10_26

Proposition 3.11: (i): Vertical angles are congruent (ii): An angle congruent to a right angle is a right angle

Part (i) of this proposition may be examined as follows: Construct two arbitrary lines {AC} and {DE}. Line {AC} exists such that A*B*C and line {DE} exists such that D*B*E. By C-5, angle ABE is congruent to angle ABE. By Definition 3.8, angle ABD is supplementary to angle ABE. Angle EBC is also supplementary to angle ABE. By Proposition 3.9, since angles ABD and EBC are both supplementary to angle ABE, angle ABD is congruent to angle __EDC__. By Definition 3.7, angles ABD and EBC are Vertical Angles. As these angles are vertical and congruent, this completes the proof that vertical angles are congruent.

Part (ii) of this proposition may similarly be examined. The main difference in the proof of this part of the proposition from part (i) lies in this part beginning with a right angle rather than any arbitrary angles. Rather than following that same argument, let us examine a different approach to the proof for this part of the proposition: Construct two lines {AC} and {BD} such that A*B*C and B*D*E. Let angle ABD be the given right angle. As angle ABD is supplementary to angle DBC, Definition 3.10 implies that angle DBC is congruent to angle ABD. By C-4, an angle, D'B'C' may be constructed such that angle D'B'C' is congruent to angle DBC. Similarly, through C-4 one may construct an angle, A'B'D' such that angle ABD is congruent to angle A'B'D'. By C-5 (Transitivity), angle ABD congruent to angle DBC and angle ABD congruent to angle A'B'D' imply that angle DBC is congruent to angle A'B'D'. This shows that an angle congruent to a right angle is also a right angle. -- What was shown here is that one can constrcut an angle congruent to a right angle, which is basiically waht C4 gives us. What was to be shown is that if we already have an angle congruent to a right angle then that angle itself is a right angle, or: If ang(ABC) is a right angle and ang(DEF) congruent to ang(ABC), then ang(DEF) is a right angle. This has not been shown here.

Proposition 3.12: For every line {l} and every point, P, there is a line through P with is perpendicular to {l}. This proposition lies on demonstrating this fact for two cases. The hypothesis for this proposition does not clearly state whether the point P is incident with line {l}. As this is the case, point P either lies on {l} or it does not. As this is not explicitly stated in the hypothesis, both cases need to be examined in order to prove the proposition. The proof of this proposition is left to the reader.

Given talk of segments, congruent angles, and the upcoming lab, questions arise on circles and what they may be used to demonstrate in terms of congruence. Before one may use circles, one must define what circles are. Given the limited vocabulary which does not include any talk of length, circles must be defined without mentioning length.

Definition of Circle: Given a segment PA and a point O, a circle, with center O and radius (congruent to) PA is the set of all points X such that OX is congruent to PA. As this definition does not mention distance or length, this definition is completely acceptable given the limited vocabulary built thus far in our geometry.

Given this new definition, various questions arise. A few of these questions may be seen below: 1: Given a segment AB, is there a point P such that A*P*B and AP is congruent to PB? If there exists such a point, P, then we may "revolve" the segment AB around P. This allows for the construction of a circle centered at P with radius AP. Despite this, the circle described prior is not the only circle through points A and B. 2: How can I find this point P? As P is not unique, this point P may not uniquely be determined. An idea that may come to mind may involve determining a midpoint for AB and then finding the perpendicular line to AB. An idea may suggest that all points on the perpendicular line may be such P. For the sake of brevity and this new concept, call the perpendicular that intersects AB at its midpoint the "Perpendicular Bisector." Is the point P here the same as the P you were looking for in 1? Because if it is, then maybe you need to rethink the "it can not be uniquely determined". If it is not, then the first sentence needs to be restated.

Given this Perpendicular Bisector, can it be proven that all points, P, lying on the bisector are equally "distant" from points A and B. This question can be proved through application of SAS. This will be left to the reader for exercise. Why? How can it be proven?

Final question: How do we know that given any circle, centered at point A, and a point not in the interior of the circle, point B, does there exist an intersection point for the segment AB and the circle?

Comments:

In regards to the final question I think we know that this is true because of continuity and those few axioms that we talked about in class. This was discussed and we decided that a circle is continuous, therefore any segment AB will intersect the circle.