## Congruence Of Triangles, Part II

This is a continuation of the article Congruence Of Triangles. Read that article first. In that part, I stated, "All three of these congruence postulates are equivalent. You can assume any one of them, and prove the other two from there." Someone named Jason phoned me to ask how to do that. So, here is how I do it. It can be done with three proofs. But, the way I do it is with four proofs. See that article (Congruence Of Triangles), for the definitions of SAS, ASA, and SSS, as well as "congruence."

1. Assume SAS and prove ASA.

We have two triangles, as shown in the diagram. Angle A=angle EDF, angle B=angle E, and AB=DE (Note that I am using=to mean "congruent"). Prove that the two triangles are congruent.

1. If BC=EF, then the two triangles are congruent, by SAS. So, assume that BC not=EF.
2. There is a point G, on EF (perhaps extended), on the same side of the line DE, so that EG=BC.
3. Triangle ABC=triangle DEG [by SAS].
4. Angle EDF=angle EDG [definition of congruent triangles].
5. So DF and DG are the same line.
6. DF and EF intersect at both F and G. So, F & G are the same point [two distinct lines intersect in only one point].
7. So EF=EG=BC, which contradicts our assumption that BC not=EF.
8. So BC=EF, and the triangles are congruent by SAS, as mentioned in step #1.

2. Assume ASA and prove SAS.

We have two triangles, as shown in the diagram. Angle A=angle D, AB=DE, and AC=DF. Prove that the two triangles are congruent.

1. If angle B=angle DEF, then the two triangles are congruent, by ASA. So, assume that angle B not=angle DEF.
2. Draw a line EG, so that angle DEG=angle B (with G on the same side of DE as F).
3. EG intersects line DEF (perhaps extended) at some point H.
4. Triangle ABC=triangle DEG [ASA].
5. AC=DH [definition of congruent triangles].
6. DH=DF [both=AC]
7. H and F are the saem point. And EH and EF are the same line. And angle DEG and angle DEF are the same angle.
8. That contradicts our assumption that angle B not=angle DEF.
9. So angle B=angle DEF, and the 2 triangles are congruent by ASA, as mentioned in step #1.

3. Assume SSS and prove SAS.

We have two triangles, as shown in the diagram. Angle A=angle EDF, AB=DE, and AC=DF. Prove that the two triangles are congruent.

1. If BC=EF, then the two triangles are congruent, by SSS. So, assume that BC not=EF.
2. Find the point G (on the same side of line DE as F is) so that DF=DG and BC=EG. We do this by drawing arcs of those radii and finding where they intersect, if they do intersect.
3. Triangle ABC=triangle DEG [SSS].
4. Angle A=angle EDG [definition of congruence of triangles].
5. Angle EDG=angle EDF [both=angle A].
6. So, they are the same angle. And F and G are the same point. And EF=BC.
7. This contradicts our assumption that BC not=EF.
8. So BC=EF, and the triangles are congruent by SSS, as mentioned in step #1.

4. Assume SAS and prove SSS.

We have two triangles, as shown in the diagram. AB=DE, AC=DF, and BC=EF. Prove that the two triangles are congruent.

1. If angle A=angle EDF, then the two triangles are congruent, by SAS. So, assume that angle A not=angle EDF.
2. Draw a line DG, so that angle BAC=angle EDG.
3. Find the point H (on the same side of DE that F is) so that DH=AC.
4. F & H are not the same point [DF and DG can only intersect at D].
5. Draw EH and FH.
6. Triangle ABC=triangle DEH [SAS].
7. EH=BC [definition of congruent triangles].
8. EH=EF [both=BC].
9. DH=DF [both=AC].
10. Both triangle DFH and triangle EFH are isosceles [definition].
11. The perpendicular bisector of FH must pass through both D and E [a Euclid theorem, which I will track down later].
12. So DE is the perpendicular bisector of FH [definition].
13. But that is impossible, since H and F are on the same side of line DE [step #3].
14. This contradicts our assumption that angle A not=angle EDF.
15. So angle A=angle EDF, and the triangles are congruent by SAS, as mentioned in step #1.

Note: Above I showed that ASA implies SAS and vice versa. I also showed that SAS implies SSS and vice versa. So, you can assume any of the three, and then go by a chain of above proofs to prove any of the other two. See the left half of this diagram.

This kind of chain can be done in three proofs, rather than four. See the right half of the diagram, to see how that could be done. I had difficulty coming up with any proof involving ASA and SSS. So, I did it with four proofs.

Notice that all four of the above proofs are fairly similar. That is further evidence that these three theorems/postulates are equivalent. And none of them is much more obvious than any of the others.