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Paragraph Proof

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,CB, C and DD are points on AE\overline{AE} such that ABDEAB \cong DE and BCCD,BC \cong CD, prove that ACCE.AC \cong CE.


First, state the given information.

It is given that points AA and EE are the endpoints of AE\overline{AE} and B,C,B, C,
and DD are points on AE.\overline{AE}. Furthermore, ABDE\overline{AB} \cong \overline{DE} and BCCD.\overline{BC} \cong \overline {CD}.

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

It is given that points AA and EE are the endpoints of AE\overline{AE} and B,C,B, C,
and DD are points on AE.\overline{AE}. Furthermore, ABDE\overline{AB} \cong \overline{DE} and BCCD.\overline{BC} \cong \overline {CD}.
Since the segments are congruent, it follows that
AB=DEAB=DE and BC=CD.BC=CD.

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

It is given that points AA and EE are the endpoints of AE\overline{AE} and B,C,B, C,
and DD are points on AE.\overline{AE}. Furthermore, ABDE\overline{AB} \cong \overline{DE} and BCCD.\overline{BC} \cong \overline {CD}.
Since the segments are congruent, it follows that AB=DEAB=DE and
BC=CD.BC=CD. From the Segment Addition Postulate,
AC=AB+BC\overline{AC}=\overline{AB}+\overline{BC} and CE=CD+DE.\overline{CE}=\overline{CD}+\overline{DE}.

The congruent segments can be substituted into the equations.

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

Use the Commutative Property of Addition to rewrite the sum.

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

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

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

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