Download Geometry Properties of Equality

Geometry Properties of Equality
REFLEXIVE
SYMMETRIC
TRANSITIVE
Segment Length
For any segment AB,
AB=AB.
If AB = CD, then
CD = AB.
If AB = CD and CD = EF,
Then AB = EF.
Solve and write a reason for each step.
1. In the diagram, AB = CD. Show that AC = BD.
Statements
Reasons
2. Given:
m1 + m2 = 66º
m1 + m2 + m3 = 99º
m3 = m1
m1 = m4
Find: m4
Statements
Reasons
1
Angle Measure
For any angle A,
mA = mA .
If mA = mB , then
mB = mA .
If mA = mB , and
mB = mC , then
mA = mC .
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A
B
C
D
2 3
4
Name _______________________
Geometry Notes
Section 2.4 Reasoning with Properties from Algebra
Warm up. Solve the following.
1) 15  3x  3
2) 10  2 x  3( x  2)  1
3)
1
3
x  4  2x 
5
5
Algebraic Properties of Equality:
Addition Property If a=b, then a+c=b+c
Subtraction Property If a=b, then a-c=b-c
Multiplication Property If a=b, then ac=bc
Division Property If a=b, then
a b
= , as long a c  0
c c
Reflexive Property a=a
Symmetric Property If a=b, then b=a
Transitive Property If a=b and b=c, then a=c
Substitution Property If a=b, then b can be substituted for a in any equation or expression
Distributive Property a(b+c)=ab+ac
Solve and write a reason for each step.
1) 5x 18  3x  2
2) 55 y  3(9 y  12)  64