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.)


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Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett