Two-column and paragraph proofs
A proof is a list of true statements. What gives each one the right to be on the list?
Where you are 18 in Geometric proof
When you reach for it
Every time you write a proof. A two-column proof lists statements down the left and, beside each one, the reason it is allowed. The statements carry the argument; the reasons make it a proof instead of a list of things that happen to be true. Anyone should be able to check a single line by reading its reason and the lines above it, without trusting the writer and without measuring the picture. A paragraph proof is the same chain of statements and reasons written out as sentences, each reason riding inside the sentence that needs it (Example 1 shows both).
The first lines are usually the Given, with the reason “given”, and the last line is the Prove, both set up as in Given, prove and the diagram. Everything between them is this skill.
The procedure
Only five kinds of reason are allowed. A statement may stand because of
- a given: a fact stated in the Given, and nothing more;
- a definition: what a word means, such as “a midpoint cuts a segment into two equal parts”;
- a postulate: a fact accepted without proof, such as the segment addition postulate or the angle addition postulate;
- a property of equality or congruence: the rules for trading equal things, listed below;
- a theorem already proved, such as the Linear Pair Theorem or the Vertical Angles Theorem.
“It looks that way”, “I measured it” and “obviously” are not on the list. Neither is the Prove itself, and neither is any theorem that was proved using the one you are proving. How facts are sorted into these kinds is definitions, postulates and theorems; why a chain of such steps settles a claim is deductive reasoning.
A definition works in both directions, because every definition is a biconditional. “M is the midpoint of \overline{AB}, so AM = MB” and “M is on \overline{AB} and AM = MB, so M is the midpoint” both cite the definition of midpoint. In the same way \angle A \cong \angle B and m\angle A = m\angle B say the same thing, and the step between them is the definition of congruent angles (for segments, of congruent segments). That step is how a proof turns marks on a figure into numbers it can add.
A theorem needs its if-part first. A theorem is a conditional. Before you cite it, its if-part must stand on an earlier line, or be a position fact read from the diagram, as what a diagram may assume allows. Then its then-part may be the new line.
Algebra steps get named. The moves below are reasons like any other. Properties of equality as reasons and reflexive, symmetric and transitive congruence treat them in depth.
| Property of equality | What it allows |
|---|---|
| reflexive | a = a |
| symmetric | if a = b, then b = a |
| transitive | if a = b and b = c, then a = c |
| substitution | if a = b, then b may replace a in any statement |
| addition, subtraction | if a = b, then a + c = b + c and a - c = b - c |
| multiplication, division | if a = b, then ac = bc, and a \div c = b \div c when c \ne 0 |
Congruence has the first three, with \cong in place of =: \overline{AB} \cong \overline{AB}; if \angle 1 \cong \angle 2, then \angle 2 \cong \angle 1; and if \angle 1 \cong \angle 2 and \angle 2 \cong \angle 3, then \angle 1 \cong \angle 3.
One more reason covers routine tidying. Simplify may be cited for arithmetic on numbers, such as 76 + 104 = 180, and for combining like terms on one side of a statement, such as AM + AM = 2AM or (2x + 14) + (4x - 20) = 6x - 6. It never covers a step that does something to both sides, which is named by its property (“subtract 12 from both sides” is the subtraction property), and it never covers multiplying out brackets, which is the distributive property: a(b + c) = ab + ac.
How a line is written. One statement per row, numbered. The reason names its kind and then, after a colon, the earlier lines it uses, as in “substitution: lines 2 and 3 into line 4”. When a postulate needs a position fact taken from the diagram, the reason says which one, as in “angle addition postulate: \overrightarrow{OB} is inside \angle AOC”. An angle measure is a number of degrees, so once a proof is solving for x the degree signs are left off: m\angle ABD = 2x + 14 means the angle measures (2x + 14)^\circ.
A reason points back, never forward. It may use any line above it and no line below.
To find the steps, work from both ends. Unpack each given with its definition to see what it offers. Then ask of the Prove: what reason could end on this, and what would that reason need? The answer is the line before last, and asking again gives the one before that. When the two ends meet, write the proof out from the top.
Worked examples
A midpoint, line by line
Prove that a midpoint makes each half of a segment half as long as the whole.
Given. M is the midpoint of \overline{AB}.
Prove. AM = \tfrac{1}{2}AB.
Plan. The midpoint gives two equal halves, and segment addition says the halves make the whole. Put those together and divide by 2.
| Statements | Reasons |
|---|---|
| 1. M is the midpoint of \overline{AB} | given |
| 2. AM = MB | definition of midpoint: line 1 |
| 3. MB = AM | symmetric property of equality: line 2 |
| 4. AM + MB = AB | segment addition postulate: M is between A and B, line 1 |
| 5. AM + AM = AB | substitution: line 3 into line 4 |
| 6. 2AM = AB | simplify: line 5 |
| 7. AM = \tfrac{1}{2}AB | division property of equality: line 6 |
Line 4 can cite line 1 for its position fact because a midpoint lies on its segment by definition. Line 5 puts AM in place of MB. Substitution lets the right side of an equation replace its left side, so the proof first turns line 2 around with the symmetric property; line 3 is exactly the equation line 5 needs.
The same proof as a paragraph. We are given that M is the midpoint of \overline{AB}, so AM = MB by the definition of midpoint, and by the symmetric property of equality MB = AM. A midpoint lies between the endpoints of its segment, so the segment addition postulate gives AM + MB = AB. Substituting AM for MB gives AM + AM = AB, which simplifies to 2AM = AB, and the division property of equality gives AM = \tfrac{1}{2}AB. Every reason from the table is still there, in the same order; only the layout changed, and the reader now has to find each reason inside its sentence.
Solving an equation as a proof
Given. \angle ABD and \angle DBC form a linear pair; m\angle ABD = (2x + 14)^\circ; m\angle DBC = (4x - 20)^\circ.
Prove. x = 31.
Plan. Solving an equation is already a proof; each move just needs its name.
| Statements | Reasons |
|---|---|
| 1. \angle ABD and \angle DBC form a linear pair | given |
| 2. m\angle ABD = 2x + 14 | given |
| 3. m\angle DBC = 4x - 20 | given |
| 4. m\angle ABD + m\angle DBC = 180 | Linear Pair Theorem: line 1 |
| 5. (2x + 14) + (4x - 20) = 180 | substitution: lines 2 and 3 into line 4 |
| 6. 6x - 6 = 180 | simplify: line 5 |
| 7. 6x = 186 | addition property of equality: line 6 |
| 8. x = 31 | division property of equality: line 7 |
Line 6 only tidies the left side, so it is simplify. Line 7 adds 6 to both sides, so it is not.
Check. 2(31) + 14 = 76 and 4(31) - 20 = 104, and 76 + 104 = 180.
Taking away a shared angle
Given. m\angle AOC = m\angle BOD.
Prove. m\angle AOB = m\angle COD.
Plan. Each large angle is a small angle plus the shaded \angle BOC. Take \angle BOC away from both.
| Statements | Reasons |
|---|---|
| 1. m\angle AOC = m\angle BOD | given |
| 2. m\angle AOB + m\angle BOC = m\angle AOC | angle addition postulate: \overrightarrow{OB} is inside \angle AOC |
| 3. m\angle BOC + m\angle COD = m\angle BOD | angle addition postulate: \overrightarrow{OC} is inside \angle BOD |
| 4. m\angle AOB + m\angle BOC = m\angle BOD | substitution: line 1 into line 2 |
| 5. m\angle BOD = m\angle BOC + m\angle COD | symmetric property of equality: line 3 |
| 6. m\angle AOB + m\angle BOC = m\angle BOC + m\angle COD | substitution: line 5 into line 4 |
| 7. m\angle AOB = m\angle COD | subtraction property of equality: line 6 |
The positions in lines 2 and 3 are not in the Given; they are read from Figure 3, which a diagram is allowed to supply. With numbers: if m\angle AOC = 100^\circ and m\angle BOC = 60^\circ, each small angle is 100^\circ - 60^\circ = 40^\circ.
Planning backwards
Given. \overleftrightarrow{AB} and \overleftrightarrow{CD} meet at E; \overrightarrow{EF} bisects \angle AEC; \overrightarrow{EG} bisects \angle BED.
Prove. \angle AEF \cong \angle BEG.
Plan. Start at the end. Two angles are congruent when their measures are equal, so the goal is m\angle AEF = m\angle BEG. Each is half of a larger angle, since a bisector splits its angle into two equal parts that add up to it. Halves of equal things are equal, so it is enough to know m\angle AEC = m\angle BED. And that is available: \angle AEC and \angle BED are vertical angles. Now write it forwards.
| Statements | Reasons |
|---|---|
| 1. \overleftrightarrow{AB} and \overleftrightarrow{CD} meet at E | given |
| 2. \angle AEC \cong \angle BED | Vertical Angles Theorem: line 1 |
| 3. m\angle AEC = m\angle BED | definition of congruent angles: line 2 |
| 4. \overrightarrow{EF} bisects \angle AEC | given |
| 5. m\angle AEF = m\angle FEC | definition of angle bisector: line 4 |
| 6. m\angle FEC = m\angle AEF | symmetric property of equality: line 5 |
| 7. m\angle AEF + m\angle FEC = m\angle AEC | angle addition postulate: \overrightarrow{EF} is inside \angle AEC, line 4 |
| 8. m\angle AEF + m\angle AEF = m\angle AEC | substitution: line 6 into line 7 |
| 9. 2m\angle AEF = m\angle AEC | simplify: line 8 |
| 10. \overrightarrow{EG} bisects \angle BED | given |
| 11. m\angle BEG = m\angle GED | definition of angle bisector: line 10 |
| 12. m\angle GED = m\angle BEG | symmetric property of equality: line 11 |
| 13. m\angle BEG + m\angle GED = m\angle BED | angle addition postulate: \overrightarrow{EG} is inside \angle BED, line 10 |
| 14. m\angle BEG + m\angle BEG = m\angle BED | substitution: line 12 into line 13 |
| 15. 2m\angle BEG = m\angle BED | simplify: line 14 |
| 16. 2m\angle AEF = m\angle BED | substitution: line 3 into line 9 |
| 17. m\angle BED = 2m\angle BEG | symmetric property of equality: line 15 |
| 18. 2m\angle AEF = 2m\angle BEG | substitution: line 17 into line 16 |
| 19. m\angle AEF = m\angle BEG | division property of equality: line 18 |
| 20. \angle AEF \cong \angle BEG | definition of congruent angles: line 19 |
Lines 7 and 13 may cite the bisector’s line because a bisector lies inside its angle; that is part of its definition. Lines 10 to 15 repeat lines 4 to 9 for the other bisector, with every reason written again. Writing “similarly” in their place would leave the reader to do that copying. When the same run of steps is needed again and again, the honest shortcut is to prove it once as a result of its own, as Exercise 16 does for a bisector, and then cite it in a single line. Lines 3 and 20 cite the same definition in opposite directions: from marks to numbers, then back.
Parallel lines inside a triangle
Given. D is on \overline{AB} and E is on \overline{AC} in \triangle ABC; \overline{DE} \parallel \overline{BC}; \angle B \cong \angle C.
Prove. \angle ADE \cong \angle AED.
Plan. Carry each base angle up to the parallel segment, then chain the congruences.
| Statements | Reasons |
|---|---|
| 1. \overline{DE} \parallel \overline{BC} | given |
| 2. \angle ADE \cong \angle B | Corresponding Angles Theorem: line 1, transversal \overleftrightarrow{AB} |
| 3. \angle AED \cong \angle C | Corresponding Angles Theorem: line 1, transversal \overleftrightarrow{AC} |
| 4. \angle B \cong \angle C | given |
| 5. \angle ADE \cong \angle C | transitive property of congruence: lines 2 and 4 |
| 6. \angle C \cong \angle AED | symmetric property of congruence: line 3 |
| 7. \angle ADE \cong \angle AED | transitive property of congruence: lines 5 and 6 |
\angle ADE and \angle ABC sit in matching positions where \overleftrightarrow{AB} crosses the two parallels: each is above its line, on the C side of the transversal. That is what makes them corresponding, and line 1 supplies the parallel lines the theorem needs.
On this map the Corresponding Angles Theorem is proved first, from the parallel postulate, and the Alternate Interior Angles Theorem is proved from it. A proof may cite either once it is proved. It may never use alternate interior angles to prove corresponding angles, since that would run the chain in a circle.
Finding the faulty reason
Given. m\angle AOB = 50^\circ; m\angle BOC = 30^\circ.
Prove. m\angle AOC = 80^\circ.
Someone offers this proof. Which reason fails?
| Statements | Reasons |
|---|---|
| 1. m\angle AOB = 50^\circ | given |
| 2. m\angle BOC = 30^\circ | given |
| 3. m\angle AOB + m\angle BOC = m\angle AOC | angle addition postulate |
| 4. 50^\circ + 30^\circ = m\angle AOC | substitution: lines 1 and 2 into line 3 |
| 5. 80^\circ = m\angle AOC | simplify: line 4 |
| 6. m\angle AOC = 80^\circ | symmetric property of equality: line 5 |
Plan. Hold each reason’s if-part up against the lines above it.
Lines 1 and 2 and lines 4 to 6 are sound. Line 3 is not. The angle addition postulate says: if \overrightarrow{OB} lies inside \angle AOC, then the two smaller measures add to the larger. Nothing above line 3 says where \overrightarrow{OB} is, and there is no diagram to read it from.
Figure 6 shows two figures that fit the Given. On the left, \overrightarrow{OB} is inside \angle AOC and m\angle AOC = 50^\circ + 30^\circ = 80^\circ. On the right, \overrightarrow{OC} is inside \angle AOB, so it is \angle AOB that splits, and m\angle AOC = 50^\circ - 30^\circ = 20^\circ. The “theorem” is false as stated, and the faulty reason is exactly where it broke. Adding “\overrightarrow{OB} lies inside \angle AOC” to the Given repairs line 3 and the proof.
Boundaries
A reason must fit the statement completely. Every condition in a definition’s or theorem’s if-part has to stand on an earlier line, or be a position fact in the diagram (Example 6). A true statement with the wrong reason is still a wrong line.
What counts as proved depends on the order. Some courses take the corresponding angles fact as a postulate and others prove it; either works. What never works is using a result in its own proof, or in the proof of anything it rests on.
The diagram gives positions, never sizes. That a point is between two others, or a ray inside an angle, may be read. That lengths or angles are equal, or lines parallel or perpendicular, needs a reason from the list.
Simplify tidies one side only, such as 2(31) + 14 = 76 or x + x = 2x. It never changes both sides of an equation, and it cannot hide a step about the figure.
Where this goes next
The first theorems written this way are about angle pairs, in proofs about angle pairs, starting with proving right angles congruent and proving vertical angles congruent. The same reasons set out as boxes and arrows make a flowchart proof, and set out as sentences they make a paragraph proof, as in Example 1; choosing a proof format compares all four, coordinate proofs included. Indirect proof uses the same reasons but argues from the opposite of the Prove. Congruent triangles bring new reasons such as SAS and ASA, in proving triangles congruent, and parallel lines get their own in proofs about parallel lines.
Exercises
Naming the reason
Give the reason for each step.
- \angle 1 and \angle 2 are complementary, so m\angle 1 + m\angle 2 = 90^\circ.
- \angle P \cong \angle Q, so m\angle P = m\angle Q.
- \overline{AB} \cong \overline{CD} and \overline{CD} \cong \overline{EF}, so \overline{AB} \cong \overline{EF}.
- x + 12 = 40, so x = 28.
- m\angle 1 = 35^\circ and m\angle 1 + m\angle 2 = 90^\circ, so 35^\circ + m\angle 2 = 90^\circ.
- \ell \parallel m, and \angle 2 and \angle 6 are corresponding angles, so \angle 2 \cong \angle 6.
- 2(x + 5) = 3x - 1, so 2x + 10 = 3x - 1.
Solving with a reason for every step
Find the unknown, giving a reason for each step.
- \overrightarrow{BD} bisects \angle ABC, m\angle ABD = (4x + 7)^\circ and m\angle DBC = (7x - 11)^\circ. Find x and m\angle ABD.
- B is between A and C, AB = 3y - 4, BC = 2y + 1 and AC = 37. Find y, AB and BC.
- \angle 1 and \angle 2 are vertical angles, m\angle 1 = (7x - 18)^\circ and m\angle 2 = (3x + 30)^\circ. Find x and m\angle 1.
- \ell \parallel m, and corresponding angles along a transversal measure (5x + 4)^\circ and (3x + 36)^\circ. Find x and the angle.
Finding the faulty reason
Each step has a wrong reason. Say what is wrong.
- \angle PQR and \angle RQS share the vertex Q and the side \overrightarrow{QR}, so they form a linear pair (definition of linear pair).
- AB = BC, so B is the midpoint of \overline{AC} (definition of midpoint).
- \overline{AB} \cong \overline{CD} (reflexive property of congruence).
- From m\angle 1 = m\angle 2 and m\angle 2 + m\angle 3 = 180^\circ, a proof writes m\angle 1 + m\angle 3 = 180^\circ (transitive property of equality).
Reasoning
- Write a two-column proof. Given. \overrightarrow{BD} bisects \angle ABC. Prove. m\angle ABD = \tfrac{1}{2}m\angle ABC.
- Two lines cross, making \angle 1, \angle 2, \angle 3 and \angle 4 in order around the point. Write a two-column proof. Given. \angle 1 \cong \angle 2. Prove. \angle 3 \cong \angle 4.
- Explain why a reason may use lines above it but never lines below it.
- A proof of the Corresponding Angles Theorem cites the Alternate Interior Angles Theorem. On this map, why is that not allowed?
- “\overline{AB} \cong \overline{AB}” seems too obvious to write down. Describe a proof that needs it, and say what it does there.
Answers
- Definition of complementary angles: their measures add to 90^\circ.
- Definition of congruent angles.
- Transitive property of congruence.
- Subtraction property of equality: subtract 12 from both sides.
- Substitution: 35^\circ replaces m\angle 1.
- Corresponding Angles Theorem. Its if-part, \ell \parallel m, is the first clause.
- Distributive property. It multiplies out brackets, which simplify does not cover.
- x = 6 and m\angle ABD = 31^\circ. m\angle ABD = m\angle DBC (definition of angle bisector); 4x + 7 = 7x - 11 (substitution); 18 = 3x (subtraction property, 4x, then addition property, 11); 6 = x (division property); x = 6 (symmetric property). Check: 4(6) + 7 = 31 = 7(6) - 11.
- y = 8, AB = 20 and BC = 17. AB + BC = AC (segment addition postulate); (3y - 4) + (2y + 1) = 37 (substitution); 5y - 3 = 37 (simplify); 5y = 40 (addition property); y = 8 (division property). Check: 20 + 17 = 37.
- x = 12 and m\angle 1 = 66^\circ. \angle 1 \cong \angle 2 (Vertical Angles Theorem); m\angle 1 = m\angle 2 (definition of congruent angles); 7x - 18 = 3x + 30 (substitution); 4x = 48 (subtraction property, 3x, then addition property, 18); x = 12 (division property). Check: 7(12) - 18 = 66 = 3(12) + 30.
- x = 16 and the angle is 84^\circ. The angles are congruent (Corresponding Angles Theorem), so their measures are equal (definition of congruent angles); 5x + 4 = 3x + 36 (substitution); 2x = 32 (subtraction property, 3x, then 4); x = 16 (division property). Check: 5(16) + 4 = 84 = 3(16) + 36.
- A linear pair also needs \overrightarrow{QP} and \overrightarrow{QS} to be opposite rays, so that P, Q and S lie on one line. Sharing a vertex and a side makes the angles only adjacent, and adjacent angles can add to anything.
- The midpoint must lie on \overline{AC}, and nothing says B does. The top corner of a triangle with AB = BC satisfies the equation and is not a midpoint.
- The reflexive property says a segment is congruent to itself. \overline{AB} and \overline{CD} are different segments, so their congruence needs a given or a theorem.
- The transitive property joins a = b and b = c into a = c, and these lines do not have that shape. The step puts m\angle 1 in place of m\angle 2 inside a sum, which is substitution, once the symmetric property has turned the first equation into m\angle 2 = m\angle 1. The statement is right; the reason is not.
- Line (1): \overrightarrow{BD} bisects \angle ABC (given). (2): m\angle ABD = m\angle DBC (definition of angle bisector: line 1). (3): m\angle DBC = m\angle ABD (symmetric property of equality: line 2). (4): m\angle ABD + m\angle DBC = m\angle ABC (angle addition postulate: \overrightarrow{BD} is inside \angle ABC, line 1). (5): m\angle ABD + m\angle ABD = m\angle ABC (substitution: line 3 into line 4). (6): 2m\angle ABD = m\angle ABC (simplify: line 5). (7): m\angle ABD = \tfrac{1}{2}m\angle ABC (division property of equality: line 6).
- Line (1): \angle 1 \cong \angle 2 (given). (2): \angle 1 \cong \angle 3 (Vertical Angles Theorem: in order around the point, \angle 1 is opposite \angle 3). (3): \angle 2 \cong \angle 4 (Vertical Angles Theorem: \angle 2 is opposite \angle 4). (4): \angle 3 \cong \angle 1 (symmetric property of congruence: line 2). (5): \angle 3 \cong \angle 2 (transitive property of congruence: lines 4 and 1). (6): \angle 3 \cong \angle 4 (transitive property of congruence: lines 5 and 3).
- A line below has not been established yet; it may itself depend on the line being justified. Citing it would let a statement support itself through a loop, and a loop proves nothing. Read top to bottom, every line rests only on what is already settled.
- On this map the Alternate Interior Angles Theorem is proved from the Corresponding Angles Theorem. Using it to prove corresponding angles would make each theorem rest on the other, so neither would rest on anything. The corresponding angles fact has to come from the parallel postulate.
- When two triangles share a side, say \triangle ABC and \triangle ABD on either side of \overline{AB}, a congruence proof needs a pair of matching sides, and the shared side supplies one: \overline{AB} \cong \overline{AB} by the reflexive property. The line is obvious, but without it the pair is not on the page for SSS or SAS to use.