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Method

# Paragraph Proof

A paragraph proof is a way of showing the reasoning behind a mathematical proof. It consists of statements and reasons written as complete sentences in a paragraph. The reasons can be postulates, theorems, or other mathematical reasoning the reader is assumed to be able to follow without difficulty.
To show how paragraph proofs are used, consider the following prompt.

If $B, C$ and $D$ are points on $\overline{AE}$ such that $AB \cong DE$ and $BC \cong CD,$ prove that $AC \cong CE.$

First, state the given information.

It is given that points $A$ and $E$ are the endpoints of $\overline{AE}$ and $B, C,$
and $D$ are points on $\overline{AE}.$ Furthermore, $\overline{AB} \cong \overline{DE}$ and $\overline{BC} \cong \overline {CD}.$

The given congruence statements imply that the congruent segments have equal lenths.

It is given that points $A$ and $E$ are the endpoints of $\overline{AE}$ and $B, C,$
and $D$ are points on $\overline{AE}.$ Furthermore, $\overline{AB} \cong \overline{DE}$ and $\overline{BC} \cong \overline {CD}.$
Since the segments are congruent, it follows that
$AB=DE$ and $BC=CD.$

The Segment Addition Postulate allows a segment to be divided into the smaller segments it contains.

It is given that points $A$ and $E$ are the endpoints of $\overline{AE}$ and $B, C,$
and $D$ are points on $\overline{AE}.$ Furthermore, $\overline{AB} \cong \overline{DE}$ and $\overline{BC} \cong \overline {CD}.$
Since the segments are congruent, it follows that $AB=DE$ and
$BC=CD.$ From the Segment Addition Postulate,
$\overline{AC}=\overline{AB}+\overline{BC}$ and $\overline{CE}=\overline{CD}+\overline{DE}.$

The congruent segments can be substituted into the equations.

It is given that points $A$ and $E$ are the endpoints of $\overline{AE}$ and $B, C,$ and $D$
are points on $\overline{AE}.$ Furthermore, $\overline{AB} \cong \overline{DE}$ and $\overline{BC} \cong \overline {CD}.$ Since the
segments are congruent, it follows that $AB=DE$ and $BC=CD.$
From the Segment Addition Postulate,$\overline{AC}=\overline{AB}+\overline{BC}$ and
$\overline{CE}=\overline{CD}+\overline{DE}.$ Because $\overline{CD}$ is congruent to $\overline{BC},$ and $\overline{DE}$
is congruent to $\overline{AB},$ $\overline{CE}$ can be written as $\overline{CE}=\overline{BC}+\overline{AB}.$

Use the Commutative Property of Addition to rewrite the sum.

It is given that points $A$ and $E$ are the endpoints of $\overline{AE}$ and $B, C,$ and $D$
are points on $\overline{AE}.$ Furthermore, $\overline{AB} \cong \overline{DE}$ and $\overline{BC} \cong \overline {CD}.$ Since the segments
are congruent, it follows that $AB=DE$ and $BC=CD.$ From the Segment
Addition Postulate, $\overline{AC}=\overline{AB}+\overline{BC}$ and $\overline{CE}=\overline{CD}+\overline{DE}.$ Because $\overline{CD}$ is
congruent to $\overline{BC},$ and $\overline{DE}$ is congruent to $\overline{AB},$ $\overline{CE}$ can be written as
$\overline{CE}=\overline{BC}+\overline{AB}.$ The Commutative Property of Addition allows
the sum of $\overline{CE}$ to be rewritten. $\overline{CE}=\overline{AB}+\overline{BC}.$

Since $\overline{AC}$ and $\overline{CE}$ are consist of segments of the same length, they are congruent.

It is given that points $A$ and $E$ are the endpoints of $\overline{AE}$ and $B, C,$ and $D$
are points on $\overline{AE}.$ Furthermore, $\overline{AB} \cong \overline{DE}$ and $\overline{BC} \cong \overline {CD}.$ Since the segments
are congruent, it follows that $AB=DE$ and $BC=CD.$ From the Segment
Addition Postulate, $\overline{AC}=\overline{AB}+\overline{BC}$ and $\overline{CE}=\overline{CD}+\overline{DE}.$ Because $\overline{CD}$
is congruent to $\overline{BC},$ and $\overline{DE}$ is congruent to $\overline{AB},$ $\overline{CE}$ can be written as
$\overline{CE}=\overline{BC}+\overline{AB}.$ The Commutative Property of Addition allows the sum
of $\overline{CE}$ to be rewritten. $\overline{CE}=\overline{AB}+\overline{BC}.$ Therefore, $\overline{AC} \cong \overline{CE}.$

It has now been shown that $\overline{AC} \cong \overline{CE}.$ Thus, the proof is complete.