Introduction to Proofs, including Algebraic, Segment, and Angle Proofs

Before getting into the proofs, we need to understand what a conditional statement is, so we can understand logical reasoning and deductive thinking.

Conditional Statements

A statement that can be written with an “if-then” form is a conditional statement. The statement after the “if” is the hypothesis, and the statement after the “then” is the conclusion. Proofs rely on logical connections between different statements in order to get to a conclusion; this is why conditional statements are typically introduced before geometric proofs.

We write a conditional statement as “if $ p$, then $ q$, or $ \displaystyle p\to q$ ($ p$ implies $ q$). Here’s an example: “If it is raining outside, then I will bring my umbrella.” In this case, the hypothesis is “it is raining outside” and the conclusion is “I will bring my umbrella.” Pretty straightforward!

Here is a table of some types of conditional statements, including some geometry-related examples. Note that if the conditional statement is true, then so is the contrapositive, but the converse and the inverse may both be true or both be false. In the example below, both the converse and inverse are false.

Type of Conditional	Symbolic Representation	Example	True or False
Conditional	$ \displaystyle p\to q$	If a figure is a triangle, then it is a polygon. (A polygon is a flat figure with straight sides that are connected; a triangle is a polygon with three sides).	True; a triangle is a type of polygon.
Converse	$ \displaystyle q\to p$	If a figure is a polygon, then it is a triangle.	False; a figure may be a polygon, but could be a quadrilateral (four sides), for example.
Inverse	$ \displaystyle \sim p\to \,\sim q$	If a figure is not a triangle, then it is not a polygon.	False; A figure could be a quadrilateral, for example, and still be a polygon. This is a counterexample.
Contrapositive	$ \displaystyle \sim q\to \,\sim p$	If a figure is not a polygon, then it is not a triangle.	True; a triangle must be a polygon, so if a figure is not a polygon, it can’t be a triangle.

Here’s another example, where, since the converse and inverse are both true, we can say that the conditional statement is an “if and only if” or “iff” statement (also known as a biconditional statement): If a polygon has three sides, then it is a triangle. Do you see why both the converse and inverse are true, in addition to the original statement and the contrapositive? We can also say that the original statement and the hypothesis have the same truth value.

It’s pretty fun, isn’t it? It’s logical; thus it’s call logic!

Proof Process

A mathematical proof uses deductive reasoning, which is based on rules, as opposed to patterns, to show something is true. The rules can be definitions, previously proven theorems, axioms and postulates, for example. Theorems are statements that can be proven, and axioms and postulates are statements that are generally known to be true, axioms being more broad to all sciences.

A proof contain a series of logical statements, beginning with what is given, and ending with what needs to be proven. It’s really as simple as that!

Here is a more detailed set of steps in the proof process:

1. List the given information and draw a picture, if possible.
2. State what needs to be proven.
3. Provide a list of logical steps to get to what you’re trying to prove.
4. For each step, include the statement and then the reason (previous theorems, postulates, and so on).
5. State what has been proven.

Postulates

Postulates are statements that are just known to be true. We don’t need to prove them (someone decided this!), since they are the fundamentals of science! We have to start somewhere, right?

Here are some examples of postulates that have to do with points, lines, and planes:

- Two Point Postulate. Through any two points, there exists only one line.
- Three Point Postulate. Through any three noncollinear points, there exists exactly one plane that contains them all.
- Line-Point Postulate. A line contains at least two points.
- Plane-Point Postulate. A plane contains at least three noncollinear points.
- Plane-Line Postulate. If a plane contains two points, it contains the line passing through those two points.
- Line Intersection Postulate. If two lines intersect, their intersection is exactly one point.
- Plane Intersection Postulate. If two planes intersect, their intersection is a line.

Paragraph and Two-Column Proofs

A paragraph proof uses a paragraph to use deductive reasoning to reach a conclusion, which is called a theorem. Again, we can use definitions, previously proven theorems, axioms and postulates to arrive at the conclusion, using a series of logical steps. Usually a picture is given to show what we are proving, and if not, it’s best to draw one for a geometric proof.

Here is an example paragraph proof that proves the Midpoint Theorem:

Given: $ B$ is the midpoint of $ \displaystyle \overline{{AC}}$.

Prove: $ \displaystyle \overline{{AB}}\approx \overline{{BC}}$

Proof: According to the definition of midpoint, since $ B$ is the midpoint of $ \displaystyle \overline{{AC}}$, it means that $ AB$ and $ BC$ have the same measurement, or $ AB=BC$. By the definition of congruence, if two segments have the same measurement, they are congruent. Thus, $ \displaystyle \overline{{AB}}\approx \overline{{BC}}$.

Now, let’s use a two-column proof to prove the Midpoint Theorem. Notice that a two-column proof uses statements and reasons that are organized in two columns. We’ll later see that flowchart proofs are commonly used. Note that abbreviations are commonly used when creating proofs.

Given: $ B$ is the midpoint of $ \displaystyle \overline{{AC}}$

Prove: $ \displaystyle \overline{{AB}}\approx \overline{{BC}}$

Proof:

Statements	Reasons
1. $ B$ is the midpoint of $ \displaystyle \overline{{AC}}$.	1. Given
2. $ AB=BC$	2. Definition of Midpoint (Def of Midpt)
3. $ \displaystyle \overline{{AB}}\approx \overline{{BC}}$	3. Definition of Congruence (Def of $ \displaystyle \cong $)

Algebraic Proofs

We learned about some of the Algebraic Properties of Real Numbers here in the Types of Numbers and Algebraic Properties section. We actually used algebraic proof methods to solve algebraic equations without even knowing it. Here, we’ll summarize the properties and demonstrate how to solve algebraic equations using “formal” proofs.

Here is a summary of the properties we’ll use:

Property of Equality and Explanation	Example
Addition Property of Equality: adding the same thing to both sides of an equation. In the example, we start with $ 5$ on each side, and then add $ 1$ to each side to get $ 6=6$. Try it for other numbers!	$ \begin{array}{c}5=5\\5+1=5+1\\\,\,\,\,\,\,\,6=6 \end{array}$ √
Subtraction Property of Equality: subtracting the same thing from both sides of an equation. Again, if we start out with $ 5$ on each side, and then subtract $ 3$ from each side, we get $ 2=2$.	$ \begin{array}{c}5=5\\5-3=5-3\\\,\,\,\,\,\,\,\,2=2\end{array}$ √
Multiplication Property of Equality: multiplying by the same thing on both sides of an equation. If we multiply each side by $ 4$, it works, too!	$ \displaystyle \begin{array}{c}5=5\\5\times 4=5\times 4\\\,\,\,\,\,\,\,\,\,20=20\end{array}$ √
Division Property of Equality: dividing by the same thing on both sides of an equation. If we divide each side by $ 2$, it works too!	$ \displaystyle \begin{array}{c}6=6\\6\div 2=6\div 2\\\,\,\,\,\,\,\,\,3=3\end{array}$ √
Reflexive Property of Equality: a number or expression always equals itself (Duh!).	$ \displaystyle 5=5$
Symmetric Property of Equality: you can switch the order of expressions that are equal.	if $ \displaystyle x=5$, then $ 5=x$.
Transitive Property of Equality: if a first number equals a second number, and that second number equals a third number, the first number equals the third number.	if $ \displaystyle a=b$, and $ b=c$, then $ a=c$.
Substitution Property of Equality: if a number equals a another number, either number can be replaced with the other.	if $ \displaystyle a=b$, then you can replace $ a$ with $ b$ in any expression or equation.
Distributive Property: when a factor is multiplied by the sum or difference of two other numbers, you can “push through” that factor, multiply by the numbers, and then add or subtract them.	$ \displaystyle n(2+3)=2n+3n=5n$

Here’s an example of a two-column algebraic proof. I know you’re thinking “this is way too much trouble to do something that’s trivial”, but it’s still important to learn how to be organized so as to perform these proofs.

Given: $ \displaystyle \frac{{2x+1}}{3}=7$

Prove: $ x=10$

Proof:

Statements	Reasons
1. $ \displaystyle \frac{{2x+1}}{3}=7$	1.Given
2. $ \displaystyle 3\left( {\frac{{2x+1}}{3}} \right)=3\left( 7 \right)$	2. Multiplication Property of Equality
3. $ \displaystyle 2x+1=21$	3. Substitution
4. $ \displaystyle 2x+1-1=21-1$	4. Subtraction Property of Equality
5. $\displaystyle 2x=20$	5. Substitution
6. $ \displaystyle \frac{{2x}}{2}=\frac{{20}}{2}$	6. Division Property of Equality
7. $ x=10$	7. Substitution

I know it seems like a lot of work, but don’t you feel organized?

Geometric Proofs

Geometric proofs use the same properties of numbers, but, as we saw above, when going back and forth between congruence (having the same size and shape) and having the same measurement (same numerical value), we typically have to use the Definition of Congruence as a step in the proof, since these are technically two different things.

Segment Relationship Proofs

For segment addition, there are two basic postulates:

- Ruler Postulate. The points on a line or segment can correspond to the Real Numbers. We don’t usually use this postulate in a proof.
- Segment Addition Postulate. If there is a point $ B$ between two points $ A$ and $ C$ on a colinear line, then $ AB+BC=AC$. Also, if a segments contains points $ A$, $ B$, and $ C$, and if $ AB+BC=AC$, then the three points are collinear, and point $ B$ is between points $ A$ and $ C$.

In the Algebraic Proof section above, we saw that measures or numbers, such as the lengths of segments, are reflexive, symmetric, and transitive. Since congruence of segments means the segments have the same measurements, congruence also has these properties:

- Reflexive Property of Congruence. $ \displaystyle \overline{{AB}}\approx \overline{{AB}}$
- Symmetric Property of Congruence. If $ \displaystyle \overline{{AB}}\approx \overline{{CD}}$, then $ \displaystyle \overline{{CD}}\approx \overline{{AB}}$.
- Transitive Property of Congruence. If $ \displaystyle \overline{{AB}}\approx \overline{{CD}}$, and $ \displaystyle \overline{{CD}}\approx \overline{{EF}}$, then $ \displaystyle \overline{{AB}}\approx \overline{{EF}}$.

Makes sense!

Here is an example two-column proof; note that there are others way to prove this. It’s probably better to err on having too many steps, if every step is correct. Notice how we can abbreviate!

Given: $ \displaystyle \overline{{EF}}\approx \overline{{BC}}$

Prove: $EF+CD=BD$

Proof:

Statements	Reasons
1. $ \displaystyle \overline{{EF}}\approx \overline{{BC}}$	1. Given
2. $ \displaystyle \overline{{BC}}\approx \overline{{EF}}$	2. Symmetric Property of Congruence (Symm Prop of $ \displaystyle \cong $)
3. $ BC=EF$	3. Definition of Congruence (Def of $ \displaystyle \cong $)
4. $BC+CD=BD$	4. Segment Addition Postulate (Seg Add Post)
5. $EF+CD=BD$	5. Substitution (Subst)

Angle Relationship Proofs

There are several postulates and theorems relating to angle relationships. It all seems overwhelming now, but you’ll see that most of these just have to do with common sense! I’ve also included common abbreviations used, where applicable.

Protractor Postulate. A unique real number between $ 0$ and $ 180$, inclusively, exists for any angle. (I’ve never seen this in a proof.)
Angle Addition Postulate. The measures of two adjacent angles can be added to find the measure of the larger angle they create together.
Supplement Theorem. Two angles that form a linear pair (share a common side and form a straight line) are supplementary. ($ \displaystyle \text{Linear Pair} \Rightarrow \text{ suppl. }\angle \,\text{ }\!\!’\!\!\text{ s}$)
Complement Theorem. If two angles form a right angle, then the two angles are complementary. ($ \displaystyle \text{Two Angles in Right Angle } \Rightarrow \text{ compl. }\angle \,\text{ }\!\!’\!\!\text{ s}$)
Reflexive, Symmetric, and Transitive Properties of Angle Congruence. These are the same properties that we saw above for algebraic proofs and congruence of segments.
Congruent Supplements Theorem. Angles supplementary to congruent angles (or the same angle) are congruent. ($ \displaystyle \angle \text{ }\!\!’\!\!\text{ s}\,\,\text{suppl}\text{. to same }\angle \,\,\text{or }\cong \,\,\,\angle \text{ }\!\!’\!\!\text{ s}\,\,\,\text{are}\,\,\cong $)
Congruent Complements Theorem. Angles complementary to congruent angles (or the same angle) are congruent. ($ \displaystyle \angle \text{ }\!\!’\!\!\text{ s}\,\,\text{compl}\text{. to same }\angle \,\,\text{or }\cong \,\,\,\angle \text{ }\!\!’\!\!\text{ s}\,\,\,\text{are}\,\,\cong $)
Vertical Angles Theorem. Vertical angles (opposite angles in the intersection of two lines) are congruent. ($ \displaystyle \text{Vertical }\angle \text{ }\!\!’\!\!\text{ s}\,\,\,\text{are}\,\,\,\cong $)
Right Angles Theorems. These all can be proved by earlier theorems; you may be asked to prove these. An example is given below.

- Intersecting perpendicular lines form four right angles.
- All right angles are congruent.
- Intersecting perpendicular lines form congruent adjacent angles.
- Two complementary, congruent angles are each right angles.
- Two congruent angles that form a linear pair are each right angles.

Don’t forget to use definitions, such as the definition of a right angle, linear pair, vertical angles, supplementary and complementary angles, and so on!

Here is an example two-column proof.

Given: $ \displaystyle \overline{{AC}}\bot \,\overline{{CD}}$, $ \displaystyle \angle \,1$ and $ \displaystyle \angle \,3$ are complementary.

Prove: $ \displaystyle \angle \,2\cong \angle \,3$

Proof:

Statements	Reasons
1. $ \displaystyle \overline{{AC}}\bot \,\overline{{CD}}$	1. Given
2. $ \angle \,ACD$ is a Right Angle	2. Def of $ \bot$
3. $ \displaystyle \text{m}\,\angle \,ACD=\,90 $	3. Def of Right $ \angle$
4. $ \displaystyle \text{m}\,\angle \,ACD=\text{m}\,\angle \,1+ \text{m}\,\angle \,2$	4. Angle Add Post
5. $ \displaystyle 90=\text{m}\,\angle \,1+ \text{m}\,\angle \,2$	5. Subst (or Transitive)
6. $ \displaystyle \angle \,1$ and $ \displaystyle \angle \,2$ are complementary.	6. Def of Comp $ \displaystyle \angle \,’s$
7. $ \displaystyle \angle \,1$ and $ \displaystyle \angle \,3$ are complementary.	7. Given
8. $ \displaystyle \angle \,2\cong \angle \,3$	8. $ \displaystyle \angle \text{ }\!\!’\!\!\text{ s}\,\,\text{compl}\text{. to same }\angle \,\,\text{or }\cong \,\,\,\angle \text{ }\!\!’\!\!\text{ s}\,\,\,\text{are}\,\,\cong $

More Problems and Solutions

Problem: Identify the hypothesis and conclusion for the following true statements, and then write the converse, inverse, and contrapositive. State whether each of the three is true or false. Find a counterexample if the statement is false.

If you live in Dallas, Texas, then you live in the United States.
If two angles are a linear pair, then they are supplementary.
If a figure is a hexagon, then it is a polygon.
If two angles are complementary, the sum of their angle measurements add up to $ 90{}^\circ $.

Solution:

	Converse	Inverse	Contrapositive
Hypothesis: you live in Dallas, Texas Conclusion: you live in the United States	If you live in the United States, then you live in Dallas, Texas. False: you could live in Oregon in the United States, but not live in Dallas.	If you don’t live in Dallas, Texas, then you don’t live in the United States. False: you could live in Oregon, which is still the United States.	If you don’t live in the United States, then you don’t live in Dallas, Texas. True
Hypothesis: two angles are a linear pair Conclusion: they are supplementary (A linear pair is a set of angles that are adjacent and form a straight line.)	If two angles are supplementary, then they are a linear pair. False: the two angles could be right angles (sum adds up to $ 180{}^\circ $ and thus they are supplementary) that are vertical.	If two angles are not a linear pair, then they are not supplementary. False: the two angles could be vertical angles (not a linear pair), and each could be a right angle. Two right angles are supplementary.	If two angles are not supplementary, then they are not a linear pair. True: the measurement of the two angles can’t be $ 180{}^\circ $, and thus, they couldn’t be a linear pair, since the measurement of angles that form a straight line up to $ 180{}^\circ $.
Hypothesis: a figure is a hexagon Conclusion: it is a polygon	If a figure is a polygon, then it is a hexagon. False: the polygon could be a triangle.	If a figure is not a hexagon, then it is not a polygon. False: the figure could be a triangle, which is still a polygon.	If the figure is not a polygon, then it is not a hexagon. True: all hexagons are polygons.
Hypothesis: two angles are complementary Conclusion: the sum of their angle measurements add up to $ 90{}^\circ $	If the sum of two angle measurements add up to $ 90{}^\circ $, then the angles are complementary True. This is a definition.	If two angles are not complementary, then the sum of their angle measurements do not add up to $ 90{}^\circ $. True. Note that since the converse and inverse are both true, this is a biconditional statement.	If the sum of the measurements of two angles don’t add up to $ 90{}^\circ $, then the two angles are not complementary. True.

Converse

Inverse

Contrapositive

Hypothesis: you live in Dallas, Texas

Conclusion: you live in the United States

If you live in the United States, then you live in Dallas, Texas.

False: you could live in Oregon in the United States, but not live in Dallas.

If you don’t live in Dallas, Texas, then you don’t live in the United States.

False: you could live in Oregon, which is still the United States.

If you don’t live in the United States, then you don’t live in Dallas, Texas.

True

Hypothesis: two angles are a linear pair

Conclusion: they are supplementary

(A linear pair is a set of angles that are adjacent and form a straight line.)

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

False: the two angles could be right angles (sum adds up to $ 180{}^\circ $ and thus they are supplementary) that are vertical.

If two angles are not a linear pair, then they are not supplementary.

False: the two angles could be vertical angles (not a linear pair), and each could be a right angle. Two right angles are supplementary.

If two angles are not supplementary, then they are not a linear pair.

True: the measurement of the two angles can’t be $ 180{}^\circ $, and thus, they couldn’t be a linear pair, since the measurement of angles that form a straight line up to $ 180{}^\circ $.

Hypothesis: a figure is a hexagon

Conclusion: it is a polygon

If a figure is a polygon, then it is a hexagon.

False: the polygon could be a triangle.

If a figure is not a hexagon, then it is not a polygon.

False: the figure could be a triangle, which is still a polygon.

If the figure is not a polygon, then it is not a hexagon.

True: all hexagons are polygons.

Hypothesis: two angles are complementary

Conclusion: the sum of their angle measurements add up to $ 90{}^\circ $

If the sum of two angle measurements add up to $ 90{}^\circ $, then the angles are complementary

True. This is a definition.

If two angles are not complementary, then the sum of their angle measurements do not add up to $ 90{}^\circ $.

True. Note that since the converse and inverse are both true, this is a biconditional statement.

If the sum of the measurements of two angles don’t add up to $ 90{}^\circ $, then the two angles are not complementary.

True.

Problem: Prove the following using a two-column proof:

Given: $ \displaystyle z=\frac{1}{2}x+5y$

Prove: $ \displaystyle x=2z-10y$

Solution:

Proof:

Statements	Reasons
1. $ \displaystyle z=\frac{1}{2}x+5y$	1. Given
2. $ \displaystyle 2z=2\left( {\frac{1}{2}x+5y} \right)$	2. Mult Prop of =
3. $ \displaystyle 2z=2\left( {\frac{1}{2}x} \right)+2\left( {5y} \right)$	3. Dist Prop
4. $ \displaystyle 2z=x+10y$	4. Subst
5. $ \displaystyle x+10y=2z$	5. Symm Prop of =
6. $ \displaystyle x=2z-10y$	6. Subt Prop of =

Problem: Prove the following using a two-column proof:

Given: $ B$ is between $ A$ and $ C$, $ C$ is between $ B$ and $ D$, $ AC=BD$.

Prove: $ \displaystyle \overline{{AB}}\approx \overline{{CD}}$

Solution:

Proof:

Statements	Reasons
1. $ B$ is between $ A$ and $ C$; $ C$ is between $ B$ and $ D$.	1. Given
2. $AB+BC=AC$; $BC+CD=BD$	2. Seg Add Post
3. $ AC=BD$	3. Given
4. $ AB+BC=BC+CD$	4. Subst (or Trans)
5. $ AB+BC-BC=BC+CD-BC$	5. Subt Prop of =
6. $ AB=CD$	6. Subst
7. $ \displaystyle \overline{{AB}}\approx \overline{{CD}}$	7. Def of $ \displaystyle \cong $

Problem: Prove the following using a paragraph proof and then a two-column proof:

Given: $ \displaystyle \angle \,1$ and $ \displaystyle \angle \,3$ form a linear pair, $ \displaystyle \text{m}\,\angle \,1+ \text{m}\,\angle \,2=180$.

Prove: $ \displaystyle \angle \,2\cong \angle \,3$

Solutions:

Paragraph Proof: It is given that $ \displaystyle \angle \,1$ and $ \displaystyle \angle \,3$ form a linear pair. By the supplement theorem, if two angles form a linear pair, the sum of their angles is $ 180$; thus, $ \displaystyle m\angle 1\,+\,m\angle \,3=180$. Then, by the definition of supplements, $ \displaystyle \angle 1$ is supplementary to $ \displaystyle \angle 3$. It is given that $ \displaystyle \text{m}\,\angle \,1+ \text{m}\,\angle \,2=180$; by the definition of supplements, $ \displaystyle \angle 1$ is supplementary to $ \displaystyle \angle 2$. Since angles supplementary to the same angles are congruent, it follows that $ \displaystyle \angle \,2\cong \angle \,3$.

Proof:

Statements	Reasons
1. $ \displaystyle \angle \,1$ and $ \displaystyle \angle \,3$ form a linear pair	1. Given
2. $ \displaystyle m\angle 1\,+\,m\angle \,3=180$	2. Suppl Theorem
3. $ \displaystyle \text{m}\,\angle \,1+ \text{m}\,\angle \,2=180$	3. Given
4. $ \displaystyle \angle 1$ is supplementary to $ \displaystyle \angle 3$; $ \displaystyle \angle 1$ is supplementary to $ \displaystyle \angle 2$	4. Def of Suppl $ \displaystyle \angle \,’s$
5. $ \displaystyle \angle \,2\cong \angle \,3$	5. $ \displaystyle \angle \text{ }\!\!’\!\!\text{ s}\,\,\text{supple}\text{. to same }\angle \,\,\text{or }\cong \,\,\,\angle \text{ }\!\!’\!\!\text{ s}\,\,\,\text{are}\,\,\cong $.

On to Parallel Lines and Transversals!