Properties of equality and congruence
Everyone can see that x = 5 follows from 2x = 10, and that a side two triangles share matches itself. So why does a proof have to say so, and what gives it the right to?
Where you are 18 in Geometric proof
Why this exists
Solving 2x = 10 is something you can do in your head. A proof will not let you. It wants every line to come with a reason, as deductive reasoning requires, because the reader has to be able to check it without trusting you.
For the algebra in a proof, the reasons are a short, fixed list. They say what you are allowed to do to an equation and still have a true equation: add the same thing to both sides, replace a thing by something equal to it, and a few more. The oldest of them open Euclid’s Elements as “common notions”: things equal to the same thing are equal to each other, and equals added to or taken from equals give equals. They are things so plain that nobody needs to be argued into them. Once they have names, a proof can say which one it is using. That is all a reason is. “Addition property of equality” is not a spell; it is a label for a move that the reader already believes.
Geometry adds a second kind of line, about figures rather than numbers. In Figure 1, \overline{CA} \cong \overline{CB} and D is the midpoint of \overline{AB}. The segment \overline{CD} splits the figure into two triangles, \triangle ACD and \triangle BCD, and it is a side of both.
To show the two triangles congruent by matching three pairs of sides, a proof needs three statements. Two come from the Given. The third is that the side of \triangle ACD called \overline{CD} is congruent to the side of \triangle BCD called \overline{CD}. Pull the triangles apart, as in the lower half of Figure 1, and it becomes clear that this is a real pair: one side in each triangle. It is also clear that the pair matches. But a proof accepts no line on the grounds that it is clear, and the line \overline{CD} \cong \overline{CD} needs a reason too.
Proofs meet two other small moves on figures just as often: turning a congruence round, and passing it along a chain. This lesson names the properties of equality, which license the algebra done on lengths and measures, then the three matching properties of congruence, which license the moves made on the figures themselves, and says why each one is allowed.
Definition
A proof meets two kinds of sameness. Two quantities are equal when they are the same amount: AB = CD says the segments have the same length, and m\angle 1 = m\angle 2 says the angles have the same measure. The sign = joins numbers. Two figures are congruent when they have the same size and shape: congruent segments have the same length and congruent angles have the same measure. The sign \cong joins figures.
The two are tied together by the definition of congruence, which works in both directions:
Each kind of sameness has its own list of properties, and a proof cites the one that matches what its line is about: an equality property for a line with =, a congruence property for a line with \cong.
Properties of equality as reasons
The properties below hold for any numbers a, b, c and d, and so for any lengths and angle measures, which are numbers.
| Property | What it says |
|---|---|
| Reflexive | Anything equals itself: a = a. |
| Symmetric | An equation may be read backwards: if a = b, then b = a. |
| Transitive | Two things equal to the same third thing are equal to each other: if a = b and b = c, then a = c. |
| Addition | Add the same amount to both sides: if a = b, then a + c = b + c. |
| Subtraction | Take the same amount from both sides: if a = b, then a - c = b - c. |
| Multiplication | Multiply both sides by the same amount: if a = b, then ac = bc. |
| Division | Divide both sides by the same amount, which is not zero: if a = b and c \ne 0, then \dfrac{a}{c} = \dfrac{b}{c}. |
| Substitution | If two things are equal, either may stand in for the other, anywhere: if a = b, then b may replace a in any statement about a. |
| Distributive | A multiplier outside brackets multiplies each term inside: a(b + c) = ab + ac. |
The last one is a fact about arithmetic rather than about equations, but proofs cite it in the same breath, so it sits in the same list. In a proof each is written as a reason: “subtraction property of equality”, or just “subtraction property”.
Reflexive, symmetric and transitive congruence
For segments and angles, congruence has three properties. Each is named after the matching property of equality.
| Property | For segments |
|---|---|
| Reflexive | \overline{AB} \cong \overline{AB} |
| Symmetric | if \overline{AB} \cong \overline{CD}, then \overline{CD} \cong \overline{AB} |
| Transitive | if \overline{AB} \cong \overline{CD} and \overline{CD} \cong \overline{EF}, then \overline{AB} \cong \overline{EF} |
| Property | For angles |
|---|---|
| Reflexive | \angle A \cong \angle A |
| Symmetric | if \angle A \cong \angle B, then \angle B \cong \angle A |
| Transitive | if \angle A \cong \angle B and \angle B \cong \angle C, then \angle A \cong \angle C |
In words: everything is congruent to itself; a congruence may be read in either direction; and two congruences that share a middle term join into one.
In a proof each is written as a reason, “reflexive property of congruence” and so on. The reflexive property is the one that justifies a shared side or a shared angle.
Intuition
Equality: a balance
Picture a balanced pair of scales. Anything you do to both pans alike leaves them balanced: add the same weight, remove the same weight, double both, halve both. That is the addition, subtraction, multiplication and division properties. The properties of equality met when solving equations are this picture, turned into rules.
Substitution has a different picture. “a = b” says that a and b are two names for one amount. If a sentence is true about the amount under one name, it is true under the other. So you may swap the names wherever you like.
The reflexive, symmetric and transitive properties are what make “equal” deserve the word: equality does not care which side you read first, and it chains.
Congruence: “has the same number as”
Congruent segments are segments with the same length, and congruent angles are angles with the same measure. Every property of congruence is a fact about “has the same number as”.
Anything has the same number as itself. “Same as” has no direction, so it reads both ways. And if the first matches the second and the second matches the third, all three carry one number, so the first matches the third.
Compare “is longer than”. It passes along a chain (if AB > CD and CD > EF, then AB > EF), but no segment is longer than itself, and it never reads both ways. So the three properties are not automatic. Congruence has them because it is defined by sameness.
Tick marks quietly rely on this. When three segments all carry one tick, the figure has already assumed that any two of them are congruent, whichever two were given.
Results
Equality in a proof
Every line of algebra in a proof is licensed by a property of equality or by plain arithmetic, and a line may need two of them, as line 5 of Example 3 does. Two consequences are used so often that they get quoted directly.
Equals added to equals. If a = b and c = d, then a + c = b + d.
Equals substituted into an equation. If a = b and the equation a + c = 7 is true, then b + c = 7 is true.
In geometry the quantities are lengths and measures, so the same properties read like this:
Congruence in a proof
Reflexive, symmetric and transitive properties of congruence. For all segments \overline{AB}, \overline{CD}, \overline{EF} and all angles \angle A, \angle B, \angle C, the six statements in the two congruence tables hold.
Congruent to the same figure. If two segments, or two angles, are each congruent to a third, they are congruent to each other.
A relation with all three properties is an equivalence relation. Equality of numbers is the first example. Congruence of segments sorts every segment into groups by length, one group for each length, and congruence of angles does the same by measure.
Why it is true
Why the equality properties hold
The eight equality properties come from one fact: an equation says two expressions name one number. The distributive property is different, a fact about how multiplication and addition interact, and is not argued here.
Reflexive and symmetric are the meaning of “the same number”, read two ways. For the addition property, suppose a = b. Then a and b are one number, and adding c to one number gives one answer, whichever name you used for it. So a + c = b + c. Subtraction, multiplication and division are the same argument with a different operation; division needs c \ne 0 only because dividing by zero gives no answer at all.
Transitive is substitution in disguise. If a = b and b = c, take the statement a = b and replace b by c, which is allowed because b = c. The statement becomes a = c.
None of this depends on what the numbers measure. A length, an angle, a price and a temperature obey the same rules, which is why the same short list serves the whole of a proof.
Why congruence inherits three of them
The definition of congruence works in both directions: \overline{AB} \cong \overline{CD} exactly when AB = CD. So each property of congruence can be carried over to the numbers, settled by the matching property of equality, and carried back.
Here is the transitive property for segments.
| Statements | Reasons |
|---|---|
| 1. \overline{AB} \cong \overline{CD} | given |
| 2. \overline{CD} \cong \overline{EF} | given |
| 3. AB = CD | definition of congruent segments: line 1 |
| 4. CD = EF | definition of congruent segments: line 2 |
| 5. AB = EF | transitive property of equality: lines 3 and 4 |
| 6. \overline{AB} \cong \overline{EF} | definition of congruent segments: line 5 |
The reflexive and symmetric properties go the same way, through AB = AB and through “if AB = CD then CD = AB”. For angles, replace each length by a measure, such as m\angle 1.
Only these three carry over as properties of \cong itself. The rest of the equality table is about arithmetic (adding, subtracting, multiplying and dividing both sides, and distributing), and \cong joins figures, which are not added or multiplied. So there is no “addition property of congruence” to cite. Substitution does not carry over either: \overline{AB} \cong \overline{CD} does not let \overline{CD} stand in for \overline{AB} in every statement, as the last of the Boundaries shows.
What looks like adding congruent segments is still true; it is just reached through the numbers. With A, B, C, D in order on a line and \overline{AB} \cong \overline{CD}, the conclusion \overline{AC} \cong \overline{BD} holds: cross over to AB = CD by the definition of congruent segments, add BC by the addition property of equality, use the segment addition postulate to reach AC = BD (this is Example 4), and cross back by the definition. Angles go the same way through the angle addition postulate (Example 5). The proof cites the definition at each crossing and the addition property of equality in the middle, never an addition property of congruence.
Some courses define congruence by rigid motions instead: one figure is congruent to another when a sequence of slides, turns and flips carries it onto the other. The three properties still hold. Doing nothing carries a figure onto itself (reflexive). Running the motions backwards, in reverse order, carries the second figure back onto the first (symmetric). Doing one sequence and then the other carries the first figure onto the third (transitive).
The result about figures congruent to the same figure uses two properties. From \angle 2 \cong \angle 3, the symmetric property gives \angle 3 \cong \angle 2; then \angle 1 \cong \angle 3 and \angle 3 \cong \angle 2 join by the transitive property.
Worked examples
Solving an equation, with a reason for each line
Solve 3x + 5 = 20, giving the property that allows each line.
Solution| Statements | Reasons |
|---|---|
| 1. 3x + 5 = 20 | given |
| 2. 3x + 5 - 5 = 20 - 5 | subtraction property: take 5 from both sides of line 1 |
| 3. 3x = 15 | arithmetic: 5 - 5 = 0 and 20 - 5 = 15 |
| 4. \dfrac{3x}{3} = \dfrac{15}{3} | division property: divide both sides of line 3 by 3, which is not zero |
| 5. x = 5 | arithmetic |
The subtraction on line 2 is chosen because it removes the 5 from the left side. Nothing in the properties tells you which move is wise; they tell you which moves are allowed.
Check. Put x = 5 into line 1: 3(5) + 5 = 20, and it is.
A distributive step, and a variable on both sides
Solve 2(x - 3) = x + 9, giving the property that allows each line.
Solution| Statements | Reasons |
|---|---|
| 1. 2(x - 3) = x + 9 | given |
| 2. 2x - 6 = x + 9 | distributive property |
| 3. 2x - 6 - x = x + 9 - x | subtraction property: take x from both sides of line 2 |
| 4. x - 6 = 9 | arithmetic |
| 5. x - 6 + 6 = 9 + 6 | addition property: add 6 to both sides of line 4 |
| 6. x = 15 | arithmetic |
Line 2 is not the equality properties at work. The two sides of line 1 are already equal, and line 2 only rewrites the left side as another expression for the same number. That is the distributive property, which is why it is named separately.
Check. Line 1 with x = 15: 2(15 - 3) = 24 and 15 + 9 = 24.
A midpoint with algebra in it
M is the midpoint of \overline{AB}, with AM = 3x + 4 and MB = 5x - 10. Find x and AB.
Solution| Statements | Reasons |
|---|---|
| 1. AM = MB | definition of midpoint |
| 2. 3x + 4 = 5x - 10 | substitution: replace AM and MB by the given expressions |
| 3. 3x + 14 = 5x | addition property: add 10 to both sides of line 2 |
| 4. 14 = 2x | subtraction property: take 3x from both sides of line 3 |
| 5. x = 7 | division property: divide both sides of line 4 by 2, then symmetric property |
| 6. AM = 3(7) + 4 = 25 | substitution: replace x by 7 |
| 7. AB = AM + MB = 25 + 25 = 50 | segment addition postulate, and line 1 |
Line 2 is substitution used on a whole expression: the given says what AM is, and the definition of a midpoint says what it is equal to. Line 5 needs two properties: division gives 7 = x, and the symmetric property turns it round to x = 7.
Check. MB = 5(7) - 10 = 25, the same as AM.
Adding the same segment to equal segments
Points A, B, C and D lie on a line in that order, as in Figure 2, and AB = CD. Prove that AC = BD.
The plan: AC is AB with BC tacked on, and BD is CD with BC tacked on. Adding the same piece to two equal lengths should keep them equal.
| Statements | Reasons |
|---|---|
| 1. AB = CD | given |
| 2. AB + BC = CD + BC | addition property: add BC to both sides of line 1 |
| 3. AC = AB + BC | segment addition postulate: B is between A and C |
| 4. AC = CD + BC | substitution: in line 3, replace AB + BC by CD + BC, which line 2 says is equal |
| 5. BD = BC + CD | segment addition postulate: C is between B and D |
| 6. BD = CD + BC | commutative property of addition: the order of a sum does not matter |
| 7. AC = BD | substitution: in line 4, replace CD + BC by BD, which line 6 says is equal |
Line 3 comes from the segment addition postulate, which is about points on a line and is not one of the equality properties. The equality properties join the geometric facts to each other.
Check. Take AB = CD = 7 and BC = 4. Then AC = 11 and BD = 11.
The same argument for angles
Rays OA, OB, OC and OD leave O in that order, as in Figure 3, and m\angle AOB = m\angle COD. Prove that m\angle AOC = m\angle BOD.
This has the same shape as the last example, with the angle addition postulate where the segment addition postulate was.
| Statements | Reasons |
|---|---|
| 1. m\angle AOB = m\angle COD | given |
| 2. m\angle AOB + m\angle BOC = m\angle COD + m\angle BOC | addition property: add m\angle BOC to both sides of line 1 |
| 3. m\angle AOC = m\angle AOB + m\angle BOC | angle addition postulate: ray OB is inside \angle AOC |
| 4. m\angle BOD = m\angle BOC + m\angle COD | angle addition postulate: ray OC is inside \angle BOD |
| 5. m\angle BOC + m\angle COD = m\angle COD + m\angle BOC | commutative property of addition |
| 6. m\angle AOC = m\angle COD + m\angle BOC | substitution: in line 3, replace the sum by its equal from line 2 |
| 7. m\angle BOD = m\angle COD + m\angle BOC | substitution: in line 4, replace m\angle BOC + m\angle COD by its equal from line 5 |
| 8. m\angle AOC = m\angle BOD | substitution: lines 6 and 7 give the two measures the same right-hand side |
Check. Take m\angle AOB = m\angle COD = 40^\circ and m\angle BOC = 50^\circ. Then m\angle AOC = 90^\circ and m\angle BOD = 90^\circ.
Halves of equal segments are equal
In Figure 4, AB = CD, M is the midpoint of \overline{AB} and N is the midpoint of \overline{CD}. Prove that AM = CN.
The idea: each segment is two of its halves, so each half is half its segment, and halves of equal things match. Multiplication and division are what carry that across.
| Statements | Reasons |
|---|---|
| 1. AM = MB | definition of midpoint, for M |
| 2. AB = AM + MB | segment addition postulate |
| 3. AB = AM + AM, so AB = 2\cdot AM | substitution: replace MB by AM in line 2, using line 1; then arithmetic |
| 4. CD = 2\cdot CN | the same three steps for N and \overline{CD} |
| 5. AB = CD | given |
| 6. 2\cdot AM = 2\cdot CN | substitution: replace AB and CD in line 5 by their equals from lines 3 and 4 |
| 7. AM = CN | division property: divide both sides of line 6 by 2 |
This is the shape of many later facts: name a length two ways, then let equality do the moving.
Equal sums force equal parts
\angle 1 and \angle 2 are supplementary, and so are \angle 3 and \angle 2. Prove that m\angle 1 = m\angle 3.
Solution| Statements | Reasons |
|---|---|
| 1. m\angle 1 + m\angle 2 = 180^\circ | given: supplementary angles add to 180^\circ |
| 2. m\angle 3 + m\angle 2 = 180^\circ | given, in the same way |
| 3. m\angle 1 + m\angle 2 = m\angle 3 + m\angle 2 | substitution: in line 1, replace 180^\circ by the sum that line 2 says equals it |
| 4. m\angle 1 = m\angle 3 | subtraction property: take m\angle 2 from both sides of line 3 |
It is worth seeing how short this is. It is the whole of the argument in the vertical angles proof, and it shows up again in proving supplements of congruent angles congruent. The skill is in spotting that two sums equal the same number, so they equal each other.
Which property, of which relation?
Name the property that justifies each statement.
(a) \angle PQR \cong \angle PQR
(b) If \overline{JK} \cong \overline{LM}, then \overline{LM} \cong \overline{JK}.
(c) m\angle PQR = m\angle PQR
(d) If \overline{ST} \cong \overline{UV} and \overline{UV} \cong \overline{WX}, then \overline{ST} \cong \overline{WX}.
Solution(a) The reflexive property of congruence: an angle is congruent to itself.
(b) The symmetric property of congruence: the same congruence, read the other way.
(c) The reflexive property of equality. m\angle PQR is a number, and = joins numbers. It looks like (a), but (a) is about a figure and (c) is about its measure, and a careful proof keeps the two apart: the equality table for lines with =, the congruence tables for lines with \cong.
(d) The transitive property of congruence: \overline{UV} is the shared middle term.
A shared side
In Figure 1, \overline{CA} \cong \overline{CB} and D is the midpoint of \overline{AB}. List three pairs of congruent sides of \triangle ACD and \triangle BCD, each with its reason.
SolutionMatch the vertices A \leftrightarrow B, C \leftrightarrow C, D \leftrightarrow D, and pair the sides in that order.
| Pair | Reason |
|---|---|
| \overline{CA} \cong \overline{CB} | given |
| \overline{AD} \cong \overline{BD} | definition of midpoint |
| \overline{CD} \cong \overline{CD} | reflexive property of congruence |
The third line is the one a first attempt leaves out, and it is the whole point here: the reflexive property is what lets a single segment supply a pair. Three pairs of congruent sides is exactly what the SSS criterion, one of the congruence criteria still to come, needs in order to conclude \triangle ACD \cong \triangle BCD. Without the third line the proof would have only two pairs to offer it.
A shared angle
In Figure 5, D lies on \overline{AB} and E lies on \overline{AC}, with \overline{AD} \cong \overline{AE} and \angle ADC \cong \angle AEB. List three pairs of congruent parts of \triangle ADC and \triangle AEB, each with its reason.
The triangles overlap, which hides the fact that they share an angle. Trace each one: \triangle ADC runs up to D and across to C, and \triangle AEB runs down to E and across to B. Both have their corner at A between ray \overrightarrow{AB} and ray \overrightarrow{AC}.
In \triangle ADC that angle is called \angle DAC, and in \triangle AEB it is called \angle EAB. The names differ, but D is on ray \overrightarrow{AB} and E is on ray \overrightarrow{AC}, so both names describe the same two rays. It is one angle.
| Pair | Reason |
|---|---|
| \angle DAC \cong \angle EAB | reflexive property of congruence (one angle, two names) |
| \overline{AD} \cong \overline{AE} | given |
| \angle ADC \cong \angle AEB | given |
In each triangle that is two angles and the side between them. That pattern is the ASA criterion, another of the congruence criteria still to come, which will turn these three lines into \triangle ADC \cong \triangle AEB. The first line is the one only the reflexive property can supply. Separating overlapping triangles like this is the work of proofs with overlapping triangles.
Two midpoints in a row
A, B, C and D lie on a line in that order. B is the midpoint of \overline{AC} and C is the midpoint of \overline{BD}. Show that \overline{AB} \cong \overline{CD}. Then, given AB = 2x + 5 and CD = 4x - 3, find x and AD.
SolutionThe two midpoints give two congruences with a shared middle term:
each by the definition of midpoint. By the transitive property of congruence, \overline{AB} \cong \overline{CD}, even though no single given mentions both segments.
Congruent segments have equal lengths, so the question passes from \cong to =, and the equality properties take over:
The second line adds 3 to both sides and takes 2x from both sides (addition and subtraction properties). The third divides both sides by 2 (division property), which gives 4 = x, and then reads it the other way round (symmetric property of equality). Then AB = 2(4) + 5 = 13 and CD = 4(4) - 3 = 13. BC is 13 as well, so AD = 13 + 13 + 13 = 39.
Congruent to the same angle
\angle P \cong \angle R and \angle Q \cong \angle R, with m\angle P = (2x + 14)^\circ and m\angle Q = (5x - 19)^\circ. Show that \angle P \cong \angle Q, then find x and m\angle R.
SolutionThe transitive property needs the middle term to sit in the middle: \angle P \cong \angle R and \angle R \cong \angle Q. The second given reads the wrong way round, so turn it first.
| Statements | Reasons |
|---|---|
| 1. \angle Q \cong \angle R | given |
| 2. \angle R \cong \angle Q | symmetric property of congruence: line 1 |
| 3. \angle P \cong \angle R | given |
| 4. \angle P \cong \angle Q | transitive property of congruence: lines 3 and 2 |
Now the measures are equal, by the definition of congruent angles: 2x + 14 = 5x - 19, so 33 = 3x and x = 11. Then m\angle P = 2(11) + 14 = 36^\circ and m\angle Q = 5(11) - 19 = 36^\circ, and m\angle R = 36^\circ too.
This is Euclid’s first Common Notion: “things which equal the same thing also equal one another”. The two-column form just shows the two properties hidden inside it.
When a relation fails the test
Decide whether “is perpendicular to”, for lines in a plane, is reflexive, symmetric and transitive.
Reflexive? No. A line lies along itself; it does not cross itself at a right angle. No line is perpendicular to itself.
Symmetric? Yes. \ell \perp m says that \ell and m meet and form a right angle. That sentence says the same thing about both lines and does not care which is named first, so whenever \ell \perp m, also m \perp \ell.
Transitive? No, and one counterexample settles it. In Figure 6 all three lines lie in one plane, with \ell \perp m and m \perp n. But \ell and n make equal corresponding angles with m, both right angles, so they are parallel: they never meet, so they form no angle at all, right or otherwise. An even quicker counterexample needs no figure. Take n to be \ell itself: \ell \perp m and, by symmetry, m \perp \ell, yet \ell is not perpendicular to \ell.
The plane matters to the picture, not to the verdict. In space, two lines each perpendicular to m need not be parallel. They can be skew, or even perpendicular to each other, like three edges meeting at one corner of a box. So in space the chain sometimes holds and sometimes does not, and “sometimes” is still a failure: the transitive property asks for “always”.
So “perpendicular” cannot be used the way congruence is. From \ell \perp m and m \perp n a proof may not write \ell \perp n. The three properties are exactly what lets congruence be passed from figure to figure, and a relation missing one of them cannot be.
“Is parallel to” is a closer call. It is symmetric, and for different lines in a plane it passes along a chain. Whether a line is parallel to itself depends on the definition. If parallel means “never meet”, then no line is parallel to itself, and the chain breaks in one case: \ell \parallel m and m \parallel \ell would give \ell \parallel \ell. Books that want a full equivalence relation count every line as parallel to itself.
Boundaries
Equality properties act on numbers, congruence properties on figures. AB = CD says two lengths are the same number; it does not say the segments are one segment. A line with = cites a property of equality, and a line with \cong cites a property of congruence. The definition of congruence is the only bridge between them, and crossing it is a step of its own, with its own reason.
Both sides must get the same thing. The addition property adds the same c to both sides. If a = b and c \ne d, then a + c \ne b + d, and no property allows the step. A line that adds c to one side and d to the other is allowed only when c = d is already known, and then its reason is “equals added to equals”, not the addition property.
Division needs a nonzero divisor. The division property says c \ne 0 for a reason. From 0 \cdot 3 = 0 \cdot 5 it would give 3 = 5, which is false. Multiplying by zero is allowed, because it keeps the equation true, but it throws the information away.
A step promises truth forwards, not backwards. Each property takes a true equation to a true equation. Whether the step can be undone depends on the move: adding c is undone by subtracting c, but multiplying both sides by 0 cannot be undone, and neither can multiplying both sides by x. From x = 3, multiplying by x and substituting gives x^2 = 9, which is true; but x^2 = 9 does not give x = 3, since x = -3 also works.
Substitution needs a known equality. Two lengths that look equal in a figure are not equal for the purposes of a proof. Only what is given, or already proved, can be substituted.
The three properties hold because congruence means “same measure”. A relation built from sameness (same length, same measure, same midpoint) has all three. A relation built from anything else must be tested, and “perpendicular”, “longer than” and “supplementary” each fail at least one.
Looking congruent is not a relation with these properties. “Within 1 mm of each other” is reflexive and symmetric, but not transitive: segments of 10, 10.8 and 11.6 mm link in a chain, yet the ends differ by 1.6 mm. A chain of near matches can drift, which is one more reason a proof never reads congruence off a ruler.
Reflexive means one object, not one letter. \angle A in two triangles is shared only when both triangles use the same two rays at A. If the triangles meet at A with their angles opposite each other, those angles are vertical angles: congruent, but by a theorem, not by the reflexive property.
Congruent is not identical. \overline{AB} \cong \overline{CD} does not make them one segment, so a fact about the position of one need not hold for the other. If M is the midpoint of \overline{AB}, it need not be the midpoint of \overline{CD}, or even lie on it.
Where this goes next
The two-column proof is the Statements and Reasons table of this lesson, used on every line: statements down the left, a reason from these lists or from geometry on the right. The properties of equality and congruence are among the reasons it writes most often, beside definitions and theorems.
Proofs about angle pairs, such as proving vertical angles congruent, are built almost entirely from Example 7, equal sums forcing equal parts.
The reflexive property of congruence matters most once triangles arrive. A shared side or angle is often the free third pair that proving triangles congruent needs, most of all in overlapping triangles. Congruence of triangles is itself reflexive, symmetric and transitive, and so is similarity.
The pattern reaches far beyond geometry. Every equivalence relation sorts things into classes of things that count as the same, and equality of numbers and congruence of segments, sorted by length, are among the first.
Exercises
Naming the reason
Each question gives a line and the line that follows it. Name the property of equality that allows the step.
- From x + 7 = 15 to x = 8.
- From \dfrac{x}{4} = 6 to x = 24.
- From 3(x + 2) = 18 to 3x + 6 = 18.
- From AB = CD and CD = EF to AB = EF.
- From m\angle 1 = m\angle 2 and m\angle 2 = 40^\circ to m\angle 1 = 40^\circ.
Solving, with a reason for every line
Solve each equation. Write it as two columns, Statements and Reasons, with a property of equality or a fact of arithmetic beside every line.
- 2x + 9 = 31
- 5x - 4 = 3x + 10
- 4(x - 2) = 2x + 6
- \dfrac{x}{3} + 5 = 11
- 3(2x + 1) = 5x + 11
- -2x + 7 = 1
Segments and angles
- Points A, B, C and D lie on a line in that order, and AC = BD. Prove that AB = CD.
- M is the midpoint of \overline{AB}, with AM = 2x + 3 and MB = 4x - 9. Find x and AB.
- B is between A and C, with AB = 3x + 1, BC = 2x + 5 and AC = 41. Find x, AB and BC.
- Ray OB lies inside \angle AOC, with m\angle AOB = (2x + 5)^\circ, m\angle BOC = (3x - 10)^\circ and m\angle AOC = 80^\circ. Find x and the two smaller angles.
- \angle 1 and \angle 2 are supplementary, with m\angle 1 = (4x + 5)^\circ and m\angle 2 = (2x + 19)^\circ. Find x and both angles.
- \angle A and \angle B are complementary, with m\angle A = (x + 10)^\circ and m\angle B = (2x - 4)^\circ. Find x and both angles.
Naming a property of congruence
Name the property that justifies each statement.
- \overline{RS} \cong \overline{RS}
- If \angle JKL \cong \angle MNP, then \angle MNP \cong \angle JKL.
- If \overline{XY} \cong \overline{YZ} and \overline{YZ} \cong \overline{ZW}, then \overline{XY} \cong \overline{ZW}.
- QR = QR
- If \angle 1 \cong \angle 2 and \angle 3 \cong \angle 2, then \angle 1 \cong \angle 3. Name both properties used.
Shared parts
- In quadrilateral ABCD, the diagonal \overline{AC} is drawn, \overline{AB} \cong \overline{AD} and \overline{CB} \cong \overline{CD}. Name the third pair of congruent sides of \triangle ABC and \triangle ADC, and its reason.
- In parallelogram ABCD, the diagonal \overline{BD} is drawn. With A \leftrightarrow C, B \leftrightarrow D and D \leftrightarrow B, write the congruence of the shared side in matching order, and name its reason.
- Segments \overline{BD} and \overline{CE} cross at A, making \triangle ABC and \triangle ADE. Why is \angle BAC \cong \angle DAE not justified by the reflexive property, and what does justify it?
- Give four different three-letter names for the shaded angle in Figure 5, and say why they all name one angle.
Chains with numbers
- \overline{AB} \cong \overline{CD} and \overline{EF} \cong \overline{CD}, with AB = 3x + 4 and EF = 5x - 6. Find x and CD.
- P, Q, R and S lie on a line in that order. Q is the midpoint of \overline{PR}, R is the midpoint of \overline{QS}, and PQ = 7. Find PS.
- \angle 1 and \angle 2 are vertical angles, and \angle 2 \cong \angle 3. Given m\angle 1 = (5y - 12)^\circ and m\angle 3 = (3y + 20)^\circ, find y and m\angle 2.
- In Exercise 23, AB = 2x + 1, AD = 3x - 4, CB = x + 9 and CD = 4x - 6. Find x and the perimeter of ABCD.
Testing a relation
For each relation, say whether it is reflexive, symmetric and transitive, with a reason or a counterexample for each.
- “is supplementary to”, for angles
- “is longer than”, for segments
- “shares an endpoint with”, for segments
- “has the same midpoint as”, for segments
Reasoning
- Given m\angle 1 = m\angle 2 and m\angle 3 = m\angle 4, prove that m\angle 1 + m\angle 3 = m\angle 2 + m\angle 4. Use only the properties of equality.
- The list of equality properties has a subtraction property and an addition property. Explain why the list would be complete without the subtraction property.
- Does x^2 = 9 follow from x = 3 by a property of equality? Does x = 3 follow from x^2 = 9? Explain both answers.
- Someone writes “0 \cdot 3 = 0 \cdot 5, so 3 = 5” and cites the division property. Say why the citation fails, and what the division property would have needed.
- \angle 1 and \angle 3 are supplementary, \angle 2 and \angle 4 are supplementary, and m\angle 1 = m\angle 2. Prove that m\angle 3 = m\angle 4.
- Prove the symmetric property of congruence for angles, giving a reason for each step.
- Both triangles in Figure 1 plainly contain \overline{CD}. Explain why a proof still writes \overline{CD} \cong \overline{CD} as its own line.
- From \angle 1 \cong \angle 2, \angle 2 \cong \angle 3 and \angle 3 \cong \angle 4, a proof writes \angle 1 \cong \angle 4 in one line, “by the transitive property”. Explain why the line is true, and how many uses of the transitive property it really contains.
Answers
- Subtraction property: take 7 from both sides.
- Multiplication property: multiply both sides by 4.
- Distributive property: 3(x + 2) = 3x + 6. It rewrites the left side as an equal expression.
- Transitive property of equality: both AB and EF equal CD.
- Substitution (the transitive property also works): replace m\angle 2 by 40^\circ in the first equation.
- x = 11. Subtract 9 from both sides (subtraction property) to get 2x = 22. Divide both sides by 2 (division property). Check: 2(11) + 9 = 31.
- x = 7. Subtract 3x from both sides (subtraction property): 2x - 4 = 10. Add 4 to both sides (addition property): 2x = 14. Divide by 2 (division property). Check: 5(7) - 4 = 31 and 3(7) + 10 = 31.
- x = 7. Distributive property: 4x - 8 = 2x + 6. Subtract 2x (subtraction property): 2x - 8 = 6. Add 8 (addition property): 2x = 14. Divide by 2 (division property). Check: 4(7 - 2) = 20 and 2(7) + 6 = 20.
- x = 18. Subtract 5 from both sides (subtraction property): \dfrac{x}{3} = 6. Multiply both sides by 3 (multiplication property). Check: \dfrac{18}{3} + 5 = 11.
- x = 8. Distributive property: 6x + 3 = 5x + 11. Subtract 5x (subtraction property): x + 3 = 11. Subtract 3 (subtraction property): x = 8. Check: 3(17) = 51 and 5(8) + 11 = 51.
- x = 3. Subtract 7 from both sides (subtraction property): -2x = -6. Divide both sides by -2 (division property), which is not zero. Check: -2(3) + 7 = 1.
- Line by line: AC = AB + BC and BD = BC + CD (segment addition postulate). The given AC = BD and substitution give AB + BC = BC + CD. Subtract BC from both sides (subtraction property): AB = CD.
- x = 6 and AB = 30. The definition of a midpoint gives AM = MB, so 2x + 3 = 4x - 9 (substitution). Then 12 = 2x by the addition and subtraction properties, and x = 6 by division. AM = 2(6) + 3 = 15 = MB, so AB = AM + MB = 30 by the segment addition postulate.
- x = 7, AB = 22 and BC = 19. The segment addition postulate gives AB + BC = AC, so (3x + 1) + (2x + 5) = 41 by substitution. Then 5x + 6 = 41, 5x = 35 by the subtraction property, and x = 7 by division. Check: 22 + 19 = 41.
- x = 17; the angles are 39^\circ and 41^\circ. The angle addition postulate gives m\angle AOB + m\angle BOC = m\angle AOC, so (2x + 5) + (3x - 10) = 80 by substitution. Then 5x - 5 = 80, 5x = 85 by the addition property, and x = 17 by division. Check: 39 + 41 = 80.
- x = 26; the angles are 109^\circ and 71^\circ. Supplementary angles add to 180^\circ, so (4x + 5) + (2x + 19) = 180 by substitution. Then 6x + 24 = 180, 6x = 156 by the subtraction property, and x = 26 by division. Check: 109 + 71 = 180.
- x = 28; the angles are 38^\circ and 52^\circ. Complementary angles add to 90^\circ, so (x + 10) + (2x - 4) = 90 by substitution. Then 3x + 6 = 90, 3x = 84 by the subtraction property, and x = 28 by division. Check: 38 + 52 = 90.
- The reflexive property of congruence.
- The symmetric property of congruence.
- The transitive property of congruence, with \overline{YZ} as the middle term.
- The reflexive property of equality: QR is a length, a number.
- Symmetric, to turn \angle 3 \cong \angle 2 into \angle 2 \cong \angle 3, then transitive.
- \overline{AC} \cong \overline{AC}, by the reflexive property of congruence.
- \overline{BD} \cong \overline{DB}, by the reflexive property. \overline{BD} and \overline{DB} are one segment, written in the order the correspondence asks for.
- The two angles are made of different rays: \overrightarrow{AB} and \overrightarrow{AD} point opposite ways, as do \overrightarrow{AC} and \overrightarrow{AE}. They are vertical angles, congruent by the Vertical Angles Theorem.
- \angle DAE, \angle DAC, \angle BAE and \angle BAC (or any of these with the outer letters swapped). D and B lie on one ray from A, and E and C on the other, so every name picks out the same two rays.
- x = 5 and CD = 19. Symmetric then transitive give \overline{AB} \cong \overline{EF}, so 3x + 4 = 5x - 6 and 2x = 10. Then AB = 19, and \overline{CD} has the same length.
- PS = 21. \overline{PQ} \cong \overline{QR} and \overline{QR} \cong \overline{RS} (midpoints), so each is 7, and PS = 7 + 7 + 7.
- y = 16 and m\angle 2 = 68^\circ. \angle 1 \cong \angle 2 (vertical angles), so \angle 1 \cong \angle 3 (transitive): 5y - 12 = 3y + 20, 2y = 32. Then m\angle 1 = 5(16) - 12 = 68^\circ, and m\angle 3 = 3(16) + 20 = 68^\circ as well.
- x = 5 and the perimeter is 50. From 2x + 1 = 3x - 4, x = 5, so AB = AD = 11. Then CB = 14 and CD = 4(5) - 6 = 14, and 11 + 11 + 14 + 14 = 50.
- Not reflexive: a 50^\circ angle is not supplementary to itself (only a 90^\circ angle is). Symmetric: the two measures add to 180^\circ in either order. Not transitive: with m\angle 1 = 50^\circ, m\angle 2 = 130^\circ and m\angle 3 = 50^\circ, \angle 1 and \angle 2 are supplementary and so are \angle 2 and \angle 3, but \angle 1 and \angle 3 add to 100^\circ.
- Not reflexive: no segment is longer than itself. Not symmetric: if AB > CD then CD > AB is false. Transitive: AB > CD and CD > EF give AB > EF.
- Reflexive: a segment shares both endpoints with itself. Symmetric: a shared endpoint is shared both ways. Not transitive: \overline{AB} and \overline{BC} share B, \overline{BC} and \overline{CD} share C, but \overline{AB} and \overline{CD} may share nothing.
- All three, since it is defined by sameness: every segment has the same midpoint as itself, a shared midpoint is shared both ways, and if \overline{AB} and \overline{CD} have one midpoint and \overline{CD} and \overline{EF} have one midpoint, it is the same point all three times.
- Add m\angle 3 to both sides of m\angle 1 = m\angle 2 (addition property): m\angle 1 + m\angle 3 = m\angle 2 + m\angle 3. Then m\angle 3 = m\angle 4 lets you replace m\angle 3 on the right by m\angle 4 (substitution), which gives m\angle 1 + m\angle 3 = m\angle 2 + m\angle 4.
- Taking c from both sides is the same as adding -c to both sides, because a - c = a + (-c). So the addition property already covers it. The subtraction property is listed because it is what proofs quote, and because it is easier to read.
- Yes to the first: multiply both sides of x = 3 by x (multiplication property) to get x \cdot x = 3x, then replace x by 3 on the right (substitution): x^2 = 9. No to the second: x^2 = 9 is also true when x = -3. Each property moves from a true equation to a true equation; none of them promises to work backwards.
- The division property needs a divisor that is not zero, and here the divisor is 0. Dividing 0 by 0 gives no answer, so neither side is a number. The equation 3 = 5 is false, which shows the step is not allowed. A legitimate use of the property would divide both sides by a nonzero amount such as 2.
- Line by line: m\angle 1 + m\angle 3 = 180^\circ and m\angle 2 + m\angle 4 = 180^\circ (supplementary). Substitution gives m\angle 1 + m\angle 3 = m\angle 2 + m\angle 4. Replace m\angle 2 by m\angle 1 (given, substitution): m\angle 1 + m\angle 3 = m\angle 1 + m\angle 4. Subtract m\angle 1 from both sides (subtraction property): m\angle 3 = m\angle 4.
- Given \angle 1 \cong \angle 2. Then m\angle 1 = m\angle 2 (definition of congruent angles), so m\angle 2 = m\angle 1 (symmetric property of equality), so \angle 2 \cong \angle 1 (definition of congruent angles).
- A congruence criterion asks for pairs of parts, one from each triangle, and every line of a proof needs a reason. \overline{CD} as a side of \triangle ACD and as a side of \triangle BCD make one of those pairs, and the reflexive property is its reason. Leave it out and the proof has only two pairs.
- Two. The transitive property joins exactly two congruences that share a middle term. \angle 1 \cong \angle 2 and \angle 2 \cong \angle 3 give \angle 1 \cong \angle 3; then \angle 1 \cong \angle 3 and \angle 3 \cong \angle 4 give \angle 1 \cong \angle 4. A chain of k congruences needs k - 1 uses, and the one-line version is shorthand for them.