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| + | 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 | ||
| + | ** Ok, I'll be more specific. Take the midpoints of CB and FE. Call them P and Q, respectively. Construct lines MP and NQ. We know that CM=FN. We know that CP=FQ (because they're both one half of CB and FE respectively, which are given as equal). And theorem sixteen says that MP is 1/2 AC, and that NQ is 1/2 DF. But AC=DF is given, so therefore MP=NQ. Now CPM and FQN are congruent by SSS. Therefore angle MCB = angle NFE. Therefore MCB is congruent to NFE. Therefore MB=NE. Should be very obvious from there. | ||
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| + | any hints for # 1? | ||
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| + | *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. | ||
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| + | Any suggestions for #3? | ||
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| + | Does anybody have suggestions for number 8? | ||
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Latest revision as of 04:13, 3 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
- Ok, I'll be more specific. Take the midpoints of CB and FE. Call them P and Q, respectively. Construct lines MP and NQ. We know that CM=FN. We know that CP=FQ (because they're both one half of CB and FE respectively, which are given as equal). And theorem sixteen says that MP is 1/2 AC, and that NQ is 1/2 DF. But AC=DF is given, so therefore MP=NQ. Now CPM and FQN are congruent by SSS. Therefore angle MCB = angle NFE. Therefore MCB is congruent to NFE. Therefore MB=NE. Should be very obvious from there.
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?
Does anybody have suggestions for number 8?
