3:36 although numerically correct, I think, because we are equating ratios between similar triangles, the second ratio should be expressed as ab/bc rather than ac/bc (because c is only present in one triangle so that is comparing two sides within one triangle). I really like the geometric proof, having come across it while trying to fathom pentagons.Thanks.
Very nice! It's also fun to see that with all of these similar triangles, we can show the relationship between cos(pi/5) and phi directly from the geometry. We know AC/BD = BC/AB. We also know AB = AC = DC and BC = BD + DC. So we can write DC/BD = (BD+DC)/DC = phi, because when that relationship holds between the lengths of the line segments it's the definition of phi. So we have cos(pi/5) = (BC/2)/DC = (BD+DC)/2DC = phi/2. Have fun! I like your videos.
3:36 although numerically correct, I think, because we are equating ratios between similar triangles, the second ratio should be expressed as ab/bc rather than ac/bc (because c is only present in one triangle so that is comparing two sides within one triangle). I really like the geometric proof, having come across it while trying to fathom pentagons.Thanks.
Very nice!
It's also fun to see that with all of these similar triangles, we can show the relationship between cos(pi/5) and phi directly from the geometry.
We know AC/BD = BC/AB.
We also know AB = AC = DC and BC = BD + DC.
So we can write DC/BD = (BD+DC)/DC = phi, because when that relationship holds between the lengths of the line segments it's the definition of phi.
So we have cos(pi/5) = (BC/2)/DC = (BD+DC)/2DC = phi/2.
Have fun! I like your videos.
I remember doing this proofs for these “special angles” as an undergrad. Thanks for reminding me of it! I’ll give it to my son as an exercise…