1. Francisco Tomasino Andy Lachler
Chapter 6
Francisco Tomasino
Andy Lachler

2. Properties of Inequality
If a>b and c≥d, then a+C > B+D
If a>b and c>0, then ac>bc and a/c > b/c
If a>b and c>0, then ac<bc and a/c <b/c
If a>b and b>c , then a>c
If a= b+c and c>0, then a>b

3. Example: Statements Reasons AC>BC; CE>CD AC+CE > BC+CD
B E
Given: AC>BC ; CE >CD
Prove: AE > BD C
Statements
Reasons
A D
AC>BC; CE>CD
AC+CE > BC+CD
AC+CE=AE; BC+CD=BD
AE>BD
Given
Prop. Of Ineq.
SAP
Substitution

4. Exterior Angle Inequality Theorem (EAIT)
The measure of an exterior angle of a triangle is grater than the measure of either remote interior angle A
A < C and B < C B C

5. Example: Given: 2 > 1 Prove: 2 > 4 Statements Reasons 2 > 1
1 > 3
2 > 3
3 = 4
2 > 4
Given
EAIT
Substitution
VAT 4 3 1 2

6. How to Write an Indirect Proof
Assume temporarily that the 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

7. Example: Given: X = 100° Prove: Y is not a right angle StatementsZ
Statements
Reasons
1. X = 100°
2. Assume Y is a right angle
3. X + Y + Z = 180°
Z= 180
5. Y is not a right angle
1. Given
2. Assumption
3. Sum of angles of a triangle is 180°
4. Substitution
5. Contradiction

8. Practice Problems Given: VY is perpendicular to YZX
Given: VY is perpendicular to YZ
Prove: angle VXZ is an obtuse angle Y Z t 1
Given: transversal “t” cuts lines “a” and “b”; angle 1 ≠ angle 2
Prove: “a” is not parallel to “b” a 3 b 2

9. Inequalities for One Triangle
In any triangle, the largest interior angle is opposite the largest side.
The smallest interior angle is opposite the smallest side
The middle-sized interior angle is opposite the middle-sized side R
Angle RST > Angle T
Because side RT is longer than RS or ST 5 3 T S 4

10. Theorems and Corollaries
If one side of a triangle is longer than a second side, then the angle opposite the first side is larger than the angle opposite the second side
Given: Triangle RST; RT>RS
Prove: m of angle RST > m of angle T
By the ruler postulate there is a point Z on segment RT such that RZ = RS. Draw segment SZ
In isosceles triangle RZS, the measure of angle 3 = the measure of angle 2
Because the measure of angle RST = the measure of angle 1 + the measure of angle 2, you have measure of angle RST > measure of angle 2
Substitution of measure of angle 3 for the measure of angle 2 yields the measure of RST > measure of angle 3
Because angle 3 is and exterior angle of triangle ZSR, you have measure of angle 3 > measure of angle T
From the m of RST > m of 3 and m of 3 > m of T, you get m RST > m T R Z 3 2 1 T S

11. Theorems and Corollaries
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
Given: Triangle RST; m of S > m of T
Prove: RT > RS R
Assume temporarily that RT is NOT > RS. The either RT = Rs or RT < RS
Case 1: If RT = RS, then m of S = m of T
Case 2: If RT< RS, then m of S < m of T
In either case there is a contradiction of the given fact that m of S > m of T
The assumption that RT is NOT > RS must be false. It follows that RT > RS T S

12. Theorems and Corollaries
Corollary 1:
The perpendicular segment from a point to a line is the shortest segment from the point to the line
Corollary 2:
The perpendicular segment from a point to a plane is the shortest segment from the point to the plane

13. Theorems and Corollaries
The sum of the length of any two sides of a triangle is greater than the length of the third side
Given: Triangle ABC
Prove: (1) AB + BC > AC
(2) AB + AC > BC
(3) AC + BC > AB
One of the sides. Say segment AB, is the longest side. Then (1) and (2) are true. To prove (3), draw a line, line CZ, through C and perpendicular to line AB.
By Corollary 1 of Theorem 6-3, segment AZ is the shortest segment from A to segment CZ. Also, Segment BZ is the shortest segment from B to segment CZ. C A B Z

14. Theorems and Corollaries Inequalities for Two Triangles
SAS Inequality Theorem:
If two sides of a triangle are congruent to two sides of another triangle and the included angle in one triangle has a greater measure than the included angle in the other, then the third side of the first triangle is longer than the third side of the second triangle.

15. Theorems and Corollaries
SSS Inequality Theorem:
If two sides of a triangle are congruent to two sides of another triangle and the third side in one triangle is longer than the third side in the other, then the angle between the pair of congruent sides in the first triangle is greater than the corresponding angle in the second triangle

16. Practice Problems A B C D E F
Given: Segment BA is congruent to segment ED; Segment BC is congruent to segment EF; m of B > m of E
Prove: AC > DF A B C
Given: Segment BA is congruent to segment ED; Segment BC is congruent to segment EF; AC > DF
Prove: m of B > m of E D E F