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Verifying Methods of Proving
Triangles Congruent

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Two figures are congruent if and only if there exists one, or more, rigid transformations 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 transformations.

Note: The locations of the triangles in the "statement of the problem" show the triangles being parallel.
This may not always be the case. Reference (1.) in the solution will mention other possible locations
of the given triangles in relation to one another.

Given three sets of congruent sides, we seek a sequence of rigid transformations that will map ΔABC onto ΔDEF proving the triangles congruent.
sssgiven
SSSpic
verifysss
1. Given sssproof1, there is a rigid transformation that will map ssproof2 onto sssproof3. A translation moving to the right and down assuming ssproof2 || sssproof3. 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 sssproof3, reflect ΔA'B'C' over sssproof3, so B' and E are on opposite sides of sssproof3.
SSpic3


2.
sssproof4 since translations (and rotations) are rigid transformations preserving length.

3. It is given that sssproof5. By the transitive property of congruence, sssproof6.

4. Draw a segment from B' to E.
sssproof3 must be the perpendicular bisector of b"e, because sssproof6. Points along a perpendicular bisector of a segment are equidistant from the endpoints of the segment. Thus, sssproof7. (B and E are the same distance from G.)

5. Points A' and C' lie on sssproof3 coinciding with points D and F.
Point B' can now be mapped onto E by a reflection in sssproof3.

6. ΔABC has been mapped onto ΔDEF through a series of rigid transformations and sssdone.

 

 

Given two sets of sides and the included angles congruent, we seek a sequence of rigid transformations that will map ΔABC onto ΔDEF proving the triangles congruent.
sasgiven
saspic
sasstick
1. Given sssproof1, there is a rigid transformation that will map ssproof2 onto sssproof3. A translation moving to the right and down assuming ssproof2 || sssproof3. 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 sssproof3, reflect ΔA'B'C' over sssproof3, so B' and E are on opposite sides of sssproof3.
sas1

2.
bcb'x' since translations (and rotations) are rigid transformations preserving length.

3. sas3 since translations (and rotations) preserve angle measure.

4. It is given that sas6. Therefore, sas7by the transitive property of congruence.

5. sssproof3 must be the bisector of sas angleas an angle bisector forms two congruent angles.

sas2
6. Map point B' onto point E by a reflection in sssproof3. Reflections will preserve angle measure so eff will coincide with bf. Since sascongruent, 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 sssdone.




Given two sets of angles and the included sides congruent, we seek a sequence of rigid transformations that will map ΔABC onto ΔDEF proving the triangles congruent.
asagiven
asapic
ASAstick
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 sssproof1, there is a rigid transformation that will map ssproof2 onto sssproof3. A translation moving to the right and down assuming ssproof2 || sssproof3. 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 sssproof3, reflect ΔA'B'C' over sssproof3, so B' and E are on opposite sides of sssproof3.
asapic2

2. asa1
since translations (and rotations) are rigid transformations preserving angle measure.

3. It is given that asa2. Therefore, asa3 by the transitive property of congruence.

4. sssproof3 must be the bisector of sas angleand asa4as an angle bisector forms two congruent angles.

asapic3
5. After a reflection in DFof 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 raya'b' will coincide with ray rayde, and ray rayc'b'will coincide with ray rayfe because reflections preserve angles, and rays coming from the same point at the same angle will coincide. Now, rays raya'b' and rayc'b' can only intersect at point B', and rays rayde and rayfe 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 sssdone.



Given the corresponding hypotenuses and legs of two right triangles congruent, we seek a sequence of rigid transformations that will map right ΔABC onto right ΔDEF proving the right triangles congruent.
hlgiven
hlpic
HLstick
1. Given bfef, there is a rigid transformation that will map bconto ef. 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 ef, reflect ΔA'B'C' over ef, so A' and D are on opposite sides of ef.
hlpic2

2. hl22
since translations (and rotations) are rigid transformations preserving angle measure.

3. hl3 since translations (and rotations) preserve length.

4. It is given that HLrepl2 Therefore, HLrepl4 by the transitive property of congruence.

5. hl6 by the Pythagorean Theorem.

hlpic3
6. Map point A' onto point D by a reflection in ef. ef is the perpendicular bisector of da' and all points lying on the perpendicular bisector of a segment are equidistant from the endpoint of the segment. So HL8and under a reflection A' will coincide with D.

7. Right ΔABC has been mapped onto right ΔDEF through a series of rigid transformations and sssdone.



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