Two figures are congruent if and only if there exists one, or more, rigid motions which will map one figure onto the other.

The following sections will verify that each of the accepted methods of proving triangles congruent (SSS, SAS, ASA, AAS, and HL) follows from the definition (shown above) of congruence in terms of rigid motions.

Given three sets of congruent sides, we seek a sequence of rigid motions that will map ΔABC onto ΔDEF proving the triangles congruent.
 1. Given , there is a rigid transformation that will map onto . A translation moving to the right and down assuming || . If not parallel, a rotation after translating point A to point D will be needed. Should B' and E be on the same side of , reflect ΔA'B'C' over , so B' and E are on opposite sides of . 2. since translations (and rotations) are rigid transformations preserving length. 3. It is given that . By the transitive property of congruence, . 4. must be the perpendicular bisector of , because . Points along a perpendicular bisector of a segment are equidistant from the endpoints of the segment. Thus, . 5. Point B' can be mapped onto E by a reflection in . 6. ΔABC has been mapped onto ΔDEF through a series of rigid transformations and .

Given two sets of sides and the included angle congruent, we seek a sequence of rigid motions that will map ΔABC onto ΔDEF proving the triangles congruent.
 1. Given , there is a rigid transformation that will map onto . A translation moving to the right and down assuming || . If not parallel, a rotation after translating point A to point D will be needed. Should B' and E be on the same side of , reflect ΔA'B'C' over , so B' and E are on opposite sides of . 2. since translations (and rotations) are rigid transformations preserving length. 3. since translations (and rotations) preserve angle measure. 4. It is given that . Therefore, by the transitive property of congruence. 5. must be the bisector of as an angle bisector forms two congruent angles. 6. Map point B' onto point E by a reflection in . Reflections will preserve angle measure so will coincide with . Since , B' and E will be the same length from point F, thus causing E to coincide with B'. 7. ΔABC has been mapped onto ΔDEF through a series of rigid transformations and .

Given two sets of angles and the included side congruent, we seek a sequence of rigid motions that will map ΔABC onto ΔDEF proving the triangles congruent.

If we verify ASA, we have also verified AAS. If two sets of angles of two triangles are congruent, their third angles are also congruent. So any AAS situation can quickly become an ASA situation.
 1. Given , there is a rigid transformation that will map onto . A translation moving to the right and down assuming || . If not parallel, a rotation after translating point A to point D will be needed. Should B' and E be on the same side of , reflect ΔA'B'C' over , so B' and E are on opposite sides of . 2. since translations (and rotations) are rigid transformations preserving angle measure. 3. It is given that . Therefore, by the transitive property of congruence. 4. must be the bisector of and as an angle bisector forms two congruent angles. 5. After a reflection in of ΔA'B'C' onto ΔDEF, point B' will coincide with point E. Points D (A') and F (C ') will not move during the reflection. Ray will coincide with ray , and ray will coincide with ray because reflections preserve angles, and rays coming from the same point at the same angle will coincide. Now, rays and can only intersect at point B', and rays and can only intersect at point E. Since the rays coincide, their intersection points, B' and E, will also coincide. 6. ΔABC has been mapped onto ΔDEF through a series of rigid transformations and .

Given the corresponding hypotenuses and legs of two right triangles congruent, we seek a sequence of rigid motions that will map right ΔABC onto right ΔDEF proving the right triangles congruent.

 1. Given , there is a rigid transformation that will map onto . A translation moving to the right and down assuming parallel segments. If not parallel, a rotation after translating point B to point E will be needed. Should A' and D be on the same side of , reflect ΔA'B'C' over , so A' and D are on opposite sides of . 2. since translations (and rotations) are rigid transformations preserving angle measure. 3. since translations (and rotations) preserve length. 4. It is given that Therefore, by the transitive property of congruence. 5. by the Pythagorean Theorem. 6. Map point A' onto point D by a reflection in . is the perpendicular bisector of and all points lying on the perpendicular bisector of a segment are equidistant from the endpoint of the segment. So and under a reflection A' will coincide with D. 7. Right ΔABC has been mapped onto right ΔDEF through a series of rigid transformations and .