Two-Column Proofs About Segments | Geometry | LevelKite

You've written two-column algebraic proofs, where each line of algebra has a reason. Now you'll use the same form for segments. A two-column proof lists statements in one column and, beside each one, the reason it is true.

Every proof runs from the Given, the facts you may assume, to the Prove, the claim you must reach. Each reason is the exact name of a definition, postulate, property or previously proved theorem, or the word Given. A statement can't be justified by its own conclusion or by how a picture looks. Figures in this lesson are not drawn to scale, so only the marks on a figure and the facts stated in the text count as givens.

Two postulates bring numbers into a proof about segments.

Ruler Postulate: the points on a line can be matched one-to-one with the real numbers, so each point gets exactly one number and each number gets exactly one point. The distance between two points is the absolute value of the difference of their numbers. A point's number is its coordinate. So if $A$ has coordinate $a$ and $C$ has coordinate $c$ , the length of the segment joining them is $AC = \lvert c - a \rvert$ .

Segment Addition Postulate: if $B$ is between $A$ and $C$ , then $AB + BC = AC$ . Between means that the three points lie on one line and $B$ lies on the part from $A$ to $C$ , not at either end.

Notice the direction this lesson uses. The fact that $B$ is between $A$ and $C$ , a betweenness fact, comes first, and the equation follows from it. So your proof must state that fact before you write the sum. A drawing can't supply it.

Not to scale. The parts

AB

and

BC

make up

AC

. The text, not the drawing, says that

B

is between

A

and

C

and gives coordinates

1

for

A

and

6

for

C

.

Worked example

Use both postulates in one proof

Given: on a number line, $A$ has coordinate $1$ and $C$ has coordinate $6$ , and $B$ is between $A$ and $C$ . Prove: $AB + BC = 5$ .

Make a plan. The Prove asks about the sum AB + BC. The Segment Addition Postulate can turn that sum into the single length AC, and the Ruler Postulate can find AC from the coordinates.
Reason: Given. Begin with the coordinates, which the problem states.
$A \text{ has coordinate } 1 \text{ and } C \text{ has coordinate } 6$
Reason: Given. The problem states this too, so you may use it. You could not use it if it only appeared to be true in the figure.
$B \text{ is between } A \text{ and } C$
Reason: Segment Addition Postulate. The betweenness fact in the row above meets the postulate's condition. You don't need to know exactly where B is.
$AB + BC = AC$
Reason: Ruler Postulate. The length AC is the absolute value of the difference of the coordinates of A and C. The result, 5, is positive, as a length must be.
$AC = \lvert 6 - 1 \rvert = \lvert 5 \rvert = 5$
Reason: Transitive Property of Equality. AB + BC equals AC, and AC equals 5, so AB + BC equals 5. This is the Prove, so the proof is done.
$AB + BC = 5$

Answer

Since $AB + BC = AC$ and $AC = 5$ , the sum $AB + BC$ equals $5$ wherever $B$ lies between $A$ and $C$ .

The finished proof

Statement	Reason
$A$ has coordinate $1$ and $C$ has coordinate $6$	Given
$B$ is between $A$ and $C$	Given
$AB + BC = AC$	Segment Addition Postulate
$AC = \lvert 6 - 1 \rvert = \lvert 5 \rvert = 5$	Ruler Postulate
$AB + BC = 5$	Transitive Property of Equality

Now look at statements about the segments themselves. $\overline{AB}$ , with a bar, names the segment, which is a figure. $AB$ , with no bar, is its length, which is a number. Congruent segments are segments with the same length, and you write $\overline{AB} \cong \overline{CD}$ .

You've met the Properties of Congruence. For segments they say:

Reflexive Property of Congruence: $\overline{AB} \cong \overline{AB}$ .
Symmetric Property of Congruence: if $\overline{AB} \cong \overline{CD}$ , then $\overline{CD} \cong \overline{AB}$ .
Transitive Property of Congruence: if $\overline{AB} \cong \overline{CD}$ and $\overline{CD} \cong \overline{EF}$ , then $\overline{AB} \cong \overline{EF}$ .

The Transitive Property needs a link: the same segment ends the first fact and starts the second. If your facts aren't lined up that way, use the Symmetric Property of Congruence to flip one.

Key concept

The bridge between segments and numbers

Segments are congruent exactly when their lengths are equal. Use this Definition of congruent segments as a bridge in both directions. Go from $\cong$ to $=$ when you want to work with numbers, using the Ruler Postulate, the Segment Addition Postulate and the Properties of Equality. Go from $=$ back to $\cong$ when you want to state a result about segments. Properties of Congruence apply to $\cong$ statements, and Properties of Equality apply to $=$ statements.

\overline{AB} \cong \overline{CD} \iff AB = CD

Worked example

Use the Symmetric and Transitive Properties

Given: $\overline{AB} \cong \overline{CD}$ and $\overline{EF} \cong \overline{CD}$ . Prove: $AB = EF$ .

Make a plan. The Given is about segments and the Prove is about lengths. First link segment AB to segment EF with the Transitive Property of Congruence. Then cross to lengths with the Definition of congruent segments.
Reason: Given. Write down the first fact you may assume.
$\overline{AB} \cong \overline{CD}$
Reason: Given. This second fact doesn't line up with the first yet, because segment CD ends both.
$\overline{EF} \cong \overline{CD}$
Reason: Symmetric Property of Congruence. The Transitive Property needs CD to start the second fact, so flip the second given.
$\overline{CD} \cong \overline{EF}$
Reason: Transitive Property of Congruence. Segment CD ends the first given and starts the flipped fact, so it links AB to EF.
$\overline{AB} \cong \overline{EF}$
Reason: Definition of congruent segments. Congruent segments have equal lengths, so AB equals EF. This is the Prove.
$AB = EF$

Answer

Because $\overline{AB}$ and $\overline{EF}$ are each congruent to $\overline{CD}$ , the proof shows that $AB = EF$ .

The finished proof

Statement	Reason
$\overline{AB} \cong \overline{CD}$	Given
$\overline{EF} \cong \overline{CD}$	Given
$\overline{CD} \cong \overline{EF}$	Symmetric Property of Congruence
$\overline{AB} \cong \overline{EF}$	Transitive Property of Congruence
$AB = EF$	Definition of congruent segments

To complete a proof, you fill in missing statements or reasons so that every row follows from the rows above it. Read the Given and the Prove first. Then ask a question about each gap.

Missing reason: what changed from the rows above? If a congruence became an equation of the same two lengths, the reason is the Definition of congruent segments. If a point's coordinates became a length, it is the Ruler Postulate. If a betweenness fact became a sum, it is the Segment Addition Postulate.
Missing statement: what does the row's reason produce when you apply it to the rows above? The row below the gap often shows what you need to reach.

Worked example

Find the missing reasons

Given: $\overline{AC} \cong \overline{EF}$ ; $B$ is between $A$ and $C$ ; on a number line, $E$ has coordinate $3$ and $F$ has coordinate $10$ . Prove: $AB + BC = 7$ . After the Given rows, a draft proof has only the five statements in the steps below. Name the reason for each one.

What changed: the given congruence became an equation of the same two lengths. Reason: Definition of congruent segments.
$AC = EF$
What changed: the coordinates of E and F became a length, and 7 is positive, as a length must be. Reason: Ruler Postulate.
$EF = \lvert 10 - 3 \rvert = \lvert 7 \rvert = 7$
What changed: AC equals EF and EF equals 7, so AC equals 7. Reason: Transitive Property of Equality.
$AC = 7$
What changed: the fact that B is between A and C became a sum. Reason: Segment Addition Postulate.
$AB + BC = AC$
What changed: AB + BC equals AC and AC equals 7, so AB + BC equals 7, which is the Prove. Reason: Transitive Property of Equality.
$AB + BC = 7$

Answer

The five reasons, in order, are the Definition of congruent segments, the Ruler Postulate, the Transitive Property of Equality, the Segment Addition Postulate and the Transitive Property of Equality, which completes the proof that $AB + BC = 7$ .

The finished proof

Statement	Reason
$\overline{AC} \cong \overline{EF}$	Given
$B$ is between $A$ and $C$	Given
$E$ has coordinate $3$ and $F$ has coordinate $10$	Given
$AC = EF$	Definition of congruent segments
$EF = \lvert 10 - 3 \rvert = \lvert 7 \rvert = 7$	Ruler Postulate
$AC = 7$	Transitive Property of Equality
$AB + BC = AC$	Segment Addition Postulate
$AB + BC = 7$	Transitive Property of Equality

Now you'll fill in missing statements in a longer proof that uses all of these ideas.

Given: $\overline{AB} \cong \overline{DC}$ ; $B$ is between $A$ and $C$ ; $C$ is between $D$ and $B$ .

Prove: $\overline{AC} \cong \overline{DB}$ .

Tick marks are small matching dashes that show segments are given as congruent. The single ticks in the figure below match the first Given. The two betweenness facts come from the text, not the drawing.

Notice that $\overline{BC}$ is part of both $\overline{AC}$ and $\overline{DB}$ . The names $\overline{BC}$ and $\overline{CB}$ are two names for this one segment, and every segment is congruent to itself, so the Reflexive Property of Congruence gives $\overline{BC} \cong \overline{CB}$ . Reading from $A$ to $C$ gives the parts $AB$ and $BC$ . Reading from $D$ to $B$ gives the parts $DC$ and $CB$ .

Not to scale. One tick each on

\overline{AB}

and

\overline{DC}

marks the Given

\overline{AB} \cong \overline{DC}

. The betweenness facts come from the text, not the drawing.

A proof with three missing statements

Statement	Reason
$\overline{AB} \cong \overline{DC}$	Given
$B$ is between $A$ and $C$	Given
$C$ is between $D$ and $B$	Given
Blank (a)	Definition of congruent segments
$\overline{BC} \cong \overline{CB}$	Reflexive Property of Congruence
$BC = CB$	Definition of congruent segments
$AB + BC = AC$	Segment Addition Postulate
Blank (b)	Segment Addition Postulate
$AB + BC = DB$	Substitution Property of Equality
Blank (c)	Substitution Property of Equality
$\overline{AC} \cong \overline{DB}$	Definition of congruent segments

Worked example

Fill in the missing statements

In the proof above, which statements belong in blanks (a), (b) and (c)?

Plan from both ends. The last row's reason is the Definition of congruent segments, so the row just before it must be an equation of the lengths AC and DB. The Given rows supply one congruence and two betweenness facts to turn into equations.
Blank (a) has the reason Definition of congruent segments. The only congruence above it is the Given, so rewrite that congruence as an equation of lengths.
$AB = DC$
Blank (b) has the reason Segment Addition Postulate. The row above it already used the fact that B is between A and C, so this row uses the other fact: C is between D and B.
$DC + CB = DB$
Blank (c) has the reason Substitution Property of Equality. The row above it replaced DC with AB and CB with BC in blank (b), giving AB + BC = DB. An earlier row says AB + BC = AC, so replace the sum with AC.
$AC = DB$
Check the chain. Each filled row uses only rows above it, no row relies on its own conclusion, and nothing depends on how long the drawn segments look.

Answer

Blank (a) is $AB = DC$ , blank (b) is $DC + CB = DB$ , and blank (c) is $AC = DB$ , so the completed proof shows that $\overline{AC} \cong \overline{DB}$ .

A proof about segments is a chain in which every row has a named reason. The Ruler Postulate turns coordinates into lengths, the Segment Addition Postulate turns a stated betweenness fact into a sum, and the Definition of congruent segments carries you between $\cong$ and $=$ . The Properties of Congruence and of Equality link and rewrite those facts. When a row has a gap, its reason and the rows around it show you what belongs there. Next, you'll write a two-column proof about segments on your own.