More on kites and rhombi today.
Ms Carpenter started an argument for 2.5, but chose to pass the chalk.
Ms Lewis gave a new construction of a rhombus. We took the opportunity to practice theorem-writing, and came to the conclusion that this is just a “better” proof of Theorem 1.4. Also, we wonder if we have stumbled on to this:
Conjecture I: (Lewis) Given a segment AB and an angle XYZ, it is possible to construct a rhombus ABCD with angle ABD congruent to angle XYZ using a compass and straightedge.
The advantage to such a statement is it looks like we can get the entire family of all rhombi into one unified construction, which would be really cool.
Ms Maass then shared work in the neighborhood of 2.1. She worked with a proof by contradiction scheme, but the logic was a little tricky to keep straight, and the statement wasn’t so clear at first. So we practiced our theorem writing again. I cannot stress enough that, “Half a proof is not proof, but progress is progress.” By taking our time and examining what we had in front of us, we found that Ms Maass has this result:
Theorem: (Maass) Let ABCD be a kite where AB is congruent BC, AD is congruent DC, and AB is not congruent to AD. Then segment AC is not the angle bisector of both angle BAD and angle BCD.
Our last discussion was led by Mr Reihman on 2.4. In an amazing turn of events, his proof had a small gap in it (using Euclid’s Proposition I.18 is tricky), but this gap is perfectly filled by Ms Maass’ theorem. I am just astounded at how well this worked out.
Theorem: (Reihman) Let ABCD be a kite where AB is congruent BC, AD is congruent DC, and AB is not congruent to AD. Then ABCD is not a parallelogram.
Again, the proof is organized as a proof by contradiction.
I am not going to set new tasks for you, as we still have several open conjectures made by the class we could discuss. We also have not quite completed our discussion of 2.2.