Chapter 1.5, Problem 28E ### Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698

#### Solutions

Chapter
Section ### Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698
Textbook Problem
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# In Exercises 27 to 30, fill in the missing reasons for each geometric proof.Given: E is the midpoint of D F ↔ Prove: D E = 1 2 ( D F ) Exercises 27, 28 PROOF Statements Reasons 1. E is the midpoint of D F ↔ 1. ? 2. D E = E F 2. ? 3. D E + E F = D F 3. ? 4. D E + D E = D F 4. ? 5. 2 ( D E ) = D F 5. ? 6. D E = 1 2 ( D F ) 6. ?

To determine

To find:

The missing reasons for the geometric proof.

Explanation

Given:

The given table is shown below.

 PROOF Statements Reasons 1. E is the midpoint of DF↔ 1. ? 2. DE=EF 2. ? 3. DE+EF=DF 3. ? 4. DE+DE=DF 4. ? 5. 2(DE)=DF 5. ? 6. DE=12(DF) 6. ?

Approach:

1. Midpoint on a line is defined as the point which divides the line into two segments of equal length.

2. According to Segment-Addition Postulate, if two points A, B are given on a line segment, then third point C lies on the line segment if and only if,

AC+CB=AB

3. According to Substitution Property of Equality, if A=B, then A can be substituted for B and B can be substituted for A in any equation.

4. The Multiplication Property of Equality states that if both the sides of an equation is multiplied by a same number then the sides of the equation remains equal.

Calculation:

The first statement of the proof is given below.

E is the midpoint of DF

Therefore, the first reason of the proof is given.

Now, go from first statement to the second statement.

E is the midpoint of DF

Use Definition of Midpoint as E is the midpoint of DF.

DE=EF

Therefore, the second reason is Definition of Midpoint.

Now, go from second statement to the third statement.

DE=EF

Use Segment-Addition Postulate as D,EandF are collinear and E is the midpoint of DF.

DE+EF=DF

Therefore, the third reason is Segment-Addition Postulate.

Now, go from third statement to fourth statement.

DE+EF=DF

Use Substitution Property of Equality and substitute DE for EF as DE=EF.

DE+DE=DF

Therefore, the fourth reason for the proof is Substitution Property of Equality

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