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* 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. | * 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. | ||
| + | * For number seven, think about theorem fifteen. That should tell you which extra line you're drawing in each triangle. Once you've done that, the solution should become pretty clear. | ||
any hints for # 1? | any hints for # 1? | ||
Revision as of 11:07, 2 December 2009
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.
- For number seven, think about theorem fifteen. That should tell you which extra line you're drawing in each triangle. Once you've done that, the solution should become pretty clear.
any hints for # 1?
- for number 1 I found three sets of similar triangles (DBI~FBI, FCI~ECI, EAI~DAI). Then when you set up the ratios you get three things equal to 1 (DB/FB, EA/DA, FC/EC). Then you can multiple those all together and change them to signed ratios.
Any suggestions for #3?
