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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.

$⎩⎪⎪⎨⎪⎪⎧ AB≅DE∠A≅∠DAC≅DF ⇒△ABC≅△DEF $

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

The primary purpose of the 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 the ways will be shown here.1

Translate $△DEF$ So That Two Corresponding Vertices Match

Translate $△DEF$ so that $D$ is mapped onto $A.$ If this translation maps $△DEF$ onto $△ABC,$ the proof is complete.

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

2

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

Rotate $△AE_{′}F_{′}$ counterclockwise about $A$ so that a pair of corresponding sides match. If the image of this transformation is $△ABC,$ the proof is complete. Note that this rotation maps $E_{′}$ onto $B.$ Therefore, the rotation maps $AE_{′}$ onto $AB.$

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

3

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

Reflect $△ABF_{′′}$ across $AB.$ Because reflections preserve angles, $AF_{′′}$ is mapped onto $AC.$ Additionally, it is given that $AC=AF_{′′}.$ Therefore, $F_{′′}$ is mapped onto $C,$ which gives that $BF_{′′}$ is mapped onto $BC.$

This time the image matches $△ABC.$

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

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

$⎩⎪⎪⎨⎪⎪⎧ ∠A≅∠DAB≅DE∠B≅∠E ⇒△ABC≅△DEF $

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

The goal of the proof is to find a rigid motion or sequence of rigid motions that maps one triangle onto the other. This can be done in several ways. One of the ways will be shown here.1

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.

2

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

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

3

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

Reflect $△ABF_{′′}$ across $AB.$ Because reflections preserve angles, $AF_{′′}$ and $BF_{′′}$ are mapped onto $AC$ and $BC,$ respectively. Then, the point of intersection of the original rays $F_{′′}$ is mapped onto the point of intersection of the image rays $C.$

This time the image matches $△ABC.$

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.

$⎩⎪⎪⎨⎪⎪⎧ AB≅DEBC≅EFAC≅DF ⇒△ABC≅△DEF $

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.1

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.

2

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

Rotate $△AE_{′}F_{′}$ counterclockwise about $A$ so that a pair of corresponding sides matches. If the image of this transformation is $△ABC,$ the proof is complete. Note that this rotation maps $E_{′}$ onto $B.$ Consequently, $AE_{′}$ is mapped onto $AB.$

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

3

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 $△ABC.$

If two angles and a non-included side of a triangle are congruent to two angles and the corresponding non-included side of another triangle, then the triangles are congruent.

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

$⎩⎪⎪⎨⎪⎪⎧ ∠A≅∠D∠B≅∠EBC≅EF ⇒△ABC≅△DEF $

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

The primary purpose of the 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 the ways will be shown here.1

Translate $△DEF$ So That Two Corresponding Vertices Match

Translate $△DEF$ so that $F$ is mapped onto $C.$ If this translation maps $△DEF$ onto $△ABC,$ the proof is complete.
Since the image of the translation does not match $△ABC,$ at least one more transformation is needed.

2

Rotate $△CD_{′}E_{′}$ So That Two Corresponding Sides Match

Rotate $△CD_{′}E_{′}$ clockwise about $C$ so that a pair of corresponding sides match. If the image of this transformation is $△ABC,$ the proof is complete. Note that this rotation maps $E_{′}$ onto $B.$ Therefore, the rotation maps $CE_{′}$ onto $CB.$

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

3

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

It is given that two angles of $△ABC$ are congruent to two angles of $△BCD_{′′}.$ Hence, by the Third Angle Theorem, $∠BCD_{′′}$ is congruent to $∠BCA.$

Reflect $△CBD_{′′}$ across $BC.$ Because reflections preserve angles, $BD_{′′}$ and $CD_{′′}$ are mapped onto $BA$ and $CA,$ respectively. Then, the point of intersection of the original segments $D_{′′}$ is mapped onto the point of intersection of the image segments $A.$
This time the image matches $△ABC.$