Write a Two-Column Proof About Segments | Geometry | LevelKite

You have already filled in missing steps of segment proofs, so you are ready to write a whole two-column proof yourself. That is a table in which each row pairs one statement with the reason it is true. The Given is what you may assume. The Prove is what you must show. A proof shows that the Prove holds in every case that fits the Given, so a picture or a few measurements cannot replace it.

A figure is not drawn to scale. Only tick, arrow and right-angle marks and the facts stated in the text are givens. You never read a length, an angle measure, parallel or perpendicular lines, or equal angles from the drawing.

Key concept

A proof is a chain of named reasons

A two-column proof links the Given to the Prove, one row at a time. Each row holds one statement and one reason. The reason is the word Given or the exact name of a definition, postulate, property of equality or congruence, or previously proved theorem. It may rest only on the Given and earlier rows, never on the row's own conclusion, an unstated fact or how the figure looks.

To write a proof:

State the Given and the Prove.
Plan backward: ask what would let you write the Prove.
Write the rows, with the Given first and the Prove last.
Check that each reason's conditions are already in the Given or in earlier rows.

Two reasons do most of the work in segment proofs, and each comes with something to watch. Here, $\overline{AB}$ names a segment and $AB$ names its length, so $\cong$ goes between segments and $=$ goes between lengths.

Definition of congruent segments: two segments are congruent exactly when their lengths are equal. It works in both directions: from $\overline{AB} \cong \overline{CD}$ to $AB = CD$ , and from $AB = CD$ to $\overline{AB} \cong \overline{CD}$ .
Segment Addition Postulate: if $B$ is between $A$ and $C$ , then $AB + BC = AC$ . Point $B$ is between $A$ and $C$ when all three points lie on one line and $B$ lies on the segment from $A$ to $C$ , not at an endpoint. Use this postulate only when the text states the order of the points.

The Properties of Equality from your earlier proofs then turn these equations into new ones.

Worked example

Switch between segments and lengths

Write a two-column proof. Given: $\overline{AB} \cong \overline{CD}$ and $CD = EF$ . Prove: $\overline{AB} \cong \overline{EF}$ .

Given. Start with what you may assume. Now plan backward: to write that two segments are congruent, you first need their lengths to be equal. So your goal is to show that length AB equals length EF.
$\overline{AB} \cong \overline{CD}$
Given. This fact is already about lengths. It links CD to EF, so you still need a link from AB to CD.
$CD = EF$
Definition of congruent segments, read from congruence to lengths. Congruent segments have equal lengths, so the first row gives this equation.
$AB = CD$
Transitive Property of Equality. AB equals CD and CD equals EF, so AB equals EF.
$AB = EF$
Definition of congruent segments, read the other way, from lengths to congruence. Segments with equal lengths are congruent. This row is the Prove, so the proof is complete.
$\overline{AB} \cong \overline{EF}$

Answer

Segment $\overline{AB}$ is congruent to segment $\overline{EF}$ , because the Given makes their lengths equal.

Finished proof: from the Given to the Prove in five rows

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

One tick on each of

\overline{AB}

and

\overline{ED}

and two ticks on each of

\overline{BC}

and

\overline{DC}

show the two given congruences. The figure is not drawn to scale.

Worked example

Add equal parts

Write a two-column proof. Given: $A$ , $B$ , $C$ , $D$ , $E$ lie on one line in that order; $\overline{AB} \cong \overline{ED}$ ; $\overline{BC} \cong \overline{DC}$ . Prove: $\overline{AC} \cong \overline{EC}$ . In the figure above, one tick marks the first congruence and two ticks mark the second.

Plan backward. The Prove is a congruence, so you need lengths AC and EC to be equal. Each of those lengths is a sum of two parts, and the Given matches the parts in pairs. So turn the congruences into equations, add the matching parts, and then replace each sum with its whole.
Given. Every proof starts from what you may assume. Write each given fact in its own row: the order of the points first, then the two congruences.
$A, B, C, D, E \text{ on one line, in order};\quad \overline{AB} \cong \overline{ED};\quad \overline{BC} \cong \overline{DC}$
Definition of congruent segments, read from congruence to lengths. You need equations because the Properties of Equality work on lengths, not on segments. Write one row for each equation.
$AB = ED,\quad BC = DC$
Addition Property of Equality. Add length BC to both sides of the first equation. Adding the same length to equal lengths keeps them equal.
$AB + BC = ED + BC$
Substitution Property of Equality. BC equals DC, so replace BC with DC on the right side. The left side stays as it is.
$AB + BC = ED + DC$
Segment Addition Postulate, used once for each whole. Its condition holds because the stated order puts B between A and C, and D between E and C. Write one row for each equation.
$AB + BC = AC,\quad ED + DC = EC$
Substitution Property of Equality. Go back to the equation with a sum on each side. Replace the left sum with AC and the right sum with EC, because each sum equals that whole.
$AC = EC$
Definition of congruent segments, read from lengths to congruence. This row is the Prove, so the proof is complete.
$\overline{AC} \cong \overline{EC}$

Answer

Segment $\overline{AC}$ is congruent to segment $\overline{EC}$ , because the Given makes their matching parts equal and the Segment Addition Postulate adds those parts to form each whole.

Finished proof: each fact and equation gets its own row

Statement	Reason
$A$ , $B$ , $C$ , $D$ , $E$ lie on one line in that order	Given
$\overline{AB} \cong \overline{ED}$	Given
$\overline{BC} \cong \overline{DC}$	Given
$AB = ED$	Definition of congruent segments
$BC = DC$	Definition of congruent segments
$AB + BC = ED + BC$	Addition Property of Equality
$AB + BC = ED + DC$	Substitution Property of Equality
$AB + BC = AC$	Segment Addition Postulate
$ED + DC = EC$	Segment Addition Postulate
$AC = EC$	Substitution Property of Equality
$\overline{AC} \cong \overline{EC}$	Definition of congruent segments

The points appear in the stated order, so

\overline{AC}

and

\overline{BD}

both contain

\overline{BC}

. The figure is not drawn to scale.

Worked example

Subtract a shared part

Write a two-column proof. Given: $A$ , $B$ , $C$ , $D$ lie on one line in that order, and $\overline{AC} \cong \overline{BD}$ . Prove: $\overline{AB} \cong \overline{CD}$ . The figure above pictures the stated order. It has no tick marks, because the congruence is stated here in words.

Plan backward. You need lengths AB and CD to be equal. The long segments from A to C and from B to D share the piece from B to C, so write each long length as a sum and then remove the shared piece.
Given. Give each fact its own row, as before.
$A, B, C, D \text{ on one line, in order};\quad \overline{AC} \cong \overline{BD}$
Definition of congruent segments, read from congruence to lengths.
$AC = BD$
Segment Addition Postulate, used twice. The stated order puts B between A and C, and C between B and D.
$AB + BC = AC,\quad BC + CD = BD$
Substitution Property of Equality. Start with the equal lengths AC and BD, then replace each one with the sum that equals it.
$AB + BC = BC + CD$
Subtraction Property of Equality. Subtract BC from both sides. The left side leaves AB, and the right side leaves CD.
$AB = CD$
Definition of congruent segments, read from lengths to congruence. This row is the Prove.
$\overline{AB} \cong \overline{CD}$

Answer

Segment $\overline{AB}$ is congruent to segment $\overline{CD}$ , because subtracting the shared length $BC$ from equal lengths leaves equal lengths.

Finished proof: the shared part is subtracted in the seventh row

Statement	Reason
$A$ , $B$ , $C$ , $D$ lie on one line in that order	Given
$\overline{AC} \cong \overline{BD}$	Given
$AC = BD$	Definition of congruent segments
$AB + BC = AC$	Segment Addition Postulate
$BC + CD = BD$	Segment Addition Postulate
$AB + BC = BC + CD$	Substitution Property of Equality
$AB = CD$	Subtraction Property of Equality
$\overline{AB} \cong \overline{CD}$	Definition of congruent segments

To write a segment proof, plan backward from the Prove, then write your rows forward from the Given. Use the Definition of congruent segments to move between segments and lengths, use the Segment Addition Postulate only when the order of the points is stated, and give every row a reason you can name exactly. If a row rests on how the figure looks, it is not yet part of your proof.