Talk:Equiangular Right Triangles are Similar

Should we raise the following?

IMO we should start with the right triangle we want to prove something about and construct the rectangle from it that we are working on. Otherwise we can be accused of saying: this just uses the fact that two halves of a rectangle are congruent right triangles. How do we know there are right triangles that do not arise as halves of a rectangle? If we start with the RHT and then construct the rectangle we don't have to plug that gap. Although that argument would be plugging that gap. --prime mover (talk) 12:52, 15 October 2023 (UTC)

I think part of the charm of the proof comes from starting with the rectangle but YMMV. Maybe a lemma proving that two copies of any right triangle can be used to construct a rectangle? --Telliott99 (talk) 13:23, 15 October 2023 (UTC)

A sketch

Let ABC be an arbitrary right triangle with the right angle at B.

Extend one line from A parallel to the base, and a second line coming from C parallel to AB, meeting at D.

By parallel lines ...

$\angle ADC = \angle ABC = 90$

$\angle BAD = \angle BCD = 90$

By ...:

$ABCD$ is a parallelogram

By ...:

ABCD is a rectangle

--Telliott99 (talk) 13:32, 15 October 2023 (UTC)