5-5 Inequalities in Triangles
1/13/17
Objective: To use inequalities involving
angles of triangles and sides of triangles.
COMPARISON PROPERTY OF INEQUALITY
If a = b + c and c > 0, then a > b
(If c were negative, b would be > a)
Do not write in notes
PROOF
Given: a = b + c, c > 0
Prove: a > b
STATEMENTS
1)
c>0
2)
b+c>b+0
3)
4)
5)
b+c>b
a=b+c
a>b
REASONS
1) Given
2) Addition Property of
Inequality (+b each side)
3) Simplify
4) Given
5) Substitution (a for b + c in
statement 3)
Do not write the proof in notes
COROLLARY TO THE TRIANGLE EXTERIOR ANGLE
THEOREM
The measure of an exterior angle of a triangle is greater than the
measure of each of its remote interior angles
m 1 > m 2 and
3
m 1>m 3
2
1
Informal Proof:
The Exterior Angle Theorem says that 1 = 2 + 3. Since
both 2 and 3 are > 0, the Comparison Property of
Inequality shows that 1 is greater than both 2 and 3.
Ex: m 2 = m 1 by Isosceles Triangle Theorem.
Explain why m 2 > m 3.
By the Corollary to the Exterior Angle Theorem,
m 1 > m 3. So, m 2 > m 3 by Substitution. O
3
P
1
4
2
T
1)
Explain why m OTY > m 3
We know 2 > 3, and
OTY =
2 + 4, so
OTY > 3
Y
Two VERY important theorems.
Do not write the proof, however.
THEOREM 5-10
If two sides of a triangle are not congruent, then the larger angle lies
opposite the longer side.
Y
If XZ > XY, then m Y > m Z
X
Z
THEOREM 5-11
If two angles of a triangle are not congruent, then the longer side lies
opposite the larger angle.
B
If m A > m B, then BC > AC
C
A
GIVEN: m A > m B
PROVE: BC > AC by CONTRADICTION!!!
STEP 1: Assume BC > AC. That is, assume BC < AC or BC = AC.
STEP 2: If BC < AC, then m A < m B (Theorem 5-10). This contradicts
the given fact that m A > m B. Therefore BC < AC must be false.
If BC = AC, then m A = m B. (Isosceles Triangle Theorem). This
also contradicts m A > m B. Therefore BC = AC must be false.
STEP 3: The assumption BC > AC is false so BC > AC.
Ex: In TUV, which side is the shortest? T
m T = 60 by Triangle Sum Theorem.
U is the smallest angle.
58o
Side TV is shortest by Theorem 5-11 U
1)
List the sides of
longest.
V
XYZ in order from shortest to
Y
80°
X
.
62o
YZ < YX < XZ
40o
60o Z
Another VERY important theorem.
THEOREM 5-12 TRIANGLE INEQUALITY THEOREM
The sum of the lengths of any two sides of a triangle is greater
than the length of the third side
XY + YZ > XZ
Y
YZ + ZX > YX
ZX + XY > ZY
X
Z
Ex: Can a triangle have sides with the given lengths? Explain.
a) 3 ft, 7 ft, 8 ft
b) 3 cm, 6 cm, 10 cm
3+7>8
10 + 3 > 6
8+7>3
10 + 6 > 3
3 + 8 > 7 YES
3 + 6 > 10
NO
1)
2)
2 m, 7 m, 9 m
4 yd, 6 yd, 9 yd
2 + 7 = 9… NO: it must be greater
4 + 6 > 9… YES
Just check the two smallest sides. If their sum is > than the third side, it’s a Δ.
Ex: A triangle has sides of lengths 8 cm and 10 cm.
Describe the lengths possible for the third side.
x = the third side
x can be 8 + 10 “at most”
x is “at least” 10 – 8
Longer than 2 cm and shorter than 18 cm
2 < x < 18 {do NOT use <}
1) A triangle has sides 3 in, 12 in. Describe the
lengths possible for the third side.
9 < x < 15
1. The exterior angle is larger
than either of the remote
interior angles.
2. Angle 3 is opposite AB.
Angle 1 is opposite BC.
AB > BC so Angle 3 > Angle 1.
A
1
25
B
12
3 4
C
27o
D
Explain why m 4 > m 1
2)
Explain why m 3 > m 1
3)
Use ABC to describe the possible lengths of AC
13 < AC < 37
_____________________________________________
1)
Can a triangle have lengths of 2 mm, 3 mm, and 6 mm?
Explain. NO: 2 + 3 < 6
2)
In XYZ, XY = 5, YZ = 8, and XZ = 7. Which angle is the
largest? Angle X, since it’s opposite the largest side, YZ.
3)
In PQT, m P = 50 and m T = 70. Which side is shortest?
1)
Angle Q = 60, so the side opposite Angle P is the
shortest. That side is QT.
USING A VARIABLE
pg 280
A
B
C
Points A, B, and C are collinear.
BC is 6 less than twice AB, and AC = 30.
What is the length of BC?
Let x = AB
AC = 30
“BC is 6 less than twice AB” so BC = 2x - 6
x + 2x – 6 = 30
3x – 6 = 30
3x = 36
x = 12, so AB = 12
BC = 2 • 12 – 6
BC = 18

Assignment:
Page 276 #4 – 21, 32, 34 – 36