Today I had fun working on a geometric proposition about trapezoids and reading some mathematical correspondence. I may do more math activities later, as well.
The proposition I worked on, which comes from Elementary College Geometry, is that the base angles of an isosceles trapezoid are equal. (That book defines a trapezoid as a quadrilateral with exactly two parallel sides. An isosceles trapezoid is a trapezoid that has legs of equal length. The legs are the non-parallel opposite sides.)
I drew a number of pictures trying to find a good way to prove this proposition.
The first one shows the approach I came up with when I first looked at the problem some time ago: construct isosceles triangles on either end of the trapezoid, show those are congruent (likely requiring some lemmas), use alternate interior angles. I think this will probably work, but it will need some details ironed out. For instance, what if $\angle ABC$ is acute? (It can’t actually be with Elementary College Geometry’s definition of a trapezoid and $AD$ equal to $BC$, but I may need to prove that.)
This method is not very elegant, even ignoring the niggly details, so before working those out, I decided to see if I could find something better. In the process, I came up with the figures below. None of these approaches has born fruit so far, though.
In the first drawing, I tried the most basic quadrilateral proof move, but just went around in circles. The second drawing didn’t really go anywhere at all. Something resembling the third drawing will be useful for proving the converse of the proposition, I think, but didn’t yield a proof in this direction.
I’m feeling much encouraged and less inclined to simply drop this project, which I had been resisting the urge to do.