Explanation Procedure used: The locus of points in a plane and equidistant from the sides of an angle is the angle bisector. Proof: 1) If a point is on the angle bisector, then it is equidistant from the sides of the angle. In the following figure, the bisector B D divides the angle ∠ A B C and the point D lies on the bisector B D . D E ¯ ⊥ A B → and D F ¯ ⊥ B C → . Since B D → bisects ∠ A B C , ∠ A B D = ∠ C B D . Since, D E ¯ ⊥ A B → and D F ¯ ⊥ B C → , ∠ D E B = ∠ D F B ≅ right triangles. By identity, B D ¯ ≅ B D ¯ . By AAS, ∠ D E B ≅ ∠ D F B then D E ¯ ≅ D F ¯ by CPCTC. 2) If a point is equidistant from the sides of an angle, then it is on the angle bisector. In the above figure, the point D is equidistant from the sides of ∠ A B C . ∠ A B C such that D E ¯ ⊥ A B → , D F ¯ ⊥ B C → and D E ¯ = D F ¯ . Since B D → bisects ∠ A B C , the point D lies on the bisector of ∠ A B C . Since, D E ¯ ⊥ A B → and D F ¯ ⊥ B C → , ∠ D E B = ∠ D F B ≅ right triangles

Elementary Geometry for College Students

6th Edition

ISBN: 9781285195698

Author: Daniel C. Alexander, Geralyn M. Koeberlein

Publisher: Cengage Learning

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Polygon with three sides, three angles, and three vertices. Based on the properties of each side, the types of triangles are scalene (triangle with three three different lengths and three different angles), isosceles (angle with two equal sides and two equal angles), and equilateral (three equal sides and three angles of 60°). The types of angles are acute (less than 90°); obtuse (greater than 90°); and right (90°).

Chapter 7.1, Problem 36E

To determine

To justify:

A perpendicular to a line from a point not on the line.

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Chapter 7.1, Problem 35E

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To justify: A perpendicular to a line from a point not on the line.

Solution Summary: The author explains how the construction of a perpendicular to the line is explained. The bisector divides the angle angle ABC and the point D lies on it.