19+Oct+-+Continuing+Congruence

Continuing Congruence:
 * I have included a link to the original document with formatting and such.

Today we started off by proving Proposition 3.3 – Segment Subtraction: If A * B * C and D * E * F, AB ≅ DE, and AC ≅ DF, then BC ≅ EF. Proof: 1. By C-1, we will say that F' lies on ray BC such that EF ≅ to BF'. 2. Because A * B * C, D * E * F, AB ≅ DE, and BF' ≅ EF, then AF' ≅ to DF by C-3. 3. Since AF' ≅ DF and AC ≅ DF and because both F' and C lie on ray AB and since AC ≅ AF (given), and because there is a uniqueness of such a point C on ray AB, we can deduce that F' = C. 4. From part 1, if EF ≅ BF' and C = F', then EF ≅ BC.

After this proof we had a question about whether or not we can use Algebraic substitution, such as in the following example:

1. Proposition 2.13 says if A * B * C, then AC = AB ⋃ BC and B is the only point in common to segments AB and BC. 2. If we know that AB ≅ DF, can we substitute DF for AB in the above equation such that AC = DF ⋃ BC? 3. We concluded that no, this is not possible as in the following case: a. If we have a line such that A * B * C and AB ≅ DF, we can see that AB ⋃ BC ≇ DF ⋃ BC. 4. This led to another question: Can we say that AB ⋃ BC ≅ DF ⋃ BC? a. We answered this by asking another question: By our definition of congruent, what may be congruent? b. We have defined congruence for segments, triangles, and angles. c. From this we concluded that we cannot state that AB ⋃ BC ≅ DF ⋃ BC because this statement has not been defined.

Continuing on with our proofs, we next set out to prove Proposition 3.4: If AC ≅ DF, then for any point B between A and C, there is a unique point E between D and F such that AB ≅ DE. Proof: 1 . There is a unique point E on ray DF such that AB ≅ DE (By C-1) 2. Hypothesis: Suppose that E were //not// between D and F. 3.  Then either E = F or D * F * E or D * E * F. 4.  If E = F and B and C are 2 distinct point on ray AC (by B-1 ) such that AC ≅ DF ≅ AB (by C-2) then there is a contradiction to the fact that B and C are unique, which was given. 5.<span style="font-family: 'Times New Roman'; font-size-adjust: none; font-size: 7pt; font-stretch: normal; font-style: normal; font-variant: normal; font-weight: normal; line-height: normal;"> If D * F * E, then by C-1, there is a point G on the ray opposite to ray CA such that FE ≅ CG. a.<span style="font-family: 'Times New Roman'; font-size-adjust: none; font-size: 7pt; font-stretch: normal; font-style: normal; font-variant: normal; font-weight: normal; line-height: normal;"> *As a side note, there was discussion as to whether it was necessary to include the phrase “on the ray opposite to ray CA.” The decision was that in order to use C‑1 there must be a given ray, so this phrase was necessary. 6.<span style="font-family: 'Times New Roman'; font-size-adjust: none; font-size: 7pt; font-stretch: normal; font-style: normal; font-variant: normal; font-weight: normal; line-height: normal;"> Then AG ≅ DE by C-3 (We know that AC ≅ DF and CG ≅ FE, and A * C * G and D * F * E, therefore we can conclude that AG ≅ DE.) 7.<span style="font-family: 'Times New Roman'; font-size-adjust: none; font-size: 7pt; font-stretch: normal; font-style: normal; font-variant: normal; font-weight: normal; line-height: normal;"> There are two points B and G on ray AC such that AG ≅ DE ≅ AB. Therefore we can conclude that B and G are the same point which contradicts B-3 (We are given that A * B * C and since B and G are the same point, we can conclude (by a substitution that is valid) A * G * C. However, we chose G such that A * C * G, and this contradicts B-3). 8.<span style="font-family: 'Times New Roman'; font-size-adjust: none; font-size: 7pt; font-stretch: normal; font-style: normal; font-variant: normal; font-weight: normal; line-height: normal;"> Therefore we can conclude that the only case that can be valid is that D * E * F.

We then discussed Definition 3.5: AB < CD if here is a point E between C and D such that AB ≅ CE. 1.<span style="font-family: 'Times New Roman'; font-size-adjust: none; font-size: 7pt; font-stretch: normal; font-style: normal; font-variant: normal; font-weight: normal; line-height: normal;"> What we thought was important about this is that we can talk about the length of a segment without actually measuring it because we can discuss whether it is “less than” another segment without assigning a number to its length.

We did not have to prove Proposition 3.6 – segment ordering: Exactly one of the following holds: AB < CD, AB ≅ CD, CD < AB (a.)<span style="font-family: 'Times New Roman'; font-size-adjust: none; font-size: 7pt; font-stretch: normal; font-style: normal; font-variant: normal; font-weight: normal; line-height: normal;"> If AB < CD and CD ≅ EF, then AB < EF. (b.)<span style="font-family: 'Times New Roman'; font-size-adjust: none; font-size: 7pt; font-stretch: normal; font-style: normal; font-variant: normal; font-weight: normal; line-height: normal;"> If AB > CD and CD ≅ EF, then AB > EF. (c.)<span style="font-family: 'Times New Roman'; font-size-adjust: none; font-size: 7pt; font-stretch: normal; font-style: normal; font-variant: normal; font-weight: normal; line-height: normal;"> If AB < CD and CD < EF, then AB < EF. 1.<span style="font-family: 'Times New Roman'; font-size-adjust: none; font-size: 7pt; font-stretch: normal; font-style: normal; font-variant: normal; font-weight: normal; line-height: normal;"> We also concluded that this last one could also be re-written as “if AB > CD and CD > EF, then AB > EF.

We next introduced Definition 3.7: Let //l// and //m// be two distinct lines meeting at B and A * B * C on //l// and D * B * E on //m//. Angles ∡ ABD and ∡ CBE are called vertical angles. We were lead to a discussion about how this statement could be confusing to students. While in some drawings, it makes sense that these are vertical angles because they are “up and down,” in many they could be “horizontal” angles. We talked about how we give our students definitions of many words, like vertical, but that they don’t always mean the same thing; sometimes vertical can mean something else. This helps us understand that when we teach new concepts to students, especially concepts that could contradict previous knowledge, we need to be aware of the confusion it could be causing our students.

Similar to our students to the above example, Definition 3.8 could cause us some confusion and discomfort. It states “if two angles, ∡ BAC and ∡ DAC have a common side, ray AC and two other sides ray AB and ray AD are opposite rays then we say that the angles are supplements of each other, or supplementary angles.” This contradicts our previous idea that supplementary angles are “angles whose measures add up to 180°.

Additional Thoughts: Test one is on Wednesday and Emina said: ·<span style="font-family: 'Times New Roman'; font-size-adjust: none; font-size: 7pt; font-stretch: normal; font-style: normal; font-variant: normal; font-weight: normal; line-height: normal;"> We have to know the definitions, but she will provide us with the propositions, theorem, etc. ·<span style="font-family: 'Times New Roman'; font-size-adjust: none; font-size: 7pt; font-stretch: normal; font-style: normal; font-variant: normal; font-weight: normal; line-height: normal;">  There won’t be any questions on congruence. ·<span style="font-family: 'Times New Roman'; font-size-adjust: none; font-size: 7pt; font-stretch: normal; font-style: normal; font-variant: normal; font-weight: normal; line-height: normal;"> There won’t be a lot of proofs, but there will be at least one. ·<span style="font-family: 'Times New Roman'; font-size-adjust: none; font-size: 7pt; font-stretch: normal; font-style: normal; font-variant: normal; font-weight: normal; line-height: normal;"> Questions include (but are not limited to): looking at models, true/false, providing definitions, etc. ·<span style="font-family: 'Times New Roman'; font-size-adjust: none; font-size: 7pt; font-stretch: normal; font-style: normal; font-variant: normal; font-weight: normal; line-height: normal;">  ***Read the Questions Carefully*** Also, the test will not take the whole time and we should be prepared to prove more propositions, including Proposition 3.9

Question about possible correction: On the proof of Proposition 3.4 #4 it states that those conditions are true by B-1 but I understand those conditions hold by C-1 (not B-1).