   Chapter 7.1, Problem 38E Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085

Solutions

Chapter
Section Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085
Textbook Problem

Note: Exercises preceded by an asterisk are of a more challenging nature.In Exercises 37 and 38, use the method of proof of Theorem 7.1.1 to justify each construction method.The construction of a perpendicular to a line from a point not on the line.

To determine

To justify:

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

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 BD divides the angle ABC and the point D lies on the bisector BD.

DE-AB and DF-BC.

Since BD bisects ABC, ABD=CBD.

Since, DE-AB and DF-BC, DEB=DFB right triangles.

By identity, BD¯BD¯.

By AAS, DEBDFB then DE¯DF¯ 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 ABC.

ABC such that DE-AB, DF-BC and DE-=DF-.

Since BD bisects ABC, the point D lies on the bisector of ABC.

Since, DE-AB and DF-BC, DEB=DFB right triangles

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