arrow_left {{ state.menu.current.label }}

arrow_left {{ state.menu.current.current.label }}

arrow_left {{ state.menu.current.current.current.label }}

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.

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.

{{ 'ml-article-textbook-solutions-description' | message }}

{{ 'ml-article-textbook-solutions-expert-solutions' | message }}

{{ 'ml-article-textbook-solutions-math-solver-scanner' | message }}

{{ 'ml-article-textbook-solutions-answers-hints-steps' | message }}

{{ 'ml-article-ecourses-description' | message }}

{{ 'ml-article-ecourses-interactive' | message }}

{{ 'ml-article-ecourses-chapter-tests' | message }}

{{ 'ml-article-ecourses-exercise-levels' | message }}

{{ 'ml-article-ecourses-rank-stats' | message }}

{{ 'ml-article-ecourses-video-lessons' | message }}

{{ 'ml-article-ecourses-course-theory' | message }}

{{ 'ml-article-ecourses-join-classroom' | message }}

{{ 'ml-article-ecourses-graphing-calculator' | message }}

{{ 'ml-article-ecourses-quiz-games' | message }}

{{ 'ml-article-ecourses-study-together' | message }}

{{ 'ml-article-community-description' | message }}

{{ 'ml-article-community-create-and-share-channels' | message }}

{{ 'ml-article-community-share-content-and-challenge' | message }}

{{ 'ml-article-community-cooperate-with-friends' | message }}

{{ 'ml-article-worksheets-description' | message }}

{{ 'ml-article-worksheets-course1' | message }}

{{ 'ml-article-worksheets-course2' | message }}

{{ 'ml-article-worksheets-course3' | message }}

{{ 'ml-article-worksheets-course4' | message }}

{{ 'ml-article-math-solver-description' | message }}

{{ 'ml-article-math-solver-photo-scan-solve' | message }}

{{ 'ml-article-math-solver-step-by-step' | message }}

{{ 'ml-article-math-solver-graph-math-problem' | message }}

close

Community rate_review

{{ r.avatar.letter }}

{{ u.avatar.letter }}

+