but over a few posts I want to explore one of the most interesting techniques: focusing on symmetry through rigid transformations. I find its very useful on many more complex problems which require auxiliary lines. Thinking about the problem in terms of transformations makes it easier to find structure/solutions. With all these examples I try to document my process in an effort to show how you often can try different approaches and experiment.
This was a problem posed by five_triangles: I think its good introductory example of the power of rigid transformations (reflection).
Square ABCD, diagonals meet at E, ∠BAC bisected, meeting BD&BC at X&Y. Prove 2XE=YC. For what school year?
To answer the question I gauge this at the advanced high school level due to its need for auxiliary lines. Here's the quick proof I sketched out:
My thinking on this problem was to increase the symmetry through reflection. That had the immediate benefit of creating a new triangle with one side of the required length (2XE). I then started to look at the implications of the 2 similar triangles AXF and AYG. It was also fairly obvious that if I could find a number for the ratio that the base could be translated into the required side since they were both on a 45-45-90 triangle. I made an early error and assumed that triangle had a side length of exactly 1/2 the square's side. That quickly led to an erroneous answer. After I rechecked my steps realized I had made a bad assumption. From there some quick angle chasing confirmed my visual impression that all 4 triangle are congruent above. Because I was aiming to use the similar triangles, to find a constant ratio and then transform that to refer to YC it was straightforward to do the computations in step 6.
Second Day: http://mymathclub.blogspot.com/2015/09/geometry-walk-though-2.html