to mAB > mCD and AB + BC and mAB + mBC will be two ways used to say;
the same thing.
This practice will be followed in order to shorten otherwise long and
A
..
B
.
C B A
(a) (b)
Fig,9.1.
cumbersome statements. The student should keep in mind, however, that- the measures of geometricfigures are being compared in these instances. ~- 9.3. Sense of inequalities. Two inequalities are of the same sense if the same
.
'
:,j '
,
:
:
'
,, '
symbol is used in the inequalities. Thus a < band c < d are inequalities of
the same sense. Two inequalities are of the opposite sense if the symbol of one"inequality is the reverse of the symbol in the other. Thus a < band c > dare>
of opposite sense. .'
A study of the basic properties of and theorems for inequalities will revearl~
processes which will transform an InequalIty to another InequalIty of the same~
sense. Some of them are:
I
'(a) Adding equal real numbers to both sides of an inequality. ) (b) Su bt,",cin~ eq ual tcal nu m bm hom both ,( de> of an inequality.l,
(c) Multiplying both sides of an inequality by equal positive real numbers.
(d) Dividing both sides of an inequality by equal positive real numbers.
(e) Substituting a number for its equal in an inequality.
The following processes will transform an inequality to another inequality, of opposite sense:
(a) Dividing the same (or equivalent) positive number by an inequality.
(b) Subtracting both sides of an inequality from the same real number.
(c) Multiplying both sides of an inequality by the same negative number.
(d) Dividing both sides of an inequality by the same negative number.
[Note: To divide by a number a is the same as to multiply by its multiplicativ>:
inverse l/a.]
i,
[
---
..
Exercises (A)
Answer each question.
possible. "
INEQUALITIES 285
If no answer is possible, indicate with "no answer
1. Bill has more money than Tom. Each earned an additional 10 dollars.
How do Bill's and Tom's total amounts compare?
2. Bill has more money than Tom and Frank has less than John. How do Bill's and John's compare?
3. Bill has the same amount of money as Alice. Alice spends more than Bill. How then do their remaining amounts compare?
4. John has more money than Tom. John loses half his money. How do their remaining amounts compare?
5. Bill has less money than Mary. Each decides to give half of his money to charity. How do the amounts they have left compare?
6. John has more money than Tom. Each doubles his amount. Who, then, has the more money?
7. Ann is older than Alice. Mary is younger than Alice. Compare Mary's and Ann's ages.
R. Mary and Alice together have as much money as Tom. Compare Tom's and Alice's amounts.
9. Ann and Bill are of different ages. Mary and Tom are also of different ages. CompoTe the ages of Ann and Tom.
10. John has twice as much mOTleY;j<;
MiI'Y, ,md Mary as Alice. Compare the amounts of John and Alice.
Exercises (B)
Copy and complete the following exercises.
write a question mark in place of the blank. If no conclusion is possible,
II. If a > band c = d, then a + c
12. Ifr < sand x > b,thenr-b.
13. If x = 2y, r = 2s, and y < s, then x 14. Ifl > kandk > m,thenl m.
15. Ifx+y=z,thenz x.
16. If AB + BC > AC, then AB
b+d.
s-x.
r.
AC-BC.
--- --
286
FUNDAMENTALS OF COLLEGE GEOMETRYc~. Y
:y H/
B~ E~
17. If Lx
==L Y
and mLr > m Ls, then
mLABC mLDEF.
18. IfBC .1 AB; EF .1 DE;
and mLf3 < mLa, then
mLa mL y.
19. If LA ==LB and CD < CE,
then AD BE.
20. If AD > BE and EC > CD,
thenAC ltL.
21. IfmLCAB = mLABC,
then BD AC.
22. !fAC = BC,
then mLABC mLBAD.
!=
l
Ex. 17.
c~ FL
B A E D
Ex. 18.
c
A B
Exs.19,20.
c
A B
Exs. 21,22.
I
;: I!I:
I
-
23. If La is an exterior L of
6ABC, then mLa mLA.
24. !fmLR > mLS, RK bisects LSRT, SK bisects LRST,
then mLa mLf3.
Ex. 24.
Ex. 23.
c
~
A B
.,.
D
R
s
Theorem 9.1
9.4. If two sides of a triangle are not congruent, the angle opposite the longer of the two sides has a greater measure than does the angle opposite the shorter side.
Given: 6ABC with BC > AC.
Conclusion: mLBAC> mLB.
Proof
STA TEMENTS
Theorem 9.1.
c
A B
REASONS
I. BC > AC.
2. On CB let D be a point such that CD = AC.
3. Draw AD.
4. mLa = mLf3.
5. mLBAC = mLBAD + mLa.
6. mLBAC > mLa.
7. mLBAC> mLf3.
8. mLf3 > mLB.
9. ". mLBAC > mLB.
--
----.--
I. Given.
2. Postulate II.
3. Postulate 2.
4. Theorem 4.16.
5. Postulate 14.
6. 0-8.
7. 0-7.
8. Theorem 4.17.
9. 0-6.
---
288
FUNDAMENTALS OF COLLEGE GEOMETRYTheorem 9.2
9.5. If two angles of a triangle are not congruent, the side opposite the larger of the two angles is greater than the side opposite the smaller of the two angles.
Given: L.ABC with mLA > mLB.
Conclusion: BC > AC.
Theorem 9.2.
Proof
STA TEMENTS
C
A~B
REASONS
1. Given.
2. Trichotomy property.
I. mLA > mLB.
2. In L.ABC, since BC and AC are real numbers, there are only the following possibilities: BC = AC, BC < AC,BC > AC.
3. Assume BC = AC.
4. Then mLA = mLB.
5. Statement 4 is false.
6. Next assume BC < .1.c.
7. Then mLA < mLB.
8. Statement 7 is false.
9. The only possibility remaining is isBC > AC.
9.6. Corollary: The shortest segment joining a point to a line perpendicular segment.
Note. Here we can prove what we stated in §1.20.
9.7. Corollary: The measure of the hypotenuse of a right triangle is greater than the measure of either leg.
Theorem 9.3
Theorem 9.3.
r-
---D
\'---
\ '...
I B
I -...
\:~
..
. . .
\:
I
1A (!
-
INEQUALITIES 289
9.8. The sum of the measures of two sides of a triangle is greater than the measure of the third side.
Given: L.ABC.
Conclusion: AB + BC > AC.
Proof
STATEMENTS REASONS
I. Let D be the point on the ray
oppositeifC
such that DB= AB.
2. Draw AD.
3. DC = DB+BC.
4. DC = AB+BC.
5. mLDAC = mLDAB + mLBAC.
6. mLDAC > mLDAB.
7. mLDAB = mLADB.
8. mLDAC > mLADB.
9. DC > AC.
IO. AB + BC > AC.
I. Postulate II.
2. Postulate 2.
3. Postulate 13.
4. £-8.
5. Postulate 14.
6. 0-8.
7. Theorem 4.16.
8. 0-7.
9. Theorem 9.2.
10. Substitution property.
This theorem may be used to show that the shortest route between two points is the straight line route.
Theorem 9.4
9.9. If two triangles have two sides of one congruent respectively to two sides of the other and the measure of the included angle of the first greater than the measure of the included angle of the second triangle, the third side of the first is greater than the third side of the second.
F
A~R D~E
c
A B
Theorem 9.4.
290 FUNDAMENT ALS OF COLLEGE GEOMETRY
9.11. Illustrative Example 1:
Given: D a point in the interior of LABC; AC = CD.
Prove: DB < AB.
Proof Given: LABC and LDEF with AC = DF, CB = FE, and mLC > mLF.
Conclusion: AB > DE.
Proof
~B
A REASONS
S1;'ATEMENTS
1. Draw
EK (see third figure) with
K on the same side ofBe
as A and such that LACK ==LDFE.
2. cK
take a point G such thatCG = FE.
3. Draw AG.
4. AC = DF.
5. LACG
==LDFE.
6. AG = DE.
7. Bisect LBCK and let H be the point where the bisector intersects
AR.
8. La
== Lf3.9. Draw HG.
10. CH = CH.
11. CB = FE.
12. CG = CB.
13. LCHG
==LCHB.
~-l4,L#u=B-ti . 15. GH+AH > AG.
16. BH+AH > AG.
17. BH+AH=AB.
18. AB > AG.
19. AB > DE.
1. Postulate 12.
Illustrative Example 1.
2. Postulate 11.
STATEMENTS REASONS
3. Postulate 2.
4. Given.
5. S.A.S.
6. §4.28.
7. 0-1.
1. AC (of LABC) = CD (of LDBC).
2. BC (of LABC) = BC (of LDBC).
3. D is in the interior of LACB.
4. mLACB = mLDCB + mLACD.
S. mLDCB < mLACB.
6. :.DB < AB.
1. Given.
2. Reflexive property.
3. Given.
4. Postulate 14.
5. 0-8.
6. Theorem 9.4.
8. §1.l9.
9. Postulate 2.
10. Reflexive property.
11. Given.
12. Theorem 3.4.
13. S.A.S.
lL§428_-
- - . . .
15. Theorem 9.3.
16. 0-7.
17. Postulate 13.
18. 0-7.
19. 0-7.
T
9.12. Illustrative Example 2:
Given: S1' = R1';
K any point on RS.
Prove: ST'?-KT.
Proof
R
s
illustrative Example 2.
Theorem 9.5 STATEMENTS REASONS
9.10. If two triangles have two sides of one congruent respectively to tWO sides of the other and the third side of the first greater than the third side of -~
the second, the measure of the angle opposite the third side of the first is_, greater than the measure of the angle opposite the third side of the second. .'
I. S1' = RT.
2. mLS = mLR.
3. mLRK1'> mLS.
4. mLRK1'> mLR.
5. R1' > KT.
6. S1' > KT.
1. Given.
2. 4.16.
3. 4.17.
4. 0-7.
5. Theorem 9.2.
6. 0-7.
(Note: This theorem is proved by the indirect method. The proof is left tOi!
the student.)