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# 2.4: Algebraic and Congruence Properties

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Understand basic properties of equality and congruence.
• Solve equations and justify each step.
• Fill in the blanks of a 2-column proof.

## Review Queue

Solve the following problems.

1. Solve 2x3=9\begin{align*}2x-3=9\end{align*}.

2. If two angles are a linear pair, they are supplementary.

If two angles are supplementary, their sum is 180\begin{align*}180^\circ\end{align*}.

What can you conclude? By which law?

3. Draw a picture with the following:

LMN is bisected by MO¯¯¯¯¯¯OMP is bisected by MN¯¯¯¯¯¯LM¯¯¯¯¯¯MP¯¯¯¯¯¯N is the midpoint of MQ¯¯¯¯¯¯

Know What? Three identical triplets are sitting next to each other. The oldest is Sara and she always tells the truth. The next oldest is Sue and she always lies. Sally is the youngest of the three. She sometimes lies and sometimes tells the truth.

Scott came over one day and didn't know who was who, so he asked each sister who was sitting in the middle. Who is who?

## Properties of Equality

Recall from Chapter 1 that the = sign and the word “equality” are used with numbers. The basic properties of equality were introduced to you in Algebra I. Here they are again:

Reflexive Property of EqualityAB=AB\begin{align*}AB = AB\end{align*}

Symmetric Property of EqualitymA=mB\begin{align*}m\angle A = m \angle B\end{align*} and mB=mA\begin{align*}m \angle B = m \angle A\end{align*}

Transitive Property of EqualityAB=CD\begin{align*}AB = CD\end{align*} and CD=EF\begin{align*}CD = EF\end{align*}, then AB=EF\begin{align*}AB = EF\end{align*}

Substitution Property of Equality — If a=9\begin{align*}a = 9\end{align*} and ac=5\begin{align*}a - c = 5\end{align*}, then 9c=5\begin{align*}9 - c = 5\end{align*}

Addition Property of Equality — If 2x=6\begin{align*}2x = 6\end{align*}, then 2x+5=6+5\begin{align*}2x + 5 = 6 + 5\end{align*} or 2x+5=11\begin{align*}2x + 5 = 11\end{align*}

Subtraction Property of Equality — If mx+15=65\begin{align*}m \angle x + 15^\circ = 65^\circ\end{align*}, then mx+1515=6515\begin{align*}m\angle x+15^\circ - 15^\circ = 65^\circ - 15^\circ\end{align*} or mx=50\begin{align*}m\angle x = 50^\circ\end{align*}

Multiplication Property of Equality — If y=8\begin{align*}y = 8\end{align*}, then 5y=58\begin{align*}5 \cdot y = 5 \cdot 8\end{align*} or 5y=40\begin{align*}5y = 40\end{align*}

Division Property of Equality — If 3b=18\begin{align*}3b=18\end{align*}, then 3b3=183\begin{align*}\frac{3b}{3}=\frac{18}{3}\end{align*} or b=6\begin{align*}b = 6\end{align*}

Distributive Property5(2x7)=5(2x)5(7)=10x35\begin{align*}5(2x-7)=5(2x)-5(7)=10x-35\end{align*}

## Properties of Congruence

Recall that AB¯¯¯¯¯CD¯¯¯¯¯\begin{align*}\overline{AB} \cong \overline{CD}\end{align*} if and only if AB=CD\begin{align*}AB = CD\end{align*} and ABCDEF\begin{align*}\angle ABC \cong \angle DEF\end{align*} if and only if mABC=mDEF\begin{align*}m \angle ABC = m \angle DEF\end{align*}. The Properties of Equality work for AB,CD,mABC\begin{align*}AB, CD, m\angle ABC\end{align*} and mDEF\begin{align*}m\angle DEF\end{align*}.

Just like the properties of equality, there are properties of congruence. These properties hold for figures and shapes.

For Line Segments For Angles

Reflexive Property

of Congruence

AB¯¯¯¯¯AB¯¯¯¯¯\begin{align*}\overline{AB} \cong \overline{AB}\end{align*} BB\begin{align*}\angle B \cong \angle B\end{align*}

Symmetric Property

of Congruence

If AB¯¯¯¯¯CD¯¯¯¯¯\begin{align*}\overline{AB} \cong \overline{CD}\end{align*}, then CD¯¯¯¯¯AB¯¯¯¯¯\begin{align*}\overline{CD} \cong \overline{AB}\end{align*} If ABCDEF\begin{align*}\angle ABC \cong \angle DEF\end{align*}, then DEFABC\begin{align*}\angle DEF \cong \angle ABC\end{align*}

Transitive Property

of Congruence

If AB¯¯¯¯¯CD¯¯¯¯¯\begin{align*}\overline{AB} \cong \overline{CD}\end{align*} and CD¯¯¯¯¯EF¯¯¯¯¯\begin{align*}\overline{CD} \cong \overline{EF}\end{align*},

then AB¯¯¯¯¯EF¯¯¯¯¯\begin{align*}\overline{AB} \cong \overline{EF}\end{align*}

If ABCDEF\begin{align*}\angle ABC \cong \angle DEF\end{align*} and DEFGHI\begin{align*}\angle DEF \cong \angle GHI\end{align*},

then ABCGHI\begin{align*}\angle ABC \cong \angle GHI\end{align*}

## Using Properties of Equality with Equations

When you solve equations in algebra you use properties of equality. You might not write out the property for each step, but you should know that there is an equality property that justifies that step. We will abbreviate “Property of Equality” “PoE\begin{align*}PoE\end{align*}” and “Property of Congruence” “PoC\begin{align*}PoC\end{align*}.”

Example 1: Solve 2(3x4)+11=x27\begin{align*}2(3x-4)+11=x-27\end{align*} and write the property for each step (also called “to justify each step”).

Solution:

2(3x4)+116x8+116x+36x+336x6xx5x5x5x=x27=x27=x27=x273=x30=xx30=30=305=6Distributive PropertyCombine like termsSubtraction PoESimplifySubtraction PoEsimplifyDivision PoESimplify

Example 2: AB=8,BC=17\begin{align*}AB = 8, BC = 17\end{align*}, and AC=20\begin{align*}AC = 20\end{align*}. Are points A,B\begin{align*}A, B\end{align*}, and \begin{align*}C\end{align*} collinear?

Solution: Set up an equation using the Segment Addition Postulate.

Because the two sides are not equal, \begin{align*}A, B\end{align*} and \begin{align*}C\end{align*} are not collinear.

Example 3: If \begin{align*}m \angle A + m\angle B = 100^\circ\end{align*} and \begin{align*}m \angle B = 40^\circ\end{align*}, prove that \begin{align*}\angle A\end{align*} is an acute angle.

Solution: We will use a 2-column format, with statements in one column and their reasons next to it, just like Example 1.

## Two-Column Proof

Examples 1 and 3 are examples of two-column proofs. They both have the left side, the statements, and on the right side are the reason for these statements. Here we will continue with more proofs and some helpful tips for completing one.

Example 4: Write a two-column proof for the following:

If \begin{align*}A, B, C\end{align*}, and \begin{align*}D\end{align*} are points on a line, in the given order, and \begin{align*}AB = CD\end{align*}, then \begin{align*}AC = BD\end{align*}.

Solution: When the statement is given in this way, the “if” part is the given and the “then” part is what we are trying to prove.

Plot the points in the order \begin{align*}A, B, C, D\end{align*} on a line.

Add the given, \begin{align*}AB = CD\end{align*}.

Statement Reason
1. \begin{align*}A, B, C\end{align*}, and \begin{align*}D\end{align*} are collinear, in that order. Given
2. \begin{align*}AB = CD\end{align*} Given
3. \begin{align*}BC = BC\end{align*} Reflexive \begin{align*}PoE\end{align*}
4. \begin{align*}AB + BC = BC + CD\end{align*} Addition \begin{align*}PoE\end{align*}

5. \begin{align*}AB + BC = AC\end{align*}

\begin{align*}BC + CD = BD\end{align*}

6. \begin{align*}AC = BD\end{align*} Substitution or Transitive \begin{align*}PoE\end{align*}

Once we reach what we wanted to prove, we are done.

When completing a proof, these keep things in mind:

• Number each step.
• Statements with the same reason can (or cannot) be combined into one step. It is up to you. For example, steps 1 and 2 above could have been one step. And, in step 5, the two statements could have been written separately.
• Draw a picture and mark it with the given information.
• You must have a reason for EVERY statement.
• The order of the statements in the proof is not fixed. For example, steps 3, 4, and 5 could have been interchanged and it would still make sense.
• Reasons will be definitions, postulates, properties and previously proven theorems. “Given” is only used as a reason if the information in the statement column was told in the problem.

Example 5: Write a two-column proof.

Given: \begin{align*}\overrightarrow{BF}\end{align*} bisects \begin{align*}\angle ABC\end{align*}; \begin{align*}\angle ABD \cong \angle CBE\end{align*}

Prove: \begin{align*}\angle DBF \cong \angle EBF\end{align*}

Solution: First, put the appropriate markings on the picture. Recall, that bisect means “to cut in half.” Therefore, \begin{align*}m \angle ABF = m \angle FBC\end{align*}.

Statement Reason
1. \begin{align*}\overrightarrow{BF}\end{align*} bisects \begin{align*}\angle ABC, \angle ABD \cong \angle CBE\end{align*} Given
2. \begin{align*}m \angle ABF = m\angle FBC\end{align*} Definition of an Angle Bisector
3. \begin{align*}m\angle ABD = m\angle CBE\end{align*} If angles are \begin{align*}\cong\end{align*}, then their measures are equal.

4. \begin{align*}m\angle ABF = m\angle ABD + m\angle DBF\end{align*}

\begin{align*}m\angle FBC = m\angle EBF + m\angle CBE\end{align*}

5. \begin{align*}m\angle ABD + m\angle DBF = m\angle EBF + m\angle CBE\end{align*} Substitution \begin{align*}PoE\end{align*}
6. \begin{align*}m\angle ABD + m\angle DBF = m\angle EBF + m\angle ABD\end{align*} Substitution \begin{align*}PoE\end{align*}
7. \begin{align*}m\angle DBF = m\angle EBF\end{align*} Subtraction \begin{align*}PoE\end{align*}
8. \begin{align*}\angle DBF \cong \angle EBF\end{align*} If measures are equal, the angles are \begin{align*}\cong\end{align*}.

Use symbols and abbreviations for words within proofs. For example, \begin{align*}\cong\end{align*} was used in place of the word congruent above. You could also use \begin{align*}\angle\end{align*} for the word angle.

Know What? Revisited Analyzing the picture and what we know the sister on the left cannot be Sara because she lied (if we take what the sister in the middle said as truth). So, let’s assume that the sister in the middle is telling the truth, she is Sally. However, we know this is impossible, because that would have to mean that the sister on the right is lying and Sarah does not lie. From this, that means that the sister on the right is Sara and she is telling the truth, the sister in the middle is Sue. So, the first sister is Sally. The order is: Sally, Sue, Sara.

## Review Questions

• Questions 1-8 are similar to Examples 1 and 3.
• Questions 9-14 use the Properties of Equality.
• Questions 15-17 are similar to Example 2.
• Questions 18 and 19 are similar to Examples 8 and 9.
• Questions 20-34 are review.

For questions 1-8, solve each equation and justify each step.

1. \begin{align*}3x+11=-16\end{align*}
2. \begin{align*}7x-3=3x-35\end{align*}
3. \begin{align*}\frac{2}{3}g+1=19\end{align*}
4. \begin{align*}\frac{1}{2} MN = 5\end{align*}
5. \begin{align*}5m \angle ABC = 540^\circ\end{align*}
6. \begin{align*}10b-2(b+3)=5b\end{align*}
7. \begin{align*}\frac{1}{4}y+\frac{5}{6}=\frac{1}{3}\end{align*}
8. \begin{align*}\frac{1}{4}AB+\frac{1}{3}AB=12+\frac{1}{2}AB\end{align*}

For questions 9-14, use the given property or properties of equality to fill in the blank. \begin{align*}x, y\end{align*}, and \begin{align*}z\end{align*} are real numbers.

1. Symmetric: If \begin{align*}x = 3\end{align*}, then ______________.
2. Distributive: If \begin{align*}4(3x - 8)\end{align*}, then ______________.
3. Transitive: If \begin{align*}y = 12\end{align*} and \begin{align*}x = y\end{align*}, then ______________.
4. Symmetric: If \begin{align*}x + y = y + z\end{align*}, then ______________.
5. Transitive: If \begin{align*}AB = 5\end{align*} and \begin{align*}AB = CD\end{align*}, then ______________.
6. Substitution: If \begin{align*}x = y - 7\end{align*} and \begin{align*}x = z + 4\end{align*}, then ______________.
7. Given points \begin{align*}E, F\end{align*}, and \begin{align*}G\end{align*} and \begin{align*}EF = 16\end{align*}, \begin{align*}FG = 7\end{align*} and \begin{align*}EG = 23\end{align*}. Determine if \begin{align*}E, F\end{align*} and \begin{align*}G\end{align*} are collinear.
8. Given points \begin{align*}H, I\end{align*} and \begin{align*}J\end{align*} and \begin{align*}HI = 9\end{align*}, \begin{align*}IJ = 9\end{align*} and \begin{align*}HJ = 16\end{align*}. Are the three points collinear? Is \begin{align*}I\end{align*} the midpoint?
9. If \begin{align*}m\angle KLM = 56^\circ\end{align*} and \begin{align*}m\angle KLM + m\angle NOP = 180^\circ\end{align*}, explain how \begin{align*}\angle NOP\end{align*} must be an obtuse angle.

Fill in the blanks in the proofs below.

1. Given: \begin{align*}\angle ABC \cong \angle DEF\end{align*} \begin{align*}\angle GHI \cong \angle JKL\end{align*} Prove: \begin{align*}m\angle ABC + m \angle GHI = m \angle DEF + m\angle JKL\end{align*}
Statement Reason
1. Given

2. \begin{align*}m\angle ABC = m\angle DEF\end{align*}

\begin{align*}m\angle GHI = m\angle JKL\end{align*}

3. Addition \begin{align*}PoE\end{align*}
4. \begin{align*}m\angle ABC + m\angle GHI = m\angle DEF + m\angle JKL\end{align*}
1. Given: \begin{align*}M\end{align*} is the midpoint of \begin{align*}\overline{AN}\end{align*}. \begin{align*}N\end{align*} is the midpoint \begin{align*}\overline{MB}\end{align*} Prove: \begin{align*}AM = NB\end{align*}
Statement Reason
1. Given
2. Definition of a midpoint
3. \begin{align*}AM = NB\end{align*}

Use the diagram to answer questions 20-25.

1. Name a right angle.
2. Name two perpendicular lines.
3. Given that \begin{align*}EF = GH\end{align*}, is \begin{align*}EG = FH\end{align*} true? Explain your answer.
4. Is \begin{align*}\angle CGH\end{align*} a right angle? Why or why not?
5. Fill in the blanks:

1. Fill in the blanks:

Use the diagram to answer questions 26-31.

Which of the following must be true from the diagram?

Take each question separately, they do not build upon each other.

1. \begin{align*}\overline{AD} \cong \overline{BC}\end{align*}
2. \begin{align*}\overline{AB} \cong \overline{CD}\end{align*}
3. \begin{align*}\overline{CD} \cong \overline{BC}\end{align*}
4. \begin{align*}\overline{AB} \bot \overline{AD}\end{align*}
5. \begin{align*}ABCD\end{align*} is a square
6. \begin{align*}\overline{AC}\end{align*} bisects \begin{align*}\angle DAB\end{align*}

Use the diagram to answer questions 32-34.

Given: \begin{align*}B\end{align*} bisects \begin{align*}\overline{AD}\end{align*}, \begin{align*}C\end{align*} is the midpoint of \begin{align*}\overline{BD}\end{align*} and \begin{align*}AD = 12\end{align*}.

What is the value of each of the following?

1. \begin{align*}AB\end{align*}
2. \begin{align*}BC\end{align*}
3. \begin{align*}AC\end{align*}

1. \begin{align*}x = 3\end{align*}
2. If 2 angles are a linear pair, then their sum is \begin{align*}180^\circ\end{align*}. Law of Syllogism.

8 , 9 , 10

## Date Created:

Feb 22, 2012

Feb 03, 2016
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