Transcript Geometry 2_4 and 2_5 Updated Version

GEOMETRY: CHAPTER 2
Ch. 2.4 Reason Using
Properties from Algebra
Ch. 2.5: Prove Statements
about Segments and
Angles
2.4: KEY CONCEPT:
Algebraic Properties of Equality
Let a, b, and c be real numbers.
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 and c≠0, then
a/c=b/c.
Substitution Property:
If a=b, then a can be substituted for b in
any equation or expression.
Distributive Property:
a (b + c ) = ab + ac.
Ex. 1 Solve: Write reasons for each step
2x + 4 = 10 – 4x
Write original
equation.
Given
2x +4 +4x = 10 - 4x+4x Add 4x to each Addition Property
side.
of Equality
6x + 4 = 10
Combine Like
terms.
Simplify.
6x = 6
Subtract four
Subtraction
from each side. Property of Equality
x=1
Divide each
side by 6.
Division Property
of Equality
Ex. 2: Use the Distributive Property and
Write reasons for each step.
Solution:
Equation
Explanation
Reason
-2 (3x + 1)=40 Write original equation Given
-6x -2 = 40
Multiply
Distribution
Property
-6x = 42
Add 2 to each side
Addition
Property of
Equality
x= -7
Divide each side by -6 Division
Property of
Equality
Key Concepts:
Reflexive Property of Equality:
Real Numbers: for any real number a, a=a.
Segment Length: for any segment AB,
AB=AB.
Angle Measure: for any angle A, mA  mA.
Symmetric Property of Equality
Real Numbers:
For any real numbers a and b, if a  b, then b  a.
Segment Length:
For any segments AB and CD, if AB  CD, then CD  AB.
Angle Measure:
For any angles A and B, if mA  mB, then mB  mA.
Transitive Property of Equality
Real Numbers:
For any real numbers a, b and c, if a  b and b  c, then a  c.
Segment Length:
For any segments AB, CD and EF , if AB  CD and CD  EF ,
then AB  EF .
Angle Measure:
For any angles A, B and C , if mA  mB and mB  mC ,
then mA  mC.
Ex. 3: The city is planning to add two stations
between the beginning and end of a
commuter train line. Use the information
given. Determine whether RS=TU.
Image taken from: Geometry. McDougal Littell: Boston, 2007. P. 107.
Given: RT=SU
Prove: RS=TU
EQUATION
RT=SU
ST=ST
RT-ST=SU-ST
RT-ST=RS
SU-ST=TU
RS=TU
Ex. 3 (cont.)
REASON
Given
Reflexive Property
Subtraction Property of Equality
Segment Addition Postulate
Segment Addition Postulate
Substitution Property of Equality
For more examples, go to CH. 2, Lesson 5,
Example 4
http://www.classzone.com/cz/books/geometry_2007_na/get_chapter_group.htm?cin=2
&rg=help_with_the_math&at=powerpoint_presentations&var=powerpoint_presentation
s
2.5 Prove Statements about Segments and Angles
A proof is a logical argument that
shows a statement is true.
A two-column proof has numbered
statements and corresponding reasons
that show an argument in a logical
order.
In a two-column proof, each statement
in the left-hand column is either
given information or the result of
applying a known property or fact to
statements already made.
The explanation for the corresponding
statement is in the right-hand
column.
Ex. 5
Image taken from: Geometry. McDougal Littell: Boston, 2007. P. 112.
You are designing
a logo to sell daffodils. Use the information given.
Determine whether the measure of angle EBA is equal
to the measure of angle DBC.
Image taken from: Geometry. McDougal Littell: Boston, 2007. P. 112.
Ex. 5 (cont.)
STATEMENTS
1. m1  m3
REASON
Given
2. mEBA  m3  m2 Angle Addition Postulate
3. mEBA  m1  m2 Substitution Property of Equality
4. m1  m2  mDBC Angle Additon Postulate
5. mEBA  mDBC
Transitive Property of Equality
Theorems—The reasons used in a
proof can include definitions,
properties, postulates, and theorems.
A theorem is a statement that can be
proven. Once you have proven a
theorem, you can use the theorem as
a reason in other proofs.
Theorem 2.1 Congruence of Segments.
Segment congruence is reflexive,
symmetric, and transitive.
Re flexive: For any segments AB, if AB  AB.
Symmetric: If AB  CD, then CD  AB.
Transitive: If AB  CD and CD  EF , then AB  EF .
Theorem 2.2 Congruence of Angles
Angle congruence is reflexive, symmetric, and
transitive.
Reflexive: For any angle A, mA  mA.
Symmetric: If A  B, then B  A.
Transitive: If A  B and B  C , then A  C.
Ex. 6: Name the property illustrated by
the statement.
a. If K  L and L  M , then K  M .
b. If EF  GH , then GH  EF .
a. Transitive Property of Angle
Congruence.
b. Symmetric Property of Segment
Congruence.
Ex. 7 Use properties of equality
Prove this property of midpoints: If you know that M
is the midpoint of segment AB, prove that AB is two
times AM and AM is one half of AB.
GIVEN: M is the midpoint of segment AB.
Prove: a. AB=2 (AM)
b. AM= ½ AB
Ex. 7 (cont.)
STATEMENTS
REASONS
1. M is the midpoint of AB. 1. Given
2. AM .  MB.
2. Definition of a Midpoint
3. AM  MB.
3. Def. of congruent segments
4. AM  MB  AB.
4. Segment Addition Postulate
5. AM  AM  AB.
5. Substitution Property of Equality
a. 6. 2 AM  AB
6. Distributive Property
b. 7. AM  1 AB
2
7. Division Property of Equality
Turn to page 103 in your book and look at the
bottom right example. Refer back to this page for
writing proofs!
Ex. 8
Go To Chapter 6, Lesson 2, Example 4:
http://www.classzone.com/cz/books/geometry_2
007_na/get_chapter_group.htm?cin=2&rg=hel
p_with_the_math&at=powerpoint_presentation
s&var=powerpoint_presentations
Write this example in your notes as Ex. 8.