11_30

Discussion began with some of the problems that were on the fourth lab assignment in class. The first problem on the lab assignment had to do with three cities that needed to build an airport in order serve the citizens of the respective cities. In order to fairly serve the three cities, it was determined that the airport was to be equally distant from each city. Where it this location? Answer: In order to find this location, one must use Perpendicular Bisectors to locate the position that the new airport should take in relation to the existent cities. Instead of describing the construction of bisectors, we will examine what is known about Perpendicular Bisectors.

What is known about Perpendicular Bisectors? Proven previously, when the Hiker/River Problem was examined, is the fact that any given point, P, is equidistant from two endpoints of any given segment, if P is on the Perpendicular Bisector to the given segment. It seems to make sense, especially when utilizing GeoGebra, that the Perpendicular Bisectors for a triangle intersect. The question now becomes whether or not the three Perpendicular Bisectors really do intersect. Let triangle ABC be the triangle formed by the segments AB, BC, and AC. Construct the Perpendicular Bisectors to segments AB and BC; Let P be the intersection point of these two bisectors. As P is on the Perpendicular Bisector of AB, segment AP is congruent to segment BP. Similarly, as P is on the Perpendicular Bisector of BC, segment BP is congruent to segment CP. Utilizing Axiom C-2, Transitivity of Congruent Segments, this implies that segment AP is congruent to segment CP. This implies that P is equidistant from points A and C, thus allowing one to conclude that point P is on the Perpendicular Bisector of segment AC. This concludes the proof that the three Perpendicular Bisectors in triangle ABC are concurrent, intersecting in one point.

Further, this proof allows one to conclude that the intersection of the Perpendicular Bisectors is equidistant from points A, B, and C. This point is called the Circumcenter for Triangle ABC. The Circumcenter is the center of a circle which has points A, B, and C all lying on the circle. One may ask if this construction of a "Circumcenter" can be extended to polygons with four or more sides. This construction can be followed for some polygons with four or more sides. But, in general, this is not repeatable for //all// polygons with four or more sides. If this construction can be followed for a Quadrilateral, then the Quadrilateral is called Cyclic.

Now that Perpendicular Bisectors have been thoroughly reviewed, one can now turn their attention to Angle Bisectors.

Similar to the construction of Perpendicular Bisectors is the construction of Angle Bisectors. Let Triangle ABC be any arbitrary triangle. We can construct the angle bisectors for Angle A, Angle B, and Angle C. Let P be the point of intersection of the bisectors of Angle A and Angle B. Let points D, E, and F exist such that D lie on Ray CA; F lie on Ray CA; and Point E lies on Ray BC. P lying on the Angle Bisector to Angle A implies that segments PD and PE are congruent. P lying on the Angle Bisector to Angle B implies that segments PE and PF are congruent. Utilizing C-2 implies that segments PD and PE are congruent. This implies that P lies on the Angle Bisector of Angle C. Note that the intersection of the Angle Bisectors is called the Incenter. The Incenter is the point that is the center of the largest circle that can be inscribed in Triangle ABC.

Now a question that may arise is in regards to what can be said about a given point and its relation to the Angle Bisector. To formalize, "If a point P is equidistant from the legs of an angle, then is P on the Angle Bisector? The proof for this can be done through application of the Pythagorean Theorem and determination of Congruent Triangles. These proofs will be left to the reader to review.

Another element of triangles is what are referred to as Medians. Medians are segments that bisect the segment opposite any vertex and connect to that vertex opposite the newly bisected segment. What can be said about the three Medians of Triangle ABC? The Medians intersect at what is referred to as the Centroid of Triangle ABC. The intersection of the Medians leads to creation of six sub-triangles contained within the original Triangle ABC. Through use of Area Axioms and the Area Formula for a triangle, it can be shown that the six sub-triangles all have equal area. This was demonstrated in class, and the demonstration of this will be left to the reader. As a side note, the Centroid is said to be the center of mass for a triangle. Meaning, that for any given Triangle ABC, the Centroid is the point where the triangle will be in balance when placed upon any finitely small point.

Finally, moving back to the Lab comes the question of where an airport should be constructed in order to have the cities drive the same distance and create the smallest amount of new highway. The explanation of this can be long, and may need to be adjusted for the type of triangle that it given (Acute, Obtuse, Right). Suffice it to say that Triangle ABC is an obtuse triangle. Let segments AB, BC, and AC be such that AB < BC < AC. Find the midpoint of the longest segment, in this case AC. Construct the Median, BD, such that D is the midpoint of segment AC. Suppose that l(AD) = l(DC) < BD. Find the point E such that E is on BD and the distance from E to AC is half of the segment length difference when examining AD and BD.