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Two figures are congruent figures if there is a rigid motion or sequence of rigid motions that maps one of the figures onto the other. As a result, congruent figures have the same size and shape. To denote algebraically that two figures are congruent, the symbol

$≅$is used.

When writing a polygon congruence, the corresponding vertices must be listed in the same order. For the polygons above, two of the possible congruence statements can be written as follows.

$ABCDE≅JKLMNorCDEAB≅LMNJK $

Two triangles are congruent if and only if their corresponding sides and angles are congruent.

Using the triangles shown, this claim can be written algebraically as follows.

$△ABC≅△DEF⇕AB≅DEBC≅EFAC≅DF and∠A≅∠D∠B≅∠E∠C≅∠F $

This proof will be developed based on the given diagram, but it is valid for any pair of triangles. The proof of this biconditional statement consists of two parts, one for each direction.

- If $△ABC$ and $△DEF$ are congruent, then their corresponding sides and angles are congruent.
- If the corresponding sides and angles of $△ABC$ and $△DEF$ are congruent, then the triangles are congruent.

Because rigid motions preserve side lengths, $AB$ and its image have the same length, that is, $AB=DE.$ Therefore, $AB≅DE.$ Similarly for the other two side lengths.

$BC≅EFandAC≅DF $

Furthermore, rigid motions preserve angle measures. Then, $∠A$ and its image have the same measure, that is, $m∠A=m∠D.$ Therefore, $∠A≅∠D.$ Similarly for the remaining angles.
$∠B≅∠Eand∠C≅∠F $

That way, it has been shown that if two triangles are congruent, then their corresponding sides and angles are congruent. To begin, mark the congruent parts on the given diagram.

The primary purpose is finding a rigid motion or sequence of rigid motions that maps one triangle onto the other. This can be done in several ways, here it is shown one of them.

1

Translate $△ABC$ so that one pair of corresponding vertices match

Apply a translation to $△ABC$ that maps $A$ to $D.$ If this translation maps $△ABC$ onto $△DEF$ the proof will be complete.

As seen, $△A_{′}B_{′}C_{′}$ did not match $△DEF.$ Therefore, a second rigid motion is needed.

2

Rotate $△DB_{′}C_{′}$ so that one pair of corresponding sides match

Apply a clockwise rotation to $△DB_{′}C_{′}$ about $D$ through $∠EDB_{′}.$ If the image matches $△DEF,$ the proof will be complete. Notice this rotation maps $B_{′}$ onto $E,$ and therefore, $DB_{′}$ onto $DE.$

As before, the image did not match $△DEF.$ Thus, a third rigid motion is required.

3

Reflect $△DEC_{′′}$ so that the corresponding sides match

Apply a reflection to $△DEC_{′′}$ across $DE.$ Because reflections preserve angles, $DC_{′′}$ is mapped onto $DF$ and $EC_{′′}$ is mapped onto $EF.$ Then, the intersection of the original rays $C_{′′},$ is mapped to the intersection of the image rays $F.$

This time the image matched $△DEF.$

Consequently, through applying different rigid motions, $△ABC$ was mapped onto $△DEF.$ This implies that $△ABC$ and $△DEF$ are congruent. Then, the proof is complete.

Two polygons are congruent if and only if their corresponding sides and angles are congruent.

Using the polygons shown, this claim can be written algebraically as follows.

$ABCD≅PQRS⇕ABBCCDAD ≅PQ ≅QR ≅RS≅PS and∠A≅∠B≅∠C≅∠D≅ ∠P∠Q∠R∠S $

This proof will be developed based on the given diagram, but it is valid for any pair of polygons. The proof of this biconditional statement consists of two parts, one for each direction.

- If $ABCD$ and $PQRS$ are congruent, then their corresponding sides and angles are congruent.
- If the corresponding sides and angles of $ABCD$ and $PQRS$ are congruent, then the polygons are congruent.

By the definition of congruent figures, if the polygons are congruent there is a rigid motion or sequence of rigid motions that maps $ABCD$ onto $PQRS.$

Because rigid motions preserve side lengths, $AB$ and its image have the same length — that is, $AB=PQ.$ Therefore, $AB$ and $PQ $ are congruent segments. Similar observations are true for the other three sides.

$BC≅QR CD≅RSAD≅PS $

Furthermore, rigid motions preserve angle measures, which means that $∠A$ and its image have the same measure. Since $m∠A=m∠P,$ $∠A$ and $∠P$ are congruent angles. Similarly, all the remaining angles can also be concluded to be congruent.
$∠B≅∠Q∠C≅∠R∠D≅∠S $

In this fashion, it has been shown that if two polygons are congruent, then their corresponding sides and angles are congruent. To begin, congruent parts on the given diagram will be marked.

The primary purpose of this part is to find a rigid motion or sequence of rigid motions that maps one polygon onto the other. This can be done in several ways, and what is shown here is only one.1

Translate $ABCD$ So That One Pair of Corresponding Vertices Match

Apply a translation that maps $D$ onto $S.$ If this translation maps $ABCD$ onto $PQRS,$ the proof will be complete.

As can be seen, $A_{′}B_{′}C_{′}D_{′}$ did not map onto $PQRS.$ Therefore, a second rigid motion is needed.

2

Rotate $A_{′}B_{′}C_{′}S$ So That One Pair of Corresponding Sides Match

Apply a clockwise rotation about $S$ through $∠RSC_{′}$ to $A_{′}B_{′}C_{′}S.$ If the image matches $PQRS,$ the proof will be complete. Note that this rotation maps $C_{′}$ onto $R$ and therefore $SC_{′}$ onto $SR.$

The image still does not match $PQRS,$ so a third rigid motion is required.

3

Reflect $A_{′′}B_{′′}RS$ So That the Corresponding Sides Match

Finally, apply a reflection across $RS$ to $A_{′′}B_{′′}RS.$ Because reflections preserve angles and lengths, $SA_{′′}$ is mapped onto $SP$ and $RB_{′′}$ is mapped onto $RQ .$ Likewise, $A_{′′}B_{′′}$ is mapped onto $PQ .$

This time the image matches $PQRS.$

Consequently, through applying a series of different rigid motions, $ABCD$ was mapped onto $PQRS.$ This implies that $ABCD$ and $PQRS$ are congruent polygons. With this, the proof is complete.