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Case 2: they are not parallel. Then they meet, let X be the meeting point. | Case 2: they are not parallel. Then they meet, let X be the meeting point. | ||
You get lots of similar triangles involving X, and with some algebra this should not be hard. | You get lots of similar triangles involving X, and with some algebra this should not be hard. | ||
| − | + | (Note that (u+v)/v =u/v + 1.) | |
[[ HW 4|Back to HW 4]] | [[ HW 4|Back to HW 4]] | ||
Latest revision as of 17:11, 22 September 2009
HW4-10MA460Fall2009
Case 1: AB and DE are parallel.
That's easy.
Case 2: they are not parallel. Then they meet, let X be the meeting point.
You get lots of similar triangles involving X, and with some algebra this should not be hard. (Note that (u+v)/v =u/v + 1.)
