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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 in the solution.
• Use a 2-column format to prove theorems.

## Review Queue

Solve the following problems.

1. Explain how you would solve \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 \begin{align*}180^\circ\end{align*}.

What can you conclude? By which law?

3. Draw a picture with the following:

\begin{align*}\angle LMN \ \text{is bisected by}\ \overline{MO} \qquad \qquad \overline{LM} \cong \overline{MP}\!\\ \angle OMP \ \text{is bisected by}\ \overline{MN} \qquad \qquad N \ \text{is the midpoint of}\ \overline{MQ}\end{align*}

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 of them one question. Scott asked the sister that was sitting on the left, “Which sister is in the middle?” and the answer he received was, “That's Sara.” Scott then asked the sister in the middle, “What is your name?” The response given was, “I'm Sally.” Scott turned to the sister on the right and asked, “Who is in the middle?” The sister then replied, “She is Sue.” Who was 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:

For all real numbers \begin{align*}a, b\end{align*}, and \begin{align*}c\end{align*}:

Examples
Reflexive Property of Equality \begin{align*}a = a\end{align*} \begin{align*}25 = 25\end{align*}
Symmetric Property of Equality \begin{align*}a = b\end{align*} and \begin{align*}b = a\end{align*} \begin{align*}m \angle P = 90^\circ\end{align*} or \begin{align*}90^\circ = m \angle P\end{align*}
Transitive Property of Equality \begin{align*}a = b\end{align*} and \begin{align*}b = c\end{align*}, then \begin{align*}a = c\end{align*} \begin{align*}a + 4 = 10\end{align*} and \begin{align*}10 = 6 + 4\end{align*}, then \begin{align*}a + 4 = 6 + 4\end{align*}
Substitution Property of Equality If \begin{align*}a = b\end{align*}, then \begin{align*}b\end{align*} can be used in place of \begin{align*}a\end{align*} and vise versa. If \begin{align*}a = 9\end{align*} and \begin{align*}a - c = 5\end{align*}, then \begin{align*}9 - c = 5\end{align*}
Addition Property of Equality If \begin{align*}a = b\end{align*}, then \begin{align*}a + c = b + c\end{align*}. If \begin{align*}2x = 6\end{align*}, then \begin{align*}2x + 5 = 6 + 11\end{align*}
Subtraction Property of Equality If \begin{align*}a = b\end{align*}, then \begin{align*}a - c = b - c\end{align*}. If \begin{align*}m \angle x + 15^\circ = 65^\circ\end{align*}, then \begin{align*}m \angle x+15^\circ-15^\circ=65^\circ-15^\circ\end{align*}
Multiplication Property of Equality If \begin{align*}a = b\end{align*}, then \begin{align*}ac = bc\end{align*}. If \begin{align*}y = 8\end{align*}, then \begin{align*}5 \cdot y=5 \cdot 8\end{align*}
Division Property of Equality If \begin{align*}a = b\end{align*}, then \begin{align*}\frac{a}{c}=\frac{b}{c}\end{align*}. If \begin{align*}3b=18\end{align*}, then \begin{align*}\frac{3b}{3}=\frac{18}{3}\end{align*}
Distributive Property \begin{align*}a(b+c)=ab+ac\end{align*} \begin{align*}5(2x-7)=5(2x)-5(7)=10x-35\end{align*}

## Properties of Congruence

Recall that \begin{align*}\overline{AB} \cong \overline{CD}\end{align*} if and only if \begin{align*}AB = CD\end{align*}. \begin{align*}\overline{AB}\end{align*} and \begin{align*}\overline{CD}\end{align*} represent segments, while \begin{align*}AB\end{align*} and \begin{align*}CD\end{align*} are lengths of those segments, which means that \begin{align*}AB\end{align*} and \begin{align*}CD\end{align*} are numbers. The properties of equality apply to \begin{align*}AB\end{align*} and \begin{align*}CD\end{align*}.

This also holds true for angles and their measures. \begin{align*}\angle ABC \cong \angle DEF\end{align*} if and only if \begin{align*}m \angle ABC = m \angle DEF\end{align*}. Therefore, the properties of equality apply to \begin{align*}m \angle ABC\end{align*} and \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 \begin{align*}\overline{AB} \cong \overline{AB}\end{align*} \begin{align*}\angle ABC \cong \angle CBA\end{align*}
Symmetric Property of Congruence If \begin{align*}\overline{AB} \cong \overline{CD}\end{align*}, then \begin{align*}\overline{CD} \cong \overline{AB}\end{align*} If \begin{align*}\angle ABC \cong \angle DEF\end{align*}, then \begin{align*}\angle DEF \cong \angle ABC\end{align*}
Transitive Property of Congruence If \begin{align*}\overline{AB} \cong \overline{CD}\end{align*} and \begin{align*}\overline{CD} \cong \overline{EF}\end{align*}, then \begin{align*}\overline{AB} \cong \overline{EF}\end{align*} If \begin{align*}\angle ABC \cong \angle DEF\end{align*} and \begin{align*}\angle DEF \cong \angle GHI\end{align*}, then \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 logical justification for each step in your solution, but you should know that there is an equality property that justifies that step. We will abbreviate “Property of Equality” “PoE” and “Property of Congruence” “PoC.”

Example 1: Solve \begin{align*}2(3x-4)+11=x-27\end{align*} and justify each step.

Solution:

\begin{align*}2(3x-4)+11 &= x-27\\ 6x-8+11 &= x-27 && \text{Distributive Property}\\ 6x+3 &= x-27 && \text{Combine like terms}\\ 6x+3-3 &= x-27-3 && \text{Subtraction PoE}\\ 6x &= x-30 && \text{Simplify}\\ 6x-x &= x-x-30 && \text{Subtraction PoE}\\ 5x &= -30 && \text{Simplify}\\ \frac{5x}{5} &= \frac{-30}{5} && \text{Division PoE}\\ x &= -6 && \text{Simplify}\end{align*}

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

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

\begin{align*}AB + BC &= AC && \text{Segment Addition Postulate}\\ 8 + 17 &= 20 && \text{Substitution PoE}\\ 25 & \neq 20 && \text{Combine like terms}\end{align*}

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 corresponding reasons in the next. This is formally called a 2-column proof.

Statement Reason
1. \begin{align*}m \angle A+m \angle B=100^\circ\end{align*} and \begin{align*}m \angle B = 40^\circ\end{align*} Given (always the reason for using facts that are told to us in the problem)
2. \begin{align*}m \angle A+40^\circ=100^\circ\end{align*} Substitution PoE
3. \begin{align*}m \angle A = 60^\circ\end{align*} Subtraction PoE
4. \begin{align*}\angle A\end{align*} is an acute angle Definition of an acute angle, \begin{align*}m \angle A < 90^\circ\end{align*}

## Two-Column Proof

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: First of all, 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 corresponding markings, \begin{align*}AB = CD\end{align*}, to the line.

Draw the 2-column proof and start with the given information. From there, we can use deductive reasoning to reach the next statement and what we want to prove. Reasons will be definitions, postulates, properties and previously proven theorems.

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 PoE
4. \begin{align*}AB + BC = BC + CD\end{align*} Addition PoE
5. \begin{align*}AB + BC = AC\!\\ BC + CD = BD\end{align*} Segment Addition Postulate
6. \begin{align*}AC = BD\end{align*} Substitution or Transitive PoE

When you reach what it is that you wanted to prove, you are done.

Prove Move: (A subsection that will help you with proofs throughout the book.) When completing a proof, a few things to keep 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.

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, if \begin{align*}\overrightarrow{BF}\end{align*} bisects \begin{align*}\angle ABC\end{align*}, then \begin{align*}m \angle ABF=m \angle FBC\end{align*}. Also, because the word “bisect” was used in the given, the definition will probably be used in the proof.

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

Prove Move: 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 The sisters, in order are: Sally, Sue, Sara. The sister on the left couldn’t have been Sara because that sister lied. The middle one could not be Sara for the same reason. So, the sister on the right must be Sara, which means she told Scott the truth and Sue is in the middle, leaving Sally to be the sister on the left.

## Review Questions

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, 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 DEF\!\\ \angle GHI \cong \angle JKL\end{align*} \begin{align*}\;\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\\ m \angle GHI = m \angle JKL\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. Using what is given in the picture AND \begin{align*}\angle EBF \cong \angle HCG\end{align*}, prove \begin{align*}\angle ABF \cong \angle DCG\end{align*}. Write a two-column proof.
6. Using what is given in the picture AND \begin{align*}AB = CD\end{align*}, prove \begin{align*}AC = BD\end{align*}. Write a two-column proof.

Use the diagram to answer questions 26-32.

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*}
7. Write a two-column proof. Given: Picture above and \begin{align*}\overline{AC}\end{align*} bisects \begin{align*}\angle DAB\end{align*} Prove: \begin{align*}m \angle BAC = 45^\circ\end{align*}
8. Draw a picture and write a two-column proof. Given: \begin{align*}\angle 1\end{align*} and \begin{align*}\angle 2\end{align*} form a linear pair and \begin{align*}m \angle 1=m \angle 2\end{align*}. Prove: \begin{align*}\angle 1\end{align*} is a right angle

1. First, subtract 3 from both sides and then divide both sides by 2. \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.

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