More discussion of rhombi today.

Mr Reihman took up his work on 1.6 from Friday. He had two gaps at that time, and he has managed to patch one of them: he gave an argument for why the interior angles of a triangle, taken together, make two right angles.

The other statement he needs is one half of this:

Conjecture D (Reihman): Suppose that we have a diagram where A and D lie on the same side of line BC. Then angles ABC and BCD taken together make two right angles if and only if lines AB and CD are parallel.

We have added this to the list of class conjectures.

Ms Maass then took up 1.6 and gave a short argument relying on Euclid I.27 to do the heavy lifting.

Mr Schulte gave an argument for Conjecture A. This relies on Ms Lewis’ Theorem 1.7, so it smuggles in a hypothesis that the diagonals of a rhombus cross. When Conjecture 1.2 becomes a theorem, then we can remove this extra hypothesis.

*Theorem (Schulte): Suppose that ABCD is a rhombus and that the diagonals AB and CD meet. Then angles BAC and BDC taken together make a right angle.

Ms Carpenter had much the same argument.

Then Ms Van Donselaar took up Conjecture 1.2. This quickly got bogged down (because it is challenging). We talked a little bit about ideas.

  • “meet” means “not parallel” so perhaps that is a line of approach,
  • We now have that a rhombus is a parallelogram, so maybe that is useful?
  • exploring the alternate case, that the lines don’t meet, requires drawing and reasoning from an “impossible figure.” Can that be used?

Keep up the good work.

Contact Prof Hitchman
Theron J Hitchman
Department of Mathematics 0506
University of Nothern Iowa
Cedar Falls, IA 50613-0506
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Office: 327 Wright Hall
Phone: 319-273-2646
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