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* For number seven, think about theorem sixteen. That should tell you which extra line you're drawing in each triangle. Once you've done that, the solution should become pretty clear. | * For number seven, think about theorem sixteen. That should tell you which extra line you're drawing in each triangle. Once you've done that, the solution should become pretty clear. | ||
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+ | -Do you have any information about #7 because we can't seem to get them congruent because we cannot compare angles between triangls, only within triangles | ||
any hints for # 1? | any hints for # 1? |
Revision as of 11:45, 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 sixteen. That should tell you which extra line you're drawing in each triangle. Once you've done that, the solution should become pretty clear.
-Do you have any information about #7 because we can't seem to get them congruent because we cannot compare angles between triangls, only within triangles
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?