Triangle Congruence Theorems

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

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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

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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 All Corresponding Sides Match

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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.

Consequently, after a sequence of rigid motions △ DEF can be mapped onto △ ABC. This means that △ DEF and △ ABC are congruent 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

expand_more

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

expand_more

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

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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.

Consequently, the application of a sequence of rigid motions allows △ DEF to be mapped onto △ ABC. This means that △ DEF and △ ABC are congruent triangles. The proof is complete.

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

expand_more

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

expand_more

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

expand_more

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.

Consequently, after a sequence of rigid motions, △ DEF can be mapped onto △ ABC. This means that △ DEF and △ ABC are congruent triangles. The proof is complete.

Let P be the point of intersection of BC and the angle bisector of ∠ A. Since AP is the angle bisector of ∠ A, then ∠ BAP ≅ ∠ CAP.

By the Reflexive Property of Congruence, AP in △ ABP is congruent to AP in △ ACP. Because of the Angle-Angle-Side Congruence Theorem, both triangles are congruent. △ ABP ≅ △ ACP Since corresponding parts of congruent triangles are congruent, it follows that AB is congruent to AC. AB ≅ AC It has been proven that if two angles of a triangle are congruent, then the sides opposite them are congruent.

Place a point D' on △ A'B'C' so that m∠ D'A'C' = m∠ BAC and A'D' = AB.

Draw △ A'C'D' and △ B'C'D'.

Note that two sides of △ ABC and their included angle are congruent to two sides of △ A'D'C' and their included angle. Because of the Side-Angle-Side Congruence Theorem, BC=D'C'. It now suffices to prove that D'C'>B'C'. To do so, note that △A'B'D is isosceles.

Because of the Isosceles Triangle Theorem, m∠ A'D'B' = m∠ A'B'D'. Furthermore, ∠ A'D'B' must measure more than ∠ C'D'B' because of the Angle Addition Postulate. m∠ A'D'B' > m∠ C'D'B' The same holds true for ∠ C'B'D' and ∠ A'B'D' m∠ C'B'D' > m∠ A'B'D' Since m∠ A'D'B' = m∠ A'B'D', the Substitution Property of Equality allows to substitute m∠ A'B'D' for m∠ A'D'B in the above inequality, and because of the Transitive Property of Inequality, the following can be concluded. m∠ C'B'D' > m∠ C'D'B' This inequality can be used along with the Triangle Larger Angle Theorem to prove that C'D' > B'C'. Finally, since C'D'= BC, use the Substitution Property of Equality one more time to obtain the desired inequality.

This theorem can be proven by contradiction. Since the goal is to prove that m∠ BAC > m∠ B'A'C' the opposite statement will be assumed, that is, m∠ BAC ≤ m∠ B'A'C'. Because this is a non-strict inequality, both m∠ BAC = m∠ B'A'C' and m∠ BAC < m∠ B'A'C' have to be considered.

m∠ BAC = m∠ B'A'C'

If m∠ BAC = m∠ B'A'C', then, △ ABC and △ A'B'C' share two congruent sides and their included angles are congruent. Because of the Side-Angle-Side Congruence Theorem, they are congruent. △ ABC ≅ △ A'B'C Since the triangles are congruent, their sides are congruent as well. This means that BC=B'C', but this contradicts the fact that the given triangle is such that BC>B'C'.

m∠ BAC < m∠ B'A'C'

If m∠ BAC < m∠ B'A'C' then the Hinge Theorem states that BCB'C'.

Conclusion

The assumption that m∠ BAC is less than or equal to m∠ B'A'C' contradicts the hypothesis. Therefore, this assumption must be false. Consequently, the initial conclusion of the theorem is true.