Writing a Paragraph Proof | Geometry | LevelKite

Earlier you decided whether a statement is always, sometimes, or never true. Now you will show why a statement is always true under its conditions. Testing examples cannot do that, because no list of examples covers every case.

A proof can. A proof is a logical argument in which every statement has a reason. A paragraph proof writes that argument as sentences in a paragraph. You already know the idea behind it from the Law of Syllogism: if $p \rightarrow q$ and $q \rightarrow r$ are true, then $p \rightarrow r$ is true. Each fact you reach becomes the starting point for the next sentence.

Every proof keeps two things apart. The Given is what you are told is true: facts stated in the text or shown by marks on a figure. The Prove statement is where you must end up, so you cannot use it as a step on the way there.

Key concept

A paragraph proof is a chain of named reasons

A paragraph proof starts from the Given and moves one statement at a time to the Prove statement. Each statement has a reason, and that reason must be the exact name of a definition, postulate, property of equality or congruence, or previously proved theorem, or the word Given.

A postulate is accepted without proof. A definition works in both directions. A theorem is a proved statement, so use one as a reason only after it has been stated and, where this course proves it, proved.

A step never rests on its own conclusion, on an unstated fact, or on how a figure looks. Figures are not drawn to scale, so only their tick, arrow and right-angle marks, along with facts stated in the text, count as given.

The reasons this lesson uses

Reason	What it lets you write
Given	A fact stated in the text, or shown by a tick, arrow or right-angle mark on a figure.
Segment Addition Postulate	If $B$ is between $A$ and $C$ on a line, then $AB + BC = AC$ .
Angle Addition Postulate	If $D$ is in the interior of $\angle ABC$ (inside its opening), then $m\angle ABD + m\angle DBC = m\angle ABC$ .
Protractor Postulate	The measure of an angle is a number between $0$ and $180$ .
Definition of congruent segments	$\overline{AB} \cong \overline{CD}$ exactly when $AB = CD$ . It works in both directions.
Definition of congruent angles	$\angle A \cong \angle B$ exactly when $m\angle A = m\angle B$ . It works in both directions.
Substitution Property of Equality	If $a = b$ , you may replace $a$ with $b$ in an equation.
Transitive Property of Equality	If $a = b$ and $b = c$ , then $a = c$ .
Subtraction Property of Equality	If $a = b$ , then $a - c = b - c$ .

Worked example

A first proof with actual lengths

Given: $B$ is between $A$ and $C$ , $AB = 5$ , $BC = 7$ , and $DE = 12$ . Prove: $\overline{DE} \cong \overline{AC}$ .

Start with a fact you can use. It is given that B is between A and C, so the Segment Addition Postulate lets you write the whole length as the sum of its two parts.
$AB + BC = AC$
It is given that AB = 5 and BC = 7. The Substitution Property of Equality lets you replace AB and BC with those values. Then add 5 and 7.
$5 + 7 = AC, \text{ so } 12 = AC$
It is given that DE = 12, and you just reached 12 = AC. The value 12 links the two equations, so the Transitive Property of Equality gives DE = AC.
$DE = 12 \text{ and } 12 = AC, \text{ so } DE = AC$
Congruent segments are segments with equal lengths. The Definition of congruent segments turns the equal lengths into the Prove statement.
$\overline{DE} \cong \overline{AC}$

Answer

Segment $DE$ is congruent to segment $AC$ , which is the Prove statement.

Now turn the steps into a paragraph. Each step becomes a sentence that states a fact and names its reason.

Paragraph proof: It is given that $B$ is between $A$ and $C$ , so $AB + BC = AC$ by the Segment Addition Postulate. It is also given that $AB = 5$ and $BC = 7$ , so $5 + 7 = AC$ , or $12 = AC$ , by the Substitution Property of Equality. It is given that $DE = 12$ , so $DE = AC$ by the Transitive Property of Equality. Therefore, $\overline{DE} \cong \overline{AC}$ by the Definition of congruent segments.

The last sentence is exactly the Prove statement, so the proof is complete.

Not to scale.

\overline{AC}

and

\overline{BD}

share the part

\overline{BC}

; their congruence is stated in the text, not marked.

Worked example

Removing a shared part

In the figure, $A$ , $B$ , $C$ and $D$ lie on line $AD$ . Given: $B$ is between $A$ and $C$ , $C$ is between $B$ and $D$ , and $\overline{AC} \cong \overline{BD}$ . Prove: $\overline{AB} \cong \overline{CD}$ .

It is given that segment AC is congruent to segment BD. A definition works in both directions, so the Definition of congruent segments takes you from congruent segments to equal lengths.
$AC = BD$
Write each long segment as a sum of its parts. It is given that B is between A and C and that C is between B and D, so use the Segment Addition Postulate twice.
$AB + BC = AC \quad \text{and} \quad BC + CD = BD$
In the equation AC = BD, replace AC and BD with their sums. The reason is the Substitution Property of Equality.
$AB + BC = BC + CD$
Now BC is on both sides. The Subtraction Property of Equality lets you subtract BC from each side, which leaves the two outer parts.
$AB + BC - BC = BC + CD - BC, \text{ so } AB = CD$
Use the Definition of congruent segments again, this time from equal lengths to congruent segments. This is the Prove statement.
$\overline{AB} \cong \overline{CD}$

Answer

Segment $AB$ is congruent to segment $CD$ , which is the Prove statement.

Paragraph proof: It is given that $\overline{AC} \cong \overline{BD}$ , so $AC = BD$ by the Definition of congruent segments. It is also given that $B$ is between $A$ and $C$ and that $C$ is between $B$ and $D$ , so $AB + BC = AC$ and $BC + CD = BD$ by the Segment Addition Postulate. By the Substitution Property of Equality, $AB + BC = BC + CD$ . Subtracting $BC$ from each side gives $AB = CD$ by the Subtraction Property of Equality. Therefore, $\overline{AB} \cong \overline{CD}$ by the Definition of congruent segments.

You never used a particular length, so this argument covers every arrangement that meets the Given. That is why the statement is always true under those conditions.

The same kind of argument works for angles. The figure below keeps the four points on line $AD$ and adds a point $E$ that is not on the line, with rays from $E$ through $A$ , $B$ , $C$ and $D$ . Its tick marks show the fact you just proved, $\overline{AB} \cong \overline{CD}$ .

The next proof uses the Angle Addition Postulate, which has a condition: the point must be in the interior of the angle, inside the opening between its two rays. The Given states that condition, so you do not have to read it from the drawing.

Not to scale. The ticks give

\overline{AB} \cong \overline{CD}

; no mark says anything about

\angle AEB

or

\angle CED

.

Worked example

An angle proof that ends with a value

Use the figure above. Given: $B$ is in the interior of $\angle AEC$ , $C$ is in the interior of $\angle BED$ , $\angle AEC \cong \angle BED$ , $m\angle AEB = (2x + 4)^\circ$ , and $m\angle CED = (x + 14)^\circ$ . Prove: $m\angle AEB = 24^\circ$ .

It is given that angle AEC is congruent to angle BED, so apply the Definition of congruent angles.
$m\angle AEC = m\angle BED$
The interior facts are given, so apply the Angle Addition Postulate to each large angle.
$m\angle AEB + m\angle BEC = m\angle AEC \quad \text{and} \quad m\angle BEC + m\angle CED = m\angle BED$
Replace the two large measures with their sums, by the Substitution Property of Equality.
$m\angle AEB + m\angle BEC = m\angle BEC + m\angle CED$
Subtract the measure of angle BEC from each side, by the Subtraction Property of Equality.
$m\angle AEB = m\angle CED$
Replace each measure with its given expression, by the Substitution Property of Equality, and compare the numbers of degrees. Then subtract x and then 4 from each side, by the Subtraction Property of Equality.
$(2x + 4)^\circ = (x + 14)^\circ,\quad 2x + 4 = x + 14,\quad x + 4 = 14,\quad x = 10$
Replace x with 10, by the Substitution Property of Equality. Check: angle CED also measures 24 degrees, and the Protractor Postulate allows this measure because 24 is between 0 and 180.
$m\angle AEB = (2(10) + 4)^\circ = 24^\circ, \quad m\angle CED = (10 + 14)^\circ = 24^\circ$

Answer

Angle $AEB$ measures $24^\circ$ , which is the Prove statement.

Paragraph proof: It is given that $\angle AEC \cong \angle BED$ , so $m\angle AEC = m\angle BED$ by the Definition of congruent angles. It is also given that $B$ is in the interior of $\angle AEC$ and $C$ is in the interior of $\angle BED$ , so $m\angle AEB + m\angle BEC = m\angle AEC$ and $m\angle BEC + m\angle CED = m\angle BED$ by the Angle Addition Postulate. By the Substitution Property of Equality, $m\angle AEB + m\angle BEC = m\angle BEC + m\angle CED$ . Subtracting $m\angle BEC$ from each side gives $m\angle AEB = m\angle CED$ by the Subtraction Property of Equality. Substituting the given expressions gives $2x + 4 = x + 14$ by the Substitution Property of Equality. Subtracting $x$ and then $4$ from each side gives $x = 10$ by the Subtraction Property of Equality. Replacing $x$ with $10$ gives $(2(10) + 4)^\circ = 24^\circ$ , a measure the Protractor Postulate allows because $24$ is between $0$ and $180$ . Therefore, $m\angle AEB = 24^\circ$ by the Substitution Property of Equality.

The ticks never appear in this proof. They are true, but the chain to the Prove statement runs only through the facts about angles.

You now have the whole method. Start from the Given, write one statement at a time, and back each statement with the exact name of a definition, postulate, property or previously proved theorem, or with the word Given. Finish when your last sentence is the Prove statement. If a step rests on how a figure looks or on what you are trying to prove, replace it with a stated fact, a mark or a named reason.