![]() Hypotenuse and construct a semi-circle (or a circle) with radius one-half the hypotenuse. ![]() Since a + b is the length of the hypotenuse, we can take the midpoint of the We will construct a triangle with hypotenuse AB of length a + b such that AD = a and BD = b. Theorem 4.10 gives a suggestion for constructing the geometric mean given two segments of length a and b. One interpretation of the geometry mean is that it is the length of the side of a square with area equal to the area of a rectangle with sides of length a' and b'.ĮXTRA: Prove the converse of Theorem 4.10 - if AB is the side of a triangle divided into segment a' and b' and h is the geometric mean of a'and b', then show that the triangle ABC with altitude h is a right triangle. In a right triangle, the altitude to the hypotenuse is the geometric mean of the segments into which it divides the hypotenuse. The next theorem is one of the most elementary and useful theorems about right triangles. We will be looking at the geometry of inversion later. This result shows that if a segment AB is divided harmonically, then the Circle of Apollonius is a circle of inversion mapping A to B and vice versa. Using the figure above (not the one in the textbook), MD is the radius of the Circle of Apollonius on segment AB, prove that Now, the we need to prove that any point P we locate on the circle will have AP = mBP. These angle bisectors for a right angle and so with the midpoint of DD' at M, we can construct a circle of with diameter DD' with will contain D, C, and D'. From that we can divide the segment AB harmonically by the ratio m and find points D and D' by the interior and exterior angle bisectors at C. We can construct a triangle ABC with AC = mBC. The locus of all point P for which the ratio of the distances from two fixed points A and B is constant is the circle of diameter DD', where D and D' divide AB harmonically in the ratio of that constant.ĭiscuss: What is to prove? Given a segment AB and a ratio m, we want to explore the set of points P such that m(AP) = BP. These two points are sometimes called the foci. Wikipedia Apollonius's Alternative Definition of a circle.Īpollonius discovered that a circle could also be defined as the set of points P that have a given ratio of distances k = d1/d2 to two given points. PROBLEM: How can this construction be accomplished if a + b < AB?Ĭut-The-Knot Locus of Points in a Given Ratio to Two Points. (An alternative construction is given in the textbook.) You may want to create a script tool and save it on your comput3er or your web site. By Theorem 4.7 and Theorem 4.8, the internal and external bisectors of the angle adjacent to sides a and b will intersect AB internally at H and externally at H' to give the desired ratio.Ĭonstruct a GSP file to implement this construction. Construct a triangle with sides a, b, and AB. Represent the ratio by two line segments of length a and b. That isĬonstruction: Given a line segment AB. Therefore the angle measures 90 degrees.Ī line segment AB is divided harmonically in a given ratio ≠ 1 by determining two points, one internal H and one external H' such that aH' = bH. The angle between the internal and external bisectors is the sum of one-half of each. Proof: The sum of the internal angle and the external angle is 180 degrees. LEMMA: The internal and external bisectors of at an angle of a triangle are perpendicular. That is, if AD' is the angle bisector of the external angle at A in triangle ABC, then If an exterior angle of a triangle is bisected, the bisector divides the opposite side externally into segments whose lengths are in the same ratio as the lengths of the other sides of the triangle. What is the locus of the vertex A if base BC is fixed and triangles ABC are constructed so that the ratio of AC to BC is fixed? That is, prepare a GSP animation such than for a base BC, AC = kAB. Open GSP file for Theorem 4.7 and its proof. ![]() That is, if AD is the angle bisector of angle A in triangle ABC, If an interior angle of a triangle is bisected, the bisector divides the opposite side into segments whose lengths are in the same ratio as the lengths of the other sides of the triangle. ![]() Overview of Section 4.2 Applications of Side-Splitting Theorem and Similarity. ![]()
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