chapter 6: inequalities in geometry sonora hospital-medina and rachel carta wagman

Chapter 6: Inequalities in Geometry

Sonora Hospital-Medina and Rachel Carta Wagman

6-1:Inequalities

6-1:Inequalities

• Segments with UNequal lengths and angles with UNequal measures

• Properties of INequality

Properties of Inequality

1. If a>b and c ≥ d, then a+c > b+d– Example: if x > 7 and 7 ≥ 5, then…– Example: if 6>5 and 5 ≥ 4, then…

2. If a > b and c > 0, then ac>bc and a/c > b/c• Example: 8 > 6 and c = 4, then…

X +7 > 7 + 5

6 + 5 > 5 + 4

(8)(4) > (6)(4) and 8/4 > 6/4

Properties of Inequality cont…

3. If a >b and C < 0, then ac < bc and a/c < b/c– Example: 8 > 6 and c = -4, then…

4. If a > b and b > c, then a > c– Example: 8 > 6 and 6 > 4, then…

5. If a = b+c and c > 0, then a>b– Example: 8 = 6+2, then…

(8)(-4) < (6)(-4) and8/-4 < 6/-4

8>4

8>6 or 8>2

Theorem 6-1Exterior Angle Inequality Theorem

• The measure of an exterior angle of a triangle is greater than the measure of either remote interior angle

EAIT cont…Proof

1E

D

F

Given: <1 is an exterior angle of DEF

Prove: m<1 > m<D; m<1 > m<E

Statements Reasons

1.) m<1 = m<D + m<E

2.) m<1 > m<D; m<1 > m<E

1.) The measure of an ext. < of a equals the sum of the measures of the 2 remote int. <s

2.) POP

6-3Indirect Proof

6-3Indirect Proof

• Same as a normal proof, BUT you are trying to contradict your assumption– How to Write an Indirect Proof:• Assume temporarily that the desired conclusion is not

true• Reason logically until you reach a contradiction of a

known fact• Point out that the temporary assumption must be

false, and that the conclusion must then be true

Indirect Proof:• Given: Trapeziod PQRS with bases PQ and SR• Prove: PQ ≠ SR S R

P Q

Statements Reasons

1.) PQ=SR2.)PQ ll SR PS ll RQ3.) PQRS = parallelogram4.) PS ll RQ5.) PQ ≠ SR

1.) assumption2.)Def. of Trapezoid3.) Theroem 5-5 (1 pair of opp. sides are ≅and ll, the quad. Is a parallelogram)4.) Def. of Parallelogram5.) Contradiction

Section 4: Inequalities for One Triangle

LSLA (Theorem 6-2)

If one side of a triangle is longer than the second side, then the angle opposite the first side is larger than the angle opposite the second side.

A

B

C

Given: AB > AC

Conclude: m<ACB > m<ABC

LSLA (Theorem 6-2)

If one angle of a triangle is larger than a second angle, then the side opposite the first angle is longer than the side opposite the second angle.

112°

50° B

A

C

Given: m<ACB = 112° m<ABC = 50°

Conclude: AB > AC

Corollaries

Corollary 1The perpendicular segment

from a point to a line is the shortest segment from the point to the line

Corollary 2The perpendicular segment

from a point to a plane is the shortest segment from the point to the plane.

P

A

B C

PA > PBPA > PC

M

T

S Q

TR > TQTR > TS

R

The Triangle Inequality (Theorem 6-4)

The sum of the lengths of any two sides of a triangle is greater than the length of the third side.

A

BC

AB + AC > BCAB + BC > ACAC + BC > AB

Section 5: Inequalities for Two Triangles

SAS Inequality

If two sides of one triangle are congruent to two sides of another triangle, but the included angle of the first triangle is larger than the included angle of the second, then the third side of the first triangle is longer than the third side of the second triangle.

Used to find which side is larger or smaller than the other.

Given: AB = DE AC = DF m<D > m<A

Conclude: EF > BC by

EAIT

D

E F

A

B C

The angle has to be the inequality and the two sides must be congruent

SSS Inequality Theorem

If two sides of one triangle are congruent to two sides of another triangle, but the third side of the first triangle is longer than the third side of the second, then the included angle of the first triangle is larger than the included angle of the second.

Used to find which angle is larger or smaller than the other.

Given: AB = DE AC = DF EF > BC

Conclude: m<D > m<A

Two sides must be congruent, and the other must be an inequality.

Practice Questions and ProofsOn a separate sheet of paper