Revision as of 08:12, 2 December 2009 by Dknott01 (Talk | contribs)


HW 12

Does anyone know how to do 2, 3, 5, 7, or 8?

  • Here's what I did for problem two. Construct along EF a length equal to BC (prop 2), then construct an angle, at E, called MEF, such that angle MEF = angle ABC (prop 23). Then construct along segment EM a length EN equal to AB. Proposition four suggests triangles NEF and ABC are congruent, and from there the argument is more or less the same.



Back to MA460 (Fall2009Walther) Homework

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett