able to do that (mostly A-A-A), but one of the statements just says that all angles are similar to $\triangle ABC$ and I didn't think that the blanket statement could be made like that. – Math1 Nov 06 '16 at 02:29

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Similarity of two objects (whether they are triangles or any other geometrical shapes) can be understood as the condition that we can get the bigger one from the smaller object by zooming in. (or the other way by zooming out). In these days google maps, and mobile phones having applications that enlarge or diminish the image, this is the best real life experience through which this mathematical process should be understood.

With this think of a right-angled triangle and another one inscribed there.

Minimum requirement is when the the smaller triangle is placed (by rotating/flipping if needed) so that one vertex and one side of it are aligned with one vertex and one side of the bigger one, the other sides should be parallel.

P Vanchinathan
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Well, $ABC$ and $NPC$ share an angle $\angle BCA$, and they are both right angled, so they both have a right angle. The third angle must be the same because two pair of angle is the same $\implies $ last pair is same as angles sum up to $180^{\circ}$. So $ABC$ is equivalent to all other triangles, as $NPC$ is.

user160738
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