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Practice C
Indirect Proof and Inequalities in One Triangle
Indirect proofs work by finding a contradiction that leads to the proof
of a statement. For Exercises 1–7, rewrite each statement. Use the
symbol  for an “if, then” statement and the symbol ~ for “not,” the
negation of a statement. Use a to stand for “The two angles are a
linear pair.” Use b to stand for “The two angles are supplementary.”
Example: If the two angles are a linear pair, then the two angles are supplementary.
ab
1. If the two angles are supplementary, then the two angles are a linear pair.
__________
2. If the two angles are not supplementary, then the two angles are a linear pair. __________
3. If the two angles are a linear pair, then the two angles are not supplementary. __________
4. If the two angles are not a linear pair, then the two angles are not
supplementary.
__________
5. If the two angles are not supplementary, then the two angles are not
a linear pair.
__________
6. If the two angles are not a linear pair, then the two angles are supplementary. __________
7. If the two angles are supplementary, then the two angles are not a linear pair. __________
8. Suppose the example statement (a  b) is to be proven. Give the number
of the statement you would begin with (knowing it would lead to a
contradiction) in order to prove the example statement by indirect proof.
__________
9. Suppose then the contradiction negates the conclusion. Give the number
of the statement that the contradiction leads you to believe must be true.
__________
10. Name the logical relationship between the answer
to Exercise 9 and the example statement.
___________________________________
11. Name the shortest segment(s) in the figure
and explain your reasoning. Do not use a ruler.
(Note: The figure may not be drawn to scale.)
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Name _______________________________________ Date __________________ Class __________________
7. yes
6.
8. yes
9. no; 12  20  36
10. 4.7 m  s  11.7 m
11. 121 ft  s  475 ft
1
1
mi  s  7 mi
2
2
13. Renaldo could travel between 8562
miles and 15,502 miles.
Practice C
12.
7. Answers will vary based on students’
choice of letters: QR, RS, QS.
8. QRS
INDIRECT PROOF AND INEQUALITIES IN
ONE TRIANGLE
Practice A
1. opposite
2. greater
3. angle
4. Q; P; R
5. GI; GH; HI
6. yes
7. yes
8. yes
9. yes
10. The segments cannot make a triangle
because 8  6  15.
11. 29  n
12. n  7
13. n  7
14. 7; 29
15. a. no
c. no
b. yes
16. between 7 and 23 feet
Practice B
1. m1  m2  m3  180°
2. Possible answer: Assume that m1 
m2  m3  180°. 4 is an exterior
angle of ABC, so by the Exterior
Angle Theorem, m1  m2  m4.
3 and 4 are a linear pair, so by the
Linear Pair Theorem, m3  m4 
180°. Substitution leads to the
conclusion that m1  m2  m3 
180°, which contradicts the assumption.
Thus the assumption is false, and the
sum of the angle measures of a triangle
cannot add to more than 180°.
3. F; D; E
4. HI;GH;GI
5. no; 8  8  16
6. yes
1. b  a
2. ~b  a
3. a  ~b
4. ~a  ~b
5. ~b  ~a
6. ~a  b
7. b  ~a
9. 5
8. 2
10. contrapositive
11. Possible answer: The shortest side in a
triangle is opposite the shortest angle.
The shortest side in AEF is AF .
ABF is equiangular, so AF has the
same length as BF . But BG is the
shortest side in BGF, so AF , AB ,
and BF cannot be the shortest
segments in the figure. CG is the
shortest segment in CHG, but BC is
the shortest segment in BCG. So BC
is shorter than CG . The shortest
segment in CDH is DH . DH has
length a and CG has length a  2, so
CG is shorter than DH . Therefore BC
is the shortest segment in the figure.
Reteach
1. Given: ABC is an obtuse , B is an
obtuse angle; Prove: ABC does not
have a right angle.
2. Assume ABC does have a right
angle. Let A be a right angle.
3. Possible answer: If A is a right angle,
then mB  mC  90°. But mB >
90°, since B is obtuse. So this is a
contradiction.
4. The assumption that ABC does have a
right angle is false. Therefore ABC
does not have a right angle.
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
Holt McDougal Geometry