{{ item.displayTitle }}

No history yet!

equalizer

rate_review

{{ r.avatar.letter }}

{{ u.avatar.letter }}

+

{{ item.displayTitle }}

{{ item.subject.displayTitle }}

{{ searchError }}

{{ courseTrack.displayTitle }} {{ statistics.percent }}% Sign in to view progress

{{ printedBook.courseTrack.name }} {{ printedBook.name }} {{ greeting }} {{userName}}

{{ 'ml-article-collection-banner-text' | message }}

{{ 'ml-article-collection-your-statistics' | message }}

{{ 'ml-article-collection-overall-progress' | message }}

{{ 'ml-article-collection-challenge' | message }} Coming Soon!

{{ 'ml-article-collection-randomly-challenge' | message }}

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

{{ 'ml-article-collection-follow-latest-updates' | message }}

To identify if two triangles are congruent, their corresponding parts can be compared. The triangles are congruent if all the angles and sides are congruent with their counterpart.

The congruent angles are the ones with the same number of arcs. In this case, they are$∠A≅∠D,∠B≅∠E,and∠C≅∠F.$

For the sides, hatch marks are used to show that they are congruent. The following pairs of sides are congruent. Determine m∠E.

Show Solution

To find the measure of angle E, we could use that the sum of a triangle's interior angles is $180_{∘}$. However, this requires that we know m∠D. Notice that angle A and angle D are corresponding and congruent.
The desired angle measure is $61_{∘}.$

∠A≅∠D

Thus, m∠A and m∠D are equal.
We can now write and solve the following equation, using that the sum of m∠D, m∠E, and m∠F is $180_{∘}.$
$m∠D+m∠E+m∠F=180_{∘}$

SubstituteII

$m∠D=29_{∘}$, $m∠F=90_{∘}$

$29_{∘}+m∠E+90_{∘}=180_{∘}$

AddTerms

Add terms

$m∠E+119_{∘}=180_{∘}$

SubEqn

$LHS−119_{∘}=RHS−119_{∘}$

$m∠E=61_{∘}$

It can be shown that two triangles are congruent through rigid motions.

Every side and angle in △ABC triangle has a corresponding congruent part in △DEF. Therefore, by applying rigid motions to one triangle, it's possible to map it onto the other, showing congruence. The triangle △DEF can be translated so that the F maps to C. The image, $△D_{′}E_{′}C,$ can then be rotated so that $D_{′}$ maps to A. This is possible because AC≅DF.Transform

If two sides and the included angle of a triangle are congruent to two sides and the included angle of another triangle, then the two triangles are congruent.

Based on the diagram above, the theorem can be written as follows.

Consider the triangles △ABC and △DEF, where
The triangle $△ABF_{′}$ can now be reflected in the line $AB.$ If the image of $F_{′}$ then falls onto C, the triangles will completely overlap. As the angles ∠CAB and $∠F_{′}AB$ are congruent, the ray $AF_{′}$ will be mapped onto the ray $AC.$ This, combined with
means that $F_{′}$ will indeed be mapped onto C when $△ABF_{′}$ gets reflected in $AB.$

$AB≅DE,∠A≅∠D,andAC≅DF.$

If either of these can be mapped onto the other using rigid motion, then they are congruent. As AB is congruent with DE, there is a rigid motion that maps one of these onto the other. This can be performed for one of the triangles, which leads to the two congruent sides overlapping.
Transform

Thus, there is a rigid motion that maps △DEF onto △ABC. Consequently, △ABC and △DEF are indeed congruent.

If the three sides of a triangle are congruent to the three sides of another triangle, then the triangles are congruent.

Based on the diagram above, the theorem can be written as follows.

This proof will be developed based on the given diagram, but it is valid for any pair of triangles.

The primary purpose of this proof is finding a rigid motion or sequence of rigid motions that maps one triangle onto the other. This can be done in several ways. One of them will be shown here.Translate △DEF So That Two Corresponding Vertices Match

Since the image of the translation does not match △ABC, at least one more transformation is needed.

Rotate $△AE_{′}F_{′}$ So That Two Corresponding Sides Match

As before, the image does not match △ABC. Therefore, a third rigid motion is required.

Reflect $△ABF_{′′}$ So That Two More Corresponding Sides Match

The points C and $F_{′′}$ are on opposite sides of $AB.$ Now, consider $CF_{′}.$ Let G denote the point of intersection between $AB$ and $CF_{′′}.$

It can be noted that $AC=AF_{′′}$ and $BC=BF_{′′}.$ By the Converse Perpendicular Bisector Theorem, $AB$ is a perpendicular bisector of $CF_{′′}.$ Points along the perpendicular bisector are equidistant from the endpoints of the segment, so $CG=GF_{′′}.$

Finally, $F_{′′}$ can be mapped onto C by a reflection across $AB$ by reflecting $△ABF_{′′}$ across $AB.$ Because reflections preserve angles, $AF_{′′}$ and $BF_{′′}$ are mapped onto $AC$ and $BC,$ respectively.
This time the image matches △ABC.

Consider the triangles △ABC and △DEF, where
Thus, ∠C and $∠F_{′}$ are congruent.
The triangles have two sides, and their included angle, that are congruent. Thus, by the SAS Congruence Theorem, the triangles are indeed congruent.

$AB≅DE,AC≅DF,andBC≅EF.$

If there exists a rigid motion that maps one of these onto the other, then they are congruent. As the sides AB and DE are congruent, there is a rigid motion that maps one of these onto the other. Performing this transformation for one of the triangles leads to the two congruent sides overlapping.
Transform

The line $CF_{′}$ can now be drawn, dividing the angle ∠C into ∠1 and ∠3, and $∠F_{′}$ into ∠2 and ∠4.

Notice that the $△BCF_{′}$ is an isosceles triangle, leading to ∠1 and ∠2 being congruent. Similarly, ∠3 and ∠4 are congruent as $△ACF_{′}$ is also an isosceles triangle. This leads tom∠1+m∠3=m∠2+m∠4,

which by construction means that
Based on the diagram above, the theorem can be written as follows.

Consider the triangles △ABC and △DEF, where

$∠A≅∠D,AB≅DE,and∠B≅∠E.$

If either of these can be mapped onto the other using rigid motion, then they are congruent. As AB is congruent with DE, there is a rigid motion that maps one of these onto the other. This can be performed for one of the triangles, which leads to the two congruent sides overlapping.
Transform

The triangle $△ABF_{′}$ can now be reflected in the line $AB.$ If the image of $F_{′}$ falls onto C, the triangles will completely overlap. As the angles ∠CAB and $∠F_{′}AB$ are congruent, the ray $AF_{′}$ will be mapped onto the ray $AC.$ Similarly, $BF_{′}$ will be mapped onto $BC.$

Thus, the intersection of $AF_{′}$ and $BF_{′},$ which is $F_{′},$ will be mapped onto the intersection of $AC$ and $BC,$ which is C.

There is a rigid motion that maps △DEF onto △ABC. Consequently, △ABC and △DEF are indeed congruent.

Show that the triangles below are congruent.

Show Solution

Two triangles are congruent only if their corresponding angles and sides are congruent. However, using one of the congruence theorems, SAS, SSS, or ASA, it's only necessary to know three congruent parts to prove complete congruence. Let's study the triangles to see if we can identify three of these parts.

It is given that ∠A≅∠D and AB≅DE. This is one side and one angle. Since SSS and SAS both require more than one congruent side, they cannot be used. To use ASA, we'll need to prove that ∠B is congruent to ∠E.

Since the measure of ∠B is given, it is necessary to show that $∠E=56_{∘}.$The angles ∠D and ∠F are known, $34_{∘}$ and $90_{∘}.$ Since the sum of the angles in a triangle is $180_{∘},$ we can write an equation that can be solved for m∠E.

$m∠D+m∠E+m∠F=180_{∘}$

Solve for m∠E

SubstituteII

$m∠D=90_{∘}$, $m∠F=34_{∘}$

$90_{∘}+m∠E+34_{∘}=180_{∘}$

AddTerms

Add terms

$m∠E+124_{∘}=180_{∘}$

SubEqn

$LHS−124_{∘}=RHS−124_{∘}$

$m∠E=56_{∘}$

Using the ASA congruence theorem, we have now proven that the triangles are congruent.

{{ 'mldesktop-placeholder-grade' | message }} {{ article.displayTitle }}!

{{ focusmode.exercise.exerciseName }}