11_2

We started out the day by proving SSS which states for triangles ABC and DEF, if we have AB congruent to DE, BC congruent to EF and AC congruent to DF, then triangle ABC is congruent to triangle DEF.

After drawing each triangle and marking the congruent sides, we created an angle CAH on triangle ABC with the properties that CAH is congruent to FDE and AH congruent to DE //(By C-1 and C-4).// By SAS, we can say that triangles CAH and FDE are congruent. We then listed the segments we knew were congruent:

AB, DE, and AH are conguent to each other BC, EF, and CH are congruent to each other AC and DF are congruent to each other

We then created segment BH which intersects AC at some point P. We then talked about where P could be. P can be between A and C, it can be point A or C, or it can be on either side of segment AC but on line AC. We then looked at the case where P is between A and C and later talked about the other cases. By creating this segment we have two new isosceles triangles, ABH and BCH. By 3.2 we know that the the base angles are congruent so:

Angles ABH and AHB are congruent and angles CBH and CHB are congruent.

By 3.15 we then know that we can add these angles and get that angles ABC and AHC are congruent. By SAS, we know that triangles AHC and ABC are congruent and, consequently, triangles ABC and DEF are congruent.

Next, we broke off into groups and talked about a handout given to us. We discussed how to find the measures of the angles on the handout. //__Some angles were off the page. We discussed what assuptions could or could not be made. We didn't know if lines were or were not parallel We didn't know if we were working in the Euclidean or Hyperbolic plane.__// The class came to the conclusion that we can find the measure of all the angles. On the picture, we drew a line perpendicular (a) to a line (l) on the page, we then drew a perpendicular to line (a) and claimed that it was parallel to the line (b). We also claimed that by doing this the angle made by line (l) and some other line (c) is the same as the angle made by the parallel line (b) and line (c). This is one way to find the measure of an angle that we can't necessarily see.//__Another way to find the measure of the anges we could not see:__// Since it looked like there were triangles made by three lines, we claimed that if we knew the measure of two angles we knew the measure of the remaining angle by adding the two angles and subtracting that from 180 degrees. We know this because the sum of the interior angles of a triangle is 180 degrees. //__This assumes there is a third angle, and therefore a triangle.__//

We were then given the following problem: Three kids were running in a circle while holding hands. One child's arms made 60angle and the other's 52. Can you determine the angle that the third child's arms made? ​We drew a picture (after a discussion about what this problem means) and decided that we can find the third angle as long as we are working in Euclidean space. It was commented that, from our lab, we know that the sum of the interior angles of a triangle is not 180 degrees in the hyperbolic plane.

Our conversation turned to trying to define a square and the following comments were made: "Parallelogram with all congruent angles." "All interior angles are right" and "all sides are congruent" "Equilateral, equiangular quadrilateral." Again, in the hyperbolic plane, a square is not what we expect it to be. The angles in the interior of a 'square' in the hyperbolic plane are not all 90 degrees. The observation was made that the lines in a square that should be parallel really aren't. This led to the comment that parallel lines should remain equidistant.

We then moved to Alternate Interior Angles (AIA) which states that if the alternate interior angles formed by a transversal and line (l) and (l') are congruent, then (l) and (l') are parallel. We looked at a few questions that are related to this theorem. First, what can we say about two two lines that are perpendicular to the same line? We decided that by AIA we know that the lines are parallel. Next, suppose we have a line (l) and a point P not on (l), is there a parallel to (l) through P? We decided that one exists and then were asked if there was more than one parallel to (l) through P. After a lot of discussion, by C-4 we know that there can only be one parallel. We were then asked about the converse of AIA and whether is was a theorem in neutral geometry. It turns out that it is in euclidean geometry but not in neutral geometry, //__which also includes the hyperbolic plane__//.

Last, we were presented with two corollaries and a few questions: Corollary 4.1.1- Two lines perpendicular to the same line are parallel. We had talked about this earlier in the class. Corollary 4.1.2- If l is any line and P a point not on the line, then there exists at least one line m through P parallel to l. We talked about how the phrase 'there exists at least one' can lead to some debate.

HOMEWORK: Prove AIA and answer the questions: What is a parallelogram? List everything you know about parallelograms.